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GENERAL CHEMISTRY
1. Scientific Notation, Significant Figures, and the Factor-Label Method of Solving
Problems
2. Naming Inorganic Compounds Naming Inorganic Compounds
3. Mole Definitions
4. Empirical and Molecular Formulas
5. Chemical Stoichiometry Problems
6. Types of Chemical Reactions; Writing Balanced Ionic Equations
7. Oxidation-Reduction Reactions
8. Working with Solutions
9. pH and Titrations
10. Ideal Gases
11. Ideal Gas Mixtures
12. Calorimetry Exercises
13. The Enthalpy of Chemical Change: Calculations using Hess's Law and Heats of
Formation
14. Electromagnetic Radiation and the Spectrum of Atomic Hydrogen
15. Quantum Numbers, Orbitals, and Electron Configurations
16. Periodic Trends
17. Hybridization of Carbon Diagram
18. Molecular Geometry Summary Chart
19. Predicting Molecular Geometry and Hybridization
20. Molecular Orbital Diagrams for the First and Second Rows
21. Intermolecular Forces
22. Chemical Kinetics: Introductory Concepts
23. How to Determine the Rate Law from Experimental Data
24. How to Work from a Mechanism to a Rate Law
25. How to Solve Equilibrium Problems
26. Le Chatelier's Principle
27. Table of Relative Strengths of Conjugate Acid-Base Pairs
28. How to Calculate the pH of an Acidic or Basic Solution
29. Strong and Weak Acids and Bases
30. Predicting the pH of Salt Solutions
31. Buffers
32. Predicting Solubility
33. Thermodynamics and Equilibrium: Important Equations
34. How to Balance Equations for Oxidation-Reduction Reactions
35. Voltaic Cells
36. Factors Affecting Corrosion
37. Electrolysis
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Scientific Notation, Significant Figures, and the
Factor-Label Method of Solving Problems
Scientific Notation
Scientific notation is a type of exponential notation in which only one digit is kept to the left
of the decimal point. Example: 8.4050 x 10-8
.
Significant Figures
It is reasonable that a calculated result can be no more precise than the least precise piece of
information that went into the calculation. Thus it is common practice to write numbers in
scientific notation with only the last place containing any uncertainty. When we do this we are
keeping only the “significant figures”.
To determine the number of significant figures in a number, you read the number from left to
right and count all digits starting with the first non-zero digit. Do not count the exponential part.
Thus the number 0.002050 contains 4 significant figures and is written in scientific notation as
2.050 x 10-3
. The trailing zeros in a non-decimal number such as 1200 may or may not be
significant: the number may be written as 1.2 x 102, 1.20 x 10
2 or 1.200 x 10
2 depending on
whether it has 2, 3, or 4 significant figures.
Significant Figures in Derived Quantities
When doing calculations, you should use all the digits allowed by your calculator in all
intermediate steps. Then in the final step, round off your answer to the appropriate number of
“significant figures” such that only the last decimal place contains any uncertainty. You do this
by following the rules:
• When adding or subtracting, first express all numbers with the same exponent. Then the
number of decimal places in the answer should be equal to the number of decimal places in
the number with the fewest decimal places.
• In multiplication or division, the number of significant figures in the answer should be the
same as that in the factor with the fewest significant figures.
When using these rules, assume that exact numbers have an infinite number of significant figures
(or decimal places). For example, there are exactly 12 inches in one foot.
Solving Problems Using Dimensional Analysis: The Factor-Label Method
Units may be used as a guide in solving problems. First decide what units you need for your
answer. Then determine what units you are given in the problem, and what conversion factors
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will take you from the given units to the desired units. If the units cancel out properly, chances
are that you are doing the right thing! The basic set up is
? units desired= units given x new units
y units given
.. ..
Conversion factors are added until the new units are the same as the units desired. Each
conversion factor has a denominator equivalent to the numerator but in different units.
Examples
1. Carry out the following mathematical operations and express your answer in scientific notation
using the proper number of significant figures.
(a) (4.28 x 10-4
) + (3.564 x 10-2
)
(b) (0.00950) x (8.501 x 107)
3.1425 x 10-11
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2. Use the factor-label method to solve the following problem. Show your work, and give your
answer in scientific notation using the proper number of significant figures.
The calorie (1 cal = 4.184 J) is a unit of energy. The burning of a sample of gasoline produces
4.0 x 102 kJ of heat. Convert this energy to calories. (10
3 J = 1 kJ.)
____________________________________________________________________
3. Use the factor-label method to solve the following problem. Show your work, and give your
answer in scientific notation using the proper number of significant figures.
The distance from the sun to the earth is 93 million miles. How many minutes does it take light
from the sun to reach earth?
Useful information: 1 km = 0.6214 mile, c = speed of light = 3.00 x 108 m/s)
Answers:
1. (a) 3.607 x 10-2
; (b) 2.57 x 1016
2. 9.6 x 104 calories
3. 8.3 minutes
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Naming Inorganic Compounds
To name a compound you must first decide whether the substance is an ionic or molecular
compound. Ionic compounds are easily recognized since they usually contain both metallic and
non-metallic elements. The most common exception to this rule are ionic compounds containing
the ammonium ion, NH4+, such as (NH4)2CO3 or NH4Br which contain no metal ions. Molecular
compounds typically contain only non-metallic atoms (and metalloids).
Note that when naming these molecular compounds, the number of atoms of a given type is
commonly indicated with a prefix (di-, tri-, tetra, etc.).
______________________________________________________________________________
Exercises
1. Complete the following chart of corresponding ion names and formulas.
Cation Name Formula Anion Name Formula
(1) potassium ion (11) nitrate ion
(2) Fe3+
(12) H2PO4-
(3) ammonium ion (13) hydrogen carbonate
(or bicarbonate) ion
(4) Ba2+
(14) MnO4-
(5) silver ion (15) perchlorate ion
(6) Cu2+
(16) S2-
(7) zinc ion (17) acetate ion
(8) Co2+
(18) dichromate ion
(9) hydrogen ion (19) CO32-
(10) chromium(III) ion (20) sulfite ion
______________________________________________________________________________
2. Complete the following chart of corresponding compound names and formulas. Circle the
names of all non-ionic (i.e., molecular) compounds.
Compound Name Formula Compound Name Formula
(1) silver nitrate (11) sodium hydrogen
phosphate
(2) Ni(CH3CO2)2 (12) SO3
(3) ammonium sulfate (13) potassium permanganate
(4) P2O5 (14) Al2S3
(5) sodium oxide (15) cobalt(III) sulfate
(6) NH4NO3 (16) Ag2CrO4
(7) nitrogen trichloride (17) SrF2
(8) NaHCO3 (18) sulfur hexafluoride
(9) iron(II) acetate (19) NH3
(10) carbon tetrachloride (20) LiClO3
Answers
1. Complete the following chart of corresponding ion names and formulas.
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Cation Name Formula Anion Name Formula
(1) potassium ion K+ (11) nitrate ion NO3
-
(2) iron(III) ion
(or ferric ion)
Fe3+
(12) dihydrogen
phosphate ion
H2PO4-
(3) ammonium ion NH4+
(13) hydrogen carbonate
(or bicarbonate) ion
HCO3-
(4) barium ion Ba2+
(14) permanganate ion MnO4-
(5) silver ion Ag+ (15) perchlorate ion ClO4
-
(6) copper(II) ion
(or cupric ion)
Cu2+
(16) sulfide ion S2-
(7) zinc ion Zn2+
(17) acetate ion CH3CO2-
(or C2H3O2-)
(8) cobalt(II) ion Co2+
(18) dichromate ion Cr2O72-
(9) hydrogen ion H+ (19) carbonate ion CO3
2-
(10) chromium(III) ion Cr3+
(20) sulfite ion SO32-
______________________________________________________________________________
2. Complete the following chart of corresponding compound names and formulas. Circle the
names of all non-ionic (i.e., molecular) compounds. (NOTE: In these answers I have used
shading (instead of circles) to indicate the molecular compounds.)
Compound Name Formula Compound Name Formula
(1) silver nitrate AgNO3 (11) sodium hydrogen
phosphate
Na2HPO4
(2) nickel(II) acetate Ni(CH3CO2)2 (12) sulfur trioxide SO3
(3) ammonium sulfate (NH4)2SO4 (13) potassium permanganate KMnO4
(4) diphosphorus
pentoxide
(or diphosphorus
pentaoxide)
P2O5 (14) aluminum sulfide Al2S3
(5) sodium oxide Na2O (15) cobalt(III) sulfate Co2(SO4)3
(6) ammonium nitrate NH4NO3 (16) silver chromate Ag2CrO4
(7) nitrogen trichloride NCl3 (17) strontium fluoride SrF2
(8) sodium hydrogen
carbonate (or sodium
bicarbonate)
NaHCO3 (18) sulfur hexafluoride
SF6
(9) iron(II) acetate Fe(CH3CO2)2
or
Fe(C2H3O2)2
(19) ammonia
NH3
(10) carbon tetrachloride CCl4 (20) lithium chlorate LiClO3
Atomic Mass, Moles, and the Periodic Table
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Atomic Mass and Molar Mass
Isotopic masses cannot be obtained by summing the masses of the elementary particles
(neutrons, protons, and electrons) from which the isotope is formed. This process would give
masses slightly too large, since mass is lost when the neutrons and protons come together to form
the nucleus.
Atomic masses (also called atomic weights) are thus assigned relative to the mass of a
particular carbon isotope, 6
12C , which is assigned the mass of 12 amu exactly. Likewise 1 mole of
6
12C has a mass of exactly 12 g. Atomic masses and molar masses of other isotopes are calculated
based on their mass relative to that of Carbon-12.
Masses of “average” atoms are found by summing isotopic masses, weighting each isotopic
mass by its abundance. Thus one “average” C atom has a mass of 12.01 amu, and the mass of 1
mole of “average” carbon atoms has a mass of 12.01 g. These average masses are what are given
on the periodic chart.
What is a Mole?
Since atoms and molecules are so tiny, it is convenient to talk about a large number of them
at a time. The chemical counting unit is known as the mole. A mole is defined as the amount of
substance that contains as many elementary entities (atoms, molecules, or other particles) as there
are atoms in exactly 12 g of the 6
12C isotope. It has been found experimentally that
1 mole of particles= 6.022 x 1023
particles
This value is known as Avogadro’s number. Just like 1 dozen of anything always contains 12
items, 1 mole of anything always contains 6.022 x 1023 items.
Molecular Masses and Compound Masses
Molecular masses are found by summing atomic masses. They are often called molecular
weights. Thus the mass of 1 mole of water, H2O, would be 2 x (molar mass of H) plus 1x (molar
mass of O) or [(2 x 1.008 g) + (1 x 16.00 g)] = 18.02 g.
Ionic compounds such as NaCl do not contain molecules. Their formulas give the relative
numbers of each kind of atom in the sample. What we mean by the molar mass (or the molecular
weight) of an ionic compound is really the formula weight. The formula weight is the sum of the
atomic masses in the formula.
Percent Composition of Compounds
The percent composition by mass is the percent by mass of each element in a compound. If
there are n moles of an element per mole of compound, the percent by mass of the element is
calculated using the equation,
% Composition of Element=n molar mass of element
molar mass of compound
100%
The sum of the % compositions of all elements in a compound is 100%.
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______________________________________________________________________________
Exercises
1. The atomic mass scale gives masses in atomic mass units (amu) relative to the mass of carbon-
12.
(a) What is the mass of one 12
C atom in atomic mass units (amu)? ____________
(b) What is the mass of an average C atom in atomic mass units (amu)? ____________
(c) What is the mass of an average Cl atom in amu? ____________
(d) What is the mass of an average Br atom in amu? ____________
______________________________________________________________________________
2. The molar mass scale gives masses in grams (g) relative to the mass of 12
C.
(a) What is the mass in grams of 1 mole (mol) of 12
C? ___________
(b) What is the mass in grams of 1 mole (mol) of carbon? ___________
(c) What is the mass in grams of 1 mole (mol) of Cl? ___________
(d) What is the mass in grams of 1 mole (mol) of Na? ___________
______________________________________________________________________________
3. How many 12
C atoms are present in a mole of 12
C ?
______________________________________________________________________________
4. Cinnamic alcohol is used mainly in perfumery, particularly in soaps and cosmetics. Its
molecular formula is C9H10O.
(a) Calculate the percent composition by mass of C, H, and O in cinnamic alcohol.
(b) How many molecules of cinnamic alcohol are contained in a sample of mass 0.469 g?
Answers:
1. (a) 12 amu exactly; (b) 12.01 amu; (c) 35.45 amu; (d) 79.90 amu.
2. (a) 12 g exactly; (b) 12.01 g; (c) 35.45 g; (d) 22.99 g.
3. 6.022 x 1023
atoms of 12
C.
4. (a) 80.56% C; 7.51% H; 11.93% O; (b) 2.11 x 1021
molecules of C9H10O.
______________________________________________________________________________
Empirical and Molecular Formulas
The empirical formula of a compound gives the simplest whole number ratio of different types of
atoms in the compound. All salt formulas are empirical formulas. On the other hand, the
molecular formula of a compound may or may not be the same as its empirical formula. For
example, the molecular formula of butane is C4H10 while its empirical formula is C2H5. The
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molecular formula gives the true number of each kind of atom in a molecule.
Empirical formulas may be easily determined from experimental data.
Usually you must first determine how many grams of each type of atom are in the compound. If
percent composition data is given, assume that you have 100.0 g of the compound; then the
number of grams of each element is equal to the percentage for that element.
The next task is to convert the grams of each element to moles of the element. Be sure to keep at
least three significant figures in your answers.
The final step is to write the molar amounts of each element as subscripts in the formula. Then
divide all molar subscripts by the smallest value in the set. At this point, the subscripts may all
be very close to whole numbers; if so, you are finished. If one (or more) of the subscripts is not
close to a whole number, multiply all molar subscripts by the simple factor which makes all
subscripts whole numbers.
Once the empirical formula is determined, the molecular formula is easily found if the molar
mass (molecular weight) of the molecule is also known. You first calculate the molar mass of the
empirical formula. Then you divide the molar mass of the molecule by the molar mass of the
empirical formula. The division should give a simple whole number. That number is the factor
by which all subscripts in the empirical formula must be multiplied to obtain the molecular
formula.
______________________________________________________________________________
Exercises
1. The molecular formula of the antifreeze ethylene glycol is C2H6O2. What is the empirical
formula?
______________________________________________________________________________
2. A well-known reagent in analytical chemistry, dimethylglyoxime, has the empirical formula
C2H4NO. If its molar mass is 116.1 g/mol, what is the molecular formula of the compound?
______________________________________________________________________________
3. Nitrogen and oxygen form an extensive series of oxides with the general formula NxOy. One
of them is a blue solid that comes apart, reversibly, in the gas phase. It contains 36.84% N.
What is the empirical formula of this oxide?
______________________________________________________________________________
4. A sample of indium chloride weighing 0.5000 g is found to contain 0.2404 g of chlorine
What is the empirical formula of the indium compound?
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______________________________________________________________________________
Answers:
1. CH3O
2. Molar mass of empirical formula is 58.06 g/mol. Thus molecular formula is C4H8N2O2.
3. The ratios are N1.00O1.50 . Since 1.50 is not close to a whole number, we multiply both
subscripts by 2. The empirical formula is thus N2O3. (The name is dinitrogen trioxide.)
4. InCl3.
Chemical Stoichiometry Problems
Calculating the yield of a chemical reaction is a process at the heart of chemistry. While
there are many ways a problem can be phrased, in all cases the stoichiometric coefficients in the
balanced reaction are used to determine the mole ratios between reactants and products. Thus
the first step is usually calculating the moles of each species available. If an amount is given in
grams, the molar mass is used as a conversion factor to change grams to moles.
Limiting Reagent Problems
In some problems, amounts of more than one species are given. In that case your first task is
to determine which species is the limiting reagent. Just as you can make only 1 bicycle from 2
wheels and 4 handlebars (with 3 handlebars left over), and only 2 bicycles from 8 wheels and 2
handlebars (with 4 wheels left over), in chemical reactions some species are limiting while others
may be present in excess.
In the case of a bicycle, we need 2 wheels
1 handlebar
. We obtain analogous information about the
relative amounts of species that react from the stoichiometric coefficients in a balanced chemical
equation. For example, in Exercise (2) below the equation
CO(g) + 2 H2(g 3OH (l)
tells us we need 2 mol H2
1 mol CO
. If we have more than 2 moles of H2 for each mole of CO, CO will
be the limiting reagent and the excess H2 will not react. Conversely, if we have more than 1
mole of CO for every 2 moles of H2, H2 will be the limiting reagent and the excess CO(g) will be
left over. In each case, the yield of CH3OH is determined by the moles of limiting reagent
available.
Calculating the Theoretical Yield
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The theoretical (maximum possible) yield is based on the amount of limiting reagent
available. The yield is calculated in steps:
• Calculate moles of all reactants available. If amounts are given in grams, convert grams to
moles using the molar mass of each reactant as your conversion factor: 1 mole reactant
# g reactant
.
• NOTE: Skip this step if you have already identified the limiting reagent. To determine
which reagent is limiting, use the mole ratio obtained from the balanced equation for the
reaction to find the moles of reactant B needed to react with the available moles of reactant A.
If the moles of B available are less than the moles of B needed, reactant B is the limiting
reagent and reactant A is in excess. Conversely, if the moles of B available are more than
the moles of B needed, A is the limiting reagent and B is in excess.
• Calculate the moles of product based on the moles of limiting reagent available; use the
stoichiometric ratio of # moles product
# moles limiting reagent
as the conversion factor.
• If you are asked for the yield in grams, convert the yield in moles to a yield in grams using
the molar mass as your conversion factor: # g product
1 mole product
Percent Yield
Most reactions do not go to completion, and so the actual yield is less than the percent yield.
The percent yield is calculated as
Percent yield=actual yield
theoretical yield
100%
_____________________________________________________________________
Exercises: Use your own paper if you want more space.
1. Ammonia is produced by the reaction
3 H2(g) + N2(g 3(g)
(a) If N2(g) is present in excess and 55.6 g of H2(g) reacts, what is the theoretical yield of
NH3(g)?
(b) What is the percent yield if the actual yield of the reaction is 159 g of NH3(g)?
Answers: #1(a) 313 g NH3(g); (b) 50.8% yield.
______________________________________________________________________
2. Methyl alcohol (wood alcohol), CH3OH, is produced via the reaction
CO(g) + 2 H2(g 3OH (l)
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A mixture of 1.20 g H2(g) and 7.45 g CO(g) are allowed to react.
(a) Which reagent is the limiting reagent?
(b) What is the yield of CH3OH? [Assume theoretical yield in g is what is wanted here.]
(c) How much of the reagent present in excess is left over?
(d) Suppose the actual yield is 7.52 g of CH3OH. What is the % yield?
Answers: # 2(a) CO is the limiting reagent; (b) 8.52 g CH3OH; (c) 0.13 g H2; (d) 88.3%
Types of Chemical Reactions
One skill that chemists learn over time is that of writing and balancing equations. The first
task is deciding what type of reaction is taking place. In this chapter we study three types:
Precipitation Reactions: In these reactions two soluble salts usually react to form to an
insoluble salt (the precipitate!) and a soluble salt. The cations of the reacting salts exchange
anions.
Acid-Base Reactions: Most commonly an acid of the type HX or H2X reacts with a basic
hydroxide to form a salt plus water. Alternatively, the acid may react with ammonia (NH3) to
form an ammonium salt (but no water). These are proton transfer reactions in which H+ (the
proton) is transferred from the acid to the base.
Oxidation-Reduction Reactions: These are reactions in which one type of atom increases in
oxidation number (is oxidized) and another type of atom decreases in oxidation number (is
reduced). A large number of oxidation-reduction (redox) reactions contain one or more
reactants or products, which are pure elements.
Note that hydroxides can react with acids in acid-base reactions, and also with other salts in
precipitation reactions.
Writing Balanced Ionic Equations
The first step in writing a balanced equation is predicting the products of the reaction as
discussed above. Then the steps below are completed in sequence:
Balance the Molecular Equation: In the “molecular” equation, nothing is broken up into
ions. Salt formulas are written so that the cation charges exactly balance out the anion
charges so that the salt is neutral. Then the equation is balanced for atoms.
Balance the Total Ionic Equation: The first step in writing an ionic equation is to decide
what species should be broken up into ions. The rules below should help!
Break up into Ions Do NOT break up! Leave “as is”!
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Strong Acids. HCl, HBr, HI, HNO3, HClO4,
and H2SO4 are the most common examples;
assume other acids are weak.
Strong Bases. NaOH, KOH, or Ba(OH)2 are the
most common examples; assume other bases are
weak.
Soluble Salts. Salts of the alkali metals, salts
containing the NH4+ ion, the NO3
- ion, and other
salts as specified in Chang, Table 4.2, p. 119.
Weak Acids. Nearly all acids are
weak.
Weak Bases. Nearly all bases are
weak.
Insoluble Salts. Most salts are
insoluble.
Non-electrolytes or Weak
Electrolytes. Examples include H2O,
gases, pure elements, hydrocarbons,
and alcohols.
Balance the Net Ionic Equation: Identify all spectator ions: these are ions that are identical
on both sides of the balanced total ionic equation. Remove the spectator ions from the
equation. What remains is the net ionic equation. Finally, simplify the stoichiometric
coefficients if all of them are divisible by a common factor.
If all the ions are spectator ions so that nothing is left for your net ionic equation, no reaction
has taken place!
______________________________________________________________________________
Exercises
For each of the following reactions, complete the chart. Be sure to balance all of your
equations.
1. Mg(OH)2(s) + HCl(aq)
(a) Reaction type:
Formulas of Products Formed:
(b) Molecular Equation
(c) Total Ionic Equation
(d) Net Ionic Equation
Answer to 1.(d): Mg(OH)2(s) + 2 H+(aq
2+(aq) + 2 H2O(l)
______________________________________________________________________________
2. AgNO3(aq) + K2Cr2O7(aq)
(a) Reaction type:
Formulas of Products Formed:
(b) Molecular Equation
(c) Total Ionic Equation
(d) Net Ionic Equation
Answer to 2.(d): 2 Ag+(aq) + Cr2O7
2-(aq 2Cr2O7(s)
______________________________________________________________________________
3. NH3(aq) + HC2H3O2(aq)
(or CH3COOH)
13
(a) Reaction type:
Formulas of Products Formed:
(b) Molecular Equation
(c) Total Ionic Equation
(d) Net Ionic Equation
Answer to 3.(d): NH3 + HC2H3O2 4+(aq) + C2H3O2
-(aq)
______________________________________________________________________________
4. NaOH(aq) + H2SO4(aq)
(a) Reaction type:
Formulas of Products Formed:
(b) Molecular Equation
(c) Total Ionic Equation
(d) Net Ionic Equation
Answer to 4.(d): OH-(aq) + H
+(aq 2O(l) (obtain this after all coefficients have been
divided by 2)
______________________________________________________________________________
5. H2S(aq) + Ba(OH)2(aq)
(a) Reaction type:
Formulas of Products Formed:
(b) Molecular Equation
(c) Total Ionic Equation
(d) Net Ionic Equation
Answer to 5.(d): H2S(aq) + Ba2+
(aq) + 2 OH-(aq s) + 2 H2O(l)
Oxidation-Reduction Reactions
Oxidation-reduction (redox) reactions are reactions in which oxidation numbers change.
Oxidation numbers are either real charges or formal charges which help chemists keep track of
electron transfer. In practice, oxidation numbers are best viewed as a bookkeeping device.
Oxidation cannot occur without reduction.
In a redox reaction, the substance oxidized contains atoms which increase in oxidation
number. Oxidation is associated with electron loss (helpful mnemonic: LEO = Loss of
Electrons, Oxidation).
The substance reduced contains atoms which decrease in oxidation number during the
reaction. Reduction is associated with electron gain (helpful mnemonic: GER = Gain of
Electrons, Reduction).
An oxidizing agent is a substance which oxidizes something else: it itself is reduced! Also, a
reducing agent is a substance that reduces another reactant: it itself is oxidized.
A disproportionation reaction is a reaction in which the same element is both oxidized and
14
reduced.
How to Assign Oxidation Numbers: The Fundamental Rules
The oxidation number of any pure element is zero. Thus the oxidation number of H in H2 is
zero.
The oxidation number of a monatomic ion is equal to its charge. Thus the oxidation number
of Cl in the Cl- ion is -1, that for Mg in the Mg+2 ion is +2, and that for oxygen in O2- ion is
-2.
The sum of the oxidation numbers in a compound is zero if neutral, or equal to the charge if
an ion.
The oxidation number of alkali metals in compounds is +1, and that of alkaline earths in
compounds is +2. The oxidation number of F is -1 in all its compounds.
The oxidation number of H is +1 in most compounds. Exceptions are H2 (where H = 0) and
the ionic hydrides, such as NaH (where H = -1).
The oxidation number of oxygen (O) is -2 in most compounds. Exceptions are O2 (where O =
0) and peroxides, such as H2O2 or Na2O2, where O = -1.
For other elements, you can usually use the sum rule above to solve for the unknown
oxidation number.
Examples:
NO(g) has O = -2, so N = +2.
NO2 (g) has two oxygen atoms and each has O = -2. Thus N + 2(-2) = 0, so N = +4.
SO42-
has O = -2. Thus S + 4(-2) = -2. Solving the equation gives S = -2 + 8 = +6.
K2Cr2O7 has K = +1 and O = -2. Thus 2(+1) + 2 Cr + 7(-2) = 0; 2 Cr = 12; Cr = +6.
Recognizing Oxidation-Reduction Reactions
Oxidation-reduction reactions are reactions in which one type of atom increases in
oxidation number (is oxidized) and another type of atom decreases in oxidation number (is
reduced). Thus to show that a reaction is a redox reaction, you need to calculate oxidation
numbers for the atoms in the reactants and products, and document that changes are taking place.
There are, however, a few useful generalizations.
A large number (but not all!) of oxidation-reduction reactions contain one or more reactants
or products which are pure elements. Why is this true? Also, all electrochemical reactions
are redox reactions.
Most acid-base reactions and most precipitation reactions are not redox reactions. Why?
Give some examples!
Working with Solutions
15
Solutions are uniform mixtures on the molecular level of two or more substances. The
substance present in largest amount is called the solvent (usually water) and any substance
dissolved in the solvent is called a solute.
Molarity, M
The molar concentration or molarity, M, of a solution is used to indicate the number of moles
of solute per liter of solution:
Molarity M mol
L
(no.of moles solute)
(no. of liters of solution)
The molarity of a solution is often used as a conversion factor between moles of solute and
volume of solution: it is a “molar density”.
Dilution
One common type of “lab assistant” problem is the preparation of a dilute solution from a
more concentrated solution. For example, we might want to prepare 250 mL of a 0.500 M
NaOH solution from a 6.00 M NaOH solution as in exercise 5 below. There is a shortcut way to
work dilution problems which is based on the knowledge that the # of moles of solute you need
for the dilute solution all come from the concentrated solution. Thus
#moles concentrated #moles dilute and since M(mol L-1
) x V(L) = n (mol), it follows that
Mconcentrated Vconcentrated Mdilute Vdilute
or, in the notation of Chang,
Minitial Vinitial Mfinal Vfinal
In the laboratory this equation is often used to determine the Vconcentrated that needs to be diluted
to give the desired volume of a more dilute solution.
Stoichiometry of Reactions in Solution
Problems involving solutions are very similar to the chemical stoichiometry problems we
have discussed earlier. The only difference is that the moles of reactant or product may need to
be calculated from a solution volume using the molarity (M = mol/L) as a conversion factor
between volume and moles.
Lab Assistant Problems
The problems below will introduce you to calculations involving molarity. I call these “lab
assistant problems” since we do this kind of calculation all the time when setting up labs! When
working these problems, it is useful to recall that 1 L = 1000 mL.
16
______________________________________________________________________
Exercises:
1. What is the molarity of a solution containing 21.0 g NaCl in 200 mL of solution?
Answer: 1.80 M NaCl
_____________________________________________________________________
2. What ions exist in a 0.245 M Na2CO3 solution? Give the molar concentration of each ion.
Answer: 0.490 M Na+ and 0.245 M CO32-
______________________________________________________________________
3. How many mL of a 0.420 M NaCl solution should be measured out to provide 1.5 g NaCl?
Answer: 61 mL of 0.420 M NaCl
_____________________________________________________________________
4. How many grams of CaCl2 are needed to make 200 mL of a 0.500 M Cl- solution? (Note:
CaCl2 is a soluble salt. The molar mass of CaCl2 is 110.98 g/mol.)
Answer: 5.55 g CaCl2
_____________________________________________________________________
5. How do you prepare 250 mL of a 0.500 M NaOH solution from a 6.00 M NaOH solution?
Answer: Take 20.8 mL of 6.00 M NaOH and dilute to a total volume of 250 mL.
_____________________________________________________________________
pH and Titration
17
Relation between pH and Hydrogen Ion Concentration
The pH scale is widely used to report the molar concentration of hydrogen ion, H+(aq), in
aqueous solution. The pH of a solution is defined as
pH log10[H] (1)
where [H+] = the molar concentration of H+(aq) in the solution. (In chemistry, square brackets
around a chemical symbol mean "the molar concentration of" whatever they enclose.) Equation
(1) above may be solved for [H+] to give
[H]10pH (2)
(Here we use the well known rule that if y log10 x , then x 10y
.) In practice, the pH scale is
only used when [H+(aq)] is less than 1.0 M. Acidic, basic, and neutral solutions can be
distinguished as shown below:
Type of Solution pH [H+] Color of litmus
Acidic < 7.00 > 1.0 107 pink
Neutral = 7.00 = 1.0 107 in between
Basic > 7.00 < 1.0 107 blue
Titration
A titration is a procedure in which a solution of known concentration is used to determine the
concentration of another solution with which it reacts. The reaction must be rapid and should go
to completion. It may be an acid-base reaction, an oxidation-reduction reaction, or a
precipitation reaction.
Typically a titration is conducted by filling a buret with one solution and transferring an exact
amount of the second solution to an Erlenmeyer (conical) flask with a pipet. Indicator is added
to the flask, and the first solution is added drop wise from the buret until the indicator changes
color. The point of color change is called the endpoint, the equivalence point, or the
stoichiometric point of the titration: all of these terms are synonymous. The indicator is chosen
so the color change occurs when stoichiometric amounts of the reactants have been added to the
flask.
The concentration of the unknown solution is calculated as illustrated in the exercise 3 below.
___________________________________________________________________________
Exercises:
18
1. Use Eq (1) and the log10 button on your calculator to determine the pH of solutions with the
specified hydrogen ion concentrations [H+]:
(a) (b) (c) (d) (e) (f)
[H+] 0.10 M 0.0010 M 10
-7 M 5.0 x 10-10 M 6.0 M 1.0 M
pH
Acidic,
basic
or
neutral
?
Answers: pH values are
(a) 1.00 (b) 3.00 (c) 7 (d) 9.30 (e) - 0.78 (f) 0.00
Ideal Gas Problems
Gases at low pressures obey the ideal gas law,
p V = n R T (1)
where R is a constant (known as the gas constant) that has the value
R = 0.08206 L atm K-1 mol-1 (2)
Appropriate units to use for p, V, n, and T in the ideal gas equation are those used for R above.
Thus the pressure (p) should be in atm, the volume (V) in L, the temperature (T) in degrees K,
and the amount of gas (n) should be in moles. Useful conversion factors are
Pressure:1 atm = 760 Torr = 760 mmHg = 101.3 kPa = 1.013 bar
Temperature:K = 273 + oC
Volume:1 L = 1000 mL = 1000 cm3
Since pV
nT R , and R is a constant, it follows that
p1V1
n1T1
p2V2
n2T2
(3)
19
where the subscript “1” represents one set of conditions, and the subscript “2” represents another
set of conditions. More specialized equations may be derived from Eq(3) when one or more of
the variables is held constant. For example, you can easily derive the familiar equations given
below in this manner (convince yourself that this works!):
Boyle‟s law: p1V1 p2V2 (obtained when n1 = n2 and T1 = T2)
Charles‟s law:
V1
T1
V2
T2
(obtained when n1 = n2 and p1 = p2)
Avogadro‟s Principle:
V1
n1
V2
n2
(obtained when T1 = T2 and p1 = p2)
STP
Often you will see gas volumes reported at STP (standard temperature and pressure). STP is
defined as T = 273 K (0oC) and p = 1 atm. Substitution of these values into Eq(1) shows that
the volume of 1 mol of any gas is approximately 22.4 L at STP. (You should verify this for
yourself using Eq(1)!).
Gas Density (d) and Molar Mass (M)
Rearranging the ideal gas equation and using the definitions of density d and molar mass M
gives
n(mol)
V(L)
p
RT and d
g
L
n(mol)
V(L) M
g
mol
pM
RT (4)
Note: M (in italics) is molar mass in g/mol, while M (no italics) is molarity in g/L
___________________________________________________________________________
1. What is the volume occupied by 35.4 g of nitrogen gas at 35oC and 735 Torr?
Answer: 33.0 L
___________________________________________________________________________
2. A scuba diver inhales a lung-full (350 mL) of air at a depth of 33 ft where the pressure is
approximately 2.0 atm and the water temperature is 18oC. If the diver holds her breath (not a
good idea!!), what volume will the same amount of air occupy at sea level where the pressure is
approximately 1.0 atm and the air temperature is 35oC?
20
Answer: 741 mL (Note: You can either use Eq(3) or you can use Eq(1) twice to solve this
problem.)
___________________________________________________________________________
3. (a) Calcium carbonate reacts with hydrochloric acid to produce carbon dioxide gas. If 35.3 g
of calcium carbonate reacts with 100 mL of 6.00 M HCl, how many liters of carbon dioxide gas
will be produced at 745 mmHg and 23.0oC?
Hints:
• Begin by writing the balanced equation for the reaction.
• This is a limiting reagent problem (why?), so you will next need to determine whether
calcium carbonate or hydrochloric acid is the limiting reagent.
• Once you have determined the identity of the limiting reagent, you can calculate the moles of
carbon dioxide produced.
• The last step is to find the volume of carbon dioxide using the ideal gas law.
(b) What volume of carbon dioxide gas would be obtained at STP?
Answers: (a) 7.43 L of carbon dioxide; (b) 6.72 L of carbon dioxide
Ideal Gas Mixtures
Dalton’s Law of Partial Pressures
The partial pressure of a gas in a mixture is the pressure it would exert if alone in the
container. Dalton’s law of partial pressures says that the total pressure of a mixture of gases is
the sum of the partial pressures. Thus, for a mixture of n gases,
pTotal p1 p2 p3 ... pi
i1
n
(1)
where
pi ni
RT
V
, pTotal nTotal
RT
V
and nTotal n1 n2 n3 ... nii1
n
(2)
It follows from Eq(2) that
21
pi
pTotal
ni(RT V )
nTotal(RT V )
ni
nTotal
Xi (3)
where Xi is the mole fraction of component i.
Vapor Pressure, Relative Humidity, and the Dew Point
A gas commonly present in gas mixtures is water vapor, H2O(g), which exerts a partial
pressure, pH 2O . The maximum value of pH 2O at a given temperature is the vapor pressure of
water. The vapor pressure of a gas is the pressure exerted by a vapor in equilibrium with its
liquid in a closed container. In experiments in which a gas is collected over water, pH 2O
contributes to the total gas pressure in the container (see exercise (1) below).
Weather reports often give the relative humidity which is the percent of the equilibrium vapor
pressure at the reported temperature which is actually present in the atmosphere:
Relative Humidity = pH2O
(vapor pressure)100% (4)
Since the vapor pressure increases as the temperature increases, a relative humidity of 90%
indicates a much higher pH 2O on a hot day in summer than on a cold day in winter! As the
temperature drops for a given pH 2O , the relative humidity increases (since the vapor pressure
decreases as T decreases); when the relative humidity reaches 100% the dew point has been
reached, and water vapor begins to condense (as “dew” or “frost”).
Average Molecular Speed (Root-Mean-Square Speed)
At a given temperature a molecule of any gas has the same average kinetic energy,
KE 1
2mu
2. This implies that gas molecules with low molar masses must have higher average
speeds u2
than those with high molar masses (why?). The average (root-mean-square) speed of a
gas molecule with molar mass M may be calculated at a temperature T :
average speed = root-mean-square speed = urms u2
3RT
M (5)
Units: We want to calculate the average speed in units of meters per second (m s-1
). To do this
the gas constant R is expressed in energy units (R = 8.314 J K-1
mol-1
where 1 J = 1 kg m2 s
-2),
and the molar mass M is expressed in units of kilograms per mole (kg mol-1
).
It follows from Eq (5) that gases with low molar masses have higher average speeds and
hence higher rates of effusion and diffusion than gases with higher molar masses. (The
spreading of one substance through another substance is called diffusion. The escape of one
substance through a small hole is called effusion.) Gases with low molar masses leak easily out
of containers and, if they have an odor, the smell will quickly permeate a room.
22
_____________________________________________________________________________
Exercises.
1. A mixture of NO(g) and NO2(g) was collected over water at 28.0oC and 745 mmHg (i.e.,
pTotal = 745 mmHg). At 28.0oC the vapor pressure of water is 28.3 mmHg.
(a) If pNO = 368 mmHg, what is the partial pressure of NO2(g) in the mixture?
(b) Calculate the mole fraction of NO2(g) in the mixture.
Answer: (a) partial pressure of NO2 XNO 2 = 0.468
_____________________________________________________________________________
2. Two flasks at the same temperature are joined by a glass tube with a stopcock. Flask A is a
4.0 L flask containing N2(g) at 2.0 atm, while flask B is a 10.0 L flask containing CO(g) at 1.4
atm. What is the final pressure in the flasks after the stopcock is opened?
Hints:
• Determine the final volume for the gases (easy!).
• Use Boyle's law (why?) to find the final partial pressure for both N2 and CO individually.
• Finally, use Dalton's law of partial pressures to find the total final pressure.
Answer: After the stopcock is opened, p(N2)= 0.57 atm, p(CO)= 1.0 atm, and pTotal = 1.57 atm
tm
______________________________________________________________________________
3. Calculate the root-mean-square speeds of He and benzene (C6H6) at 25oC. Give two reasons
why He will leak more readily than benzene out of a minute opening in a container.
Answer: He: u2
= 1.36 x 103 m s
-1; Benzene: u
2= 3.08 x 10
2 m s
-1; He atoms are much smaller
than benzene molecules and will leak out of smaller openings, and u2
is much larger for He than
for benzene.
Calorimetry Exercises ____________________________________________________________________________
1. When 12.29 g of finely divided brass (60% Cu, 40% Zn) at 95.0oC is quickly stirred into
40.00 g of water at 22.0oC in a calorimeter, the water temperature rises to 24.0
oC. Find the
specific heat of brass.
Hints:
• The heat lost by the brass is gained by the surroundings (the water plus the calorimeter).
What relation can you therefore write between qbrass and qsurr?
• Since no information is given about the heat capacity of the calorimeter, you should assume it
23
is negligible.
• The final temperature of the brass is the same as the final temperature of the water. The
specific heat of water, s(H2O), is 4.184 J g-1
oC
-1.
Answers: qsurr = q(H2O) = 334.7 J; qbrass = - 334.7 J; sbrass = 0.38 J g-1
oC
-1.
______________________________________________________________________________
2. In an experiment, 400. mL of 0.600 M HNO3(aq) is mixed with 400. mL of 0.300 M
Ba(OH)2(aq) in a constant-pressure calorimeter having a heat capacity of 387 J/oC. The initial
temperature of both solutions is the same at 18.88oC, and the final temperature of the mixed
solution is 22.49oC. Calculate the heat of neutralization in kJ per mole of HNO3.
Hints:
The heat evolved in the neutralization reaction is gained by the surroundings (the mixed
solution plus the calorimeter). What relation can you therefore write between qrxn and qsurr?
There are two contributions to qsurr. What are they? What assumptions (if any) need to be
made in calculating these contributions?
Is this a limiting reagent problem, or are reactants supplied in the stoichiometric ratio given
by the equation? (Why do we care about this?)
We want our answer in kJ per mole of HNO3. How do we calculate that?
Answers: ∆T = 3.61oC; qsolution = 12083.4 J; qcalorimeter = 1397.1 J; qsurr = 13481 J; qrxn = - 13481 J;
neut(kJ/mol HNO3) = - 56.2 kJ/ mol HNO3.
The Enthalpy of Chemical Change:
Calculations using Hess's Law and Heats of Formation
Enthalpy of reaction values have been determined experimentally for numerous reactions,
and these ∆H values may be used to calculate ∆H values for other reactions involving the same
chemical species. The reason this is possible is that enthalpy H is a state property so ∆H is
independent of path. (Similarly, the height of a mountain above sea level is independent of the
path you follow to climb the mountain.) Because ∆H is independent of path, we can determine
the enthalpy of foods by burning them in a bomb calorimeter in the laboratory to produce the
same products that are obtained by the complicated metabolic pathways in our body!
There are two principle methods used to calculate ∆H values for a reaction, both of which are
based on the idea that ∆H for a reaction is independent of the path used to go from reactants to
products. The first makes use of Hess's Law while the second employs tabulated heats of
formation H fo
(kJ/mol).
Use of Hess's Law to Calculate ∆H
Hess's Law states that ∆H for a reaction can be found indirectly by summing ∆H values for
any set of reactions which sum to the desired reaction. Usually before reactions are added
together, some of them must be reversed and/or multiplied by a factor n in order that they sum to
the desired reaction. In this process the rules are:
24
• Whenever you multiply a reaction by n, ∆H for the reaction is also multiplied by n.
• If you reverse a reaction, ∆H changes sign.
Problem (1) below is an example of how this procedure is used.
Use of Tabulated Heats of Formation to Calculate ∆H
The Standard Heat of Formation H fo
(kJ/mol) for a compound is the heat absorbed (or
released) in forming one mole of the compound from its elements in their standard states at 1 bar
(≈ 1 atm) pressure and the specified temperature (usually 25oC). Thus H fo
(kJ/mol) for acetone
CH3COCH3 is the heat of the reaction
3 C(s) + 3 H2(g) + 1/2 O2(g) CH3COCH3(l) ∆H = – 246.8 kJ (1)
and so H fo
(CH3COCH3(l)) = – 246.8 kJ/mol). By definition, H fo
(kJ/mol) = 0 for any
element in its standard state at 25oC and 1 bar.
Tabulated heats of formation H fo
(kJ/mol) are given in Table 6.4, p. 238, and Appendix 3, pp.
A-8 to A-12, in Chang. These may be used to calculate the standard enthalpy change, Hrxn
o, for
any reaction for which the heats of formation of all reactants and products are known:
Hrxn
o nprodH f
o( prod)
products
nreactH f
o(react)
reactants
(2)
This equation tells us to sum the enthalpies of formation of each product multiplied by its
stoichiometric coefficient in the reaction equation and then to subtract the enthalpy of formation
of each reactant multiplied by its stoichiometric coefficient. We use this equation to work
problem (2).
Hrxn
o applies for a balanced equation with specific stoichiometric amounts. If a different
number of moles reacts, the heat absorbed or evolved will change proportionately (problem 3).
______________________________________________________________________________
1. From the following heats of reaction
2 SO2(g) + O2(g) 2 SO3(g) ∆H = – 196 kJ (a)
2 S(s) + 3 O2(g) 2 SO3 (g) ∆H = – 790 kJ (b)
calculate the heat of reaction for
S(s) + O2(g) SO2(g) ∆H = ? kJ (c)
Method: Use Hess‟s Law to solve this problem:
• Identify a species in the target equation (c) which is on the correct side in only one of the
listed equations, (a) or (b). Multiply the entire equation, and its ∆H value by the factor n
necessary to make the stoichiometric coefficient for the species identical to that in equation
(c).
• Reverse a listed equation, (a) or (b), and change the sign of its ∆H value if it contains a
species which is on the wrong side of the target equation (c); next multiply the entire reversed
25
equation by the factor n necessary to make the stoichiometric coefficient for the species
identical to that in equation (c). The ∆H value for the rewritten equation is ( - n) times that of
the original equation. (Ignore any species present in both equations (a) and (b).)
• Test to see if your rewritten equations now sum to the desired equation (c). If they do, the ∆H
value for equation (c) is the sum of the values of the rewritten equations.
Answer: (1/2) Eq(b) + ( - 1/2) Eq(a) = Eq(c); Thus, (1/2) ∆H (b) + ( – 1/2) ∆H (a) = ∆H (c) so
∆H(c) =(1/2)( – 790 kJ) + ( – 1/2)( – 196 kJ) = kJ
______________________________________________________________________________
2. Calculate the standard reaction enthalpy for the photosynthesis reaction,
6 CO2(g) + 6 H2O(l 6H12O6(s) + 6 O2(g) Hrxn
o = ? kJ
Note: The heat of formation of glucose, C6H12O6(s), is given in Appendix 3 on pg A-12.
Answer: Use Eq (2) with H fo
(CO2(g)) = – 393.5 kJ/mol, H fo
(H2O(l)) = – 285.8 kJ/mol, H fo
(C6H12O6(s)) = – 1274.5 kJ/mol, and H fo
(O2(g)) = 0 kJ/mol. Thus ∆Ho = 2801.3 kJ ≈ 2801
kJ for the photosynthesis reaction.
______________________________________________________________________________
3. Is the photosynthesis reaction above endothermic or exothermic? How much heat is absorbed
or evolved if 11.0 g of CO2(g) reacts completely with excess water to form glucose and oxygen
gas?
Answers: Endothermic. 117 kJ of heat is absorbed.
Electromagnetic Radiation and the Spectrum of Atomic Hydrogen
To make sense of the chemistry of the elements we need to understand the electronic
structure of atoms. It is the atom‟s electronic structure which governs everything from molecular
geometry to chemical reactivity. Electromagnetic radiation is the probe we use to obtain
knowledge of electronic structure, so we begin by looking at some of its properties.
Light has a dual nature – it is both wave-like and particle-like. Thus light and all other forms
of electromagnetic radiation obey two equations, one of which shows the inverse relation
between wavelength and frequency (both properties of waves) and the other which relates the
energy of light photons (“particles”) to their frequency :
Wave model (m) (s1
) c(m s) (1)
Particle model Ephoton(J) h(J s) (s1
) (2)
The constants and variables in these equations are
26
Speed of light = c = 3.00 x 108 m/s = 3.00 x 108 m s-1
Planck's constant = h = 6.626 x 10-34 J s
Wavelength = = Greek lower case lambda
Frequency = = Greek lower case nu
Photon Energy = Ephoto n
Common units include
Wavelength : 1 nm = 10-9 m = 10 Å
Frequency : 1 s-1 = 1 Hz
The H Atom Emission Spectrum
Our modern theory of the electronic structure of atoms began with attempts to understand the
emission spectrum of the simplest atom – hydrogen. By the end of the nineteenth century it was
established that the frequencies of the spectral lines in the H atom emission spectrum all fit a
very simple formula (see Eq (6) ahead). The famous Bohr model of the H atom proposed by Bohr
in 1914 was an attempt to understand where this formula came from. Bohr recognized that the
frequencies would be predicted correctly if the H atom energy levels obeyed the equation
En RH
1
n2
for n = 1, 2, 3, ... (Applies only to H atom!!) (3)
where
RH = Rydberg constant = 2.18 x 10-18 J
While the Bohr model was incorrect in assuming that the electrons orbit the nucleus like planets
orbit the sun, it was correct in predicting the energies of the H atom energy levels.
Absorption and Emission
Absorption of electromagnetic radiation occurs when electrons make transitions from lower
to higher energy levels. Photons provide the energy required for the jumps. Similarly, when
electrons move from higher to lower energy levels, photons carry off the excess energy:
electromagnetic radiation is emitted and an emission spectrum is obtained. In absorption and
emission, photon energies and frequencies obey the equations
Energy Gained in Absorption Ephoton h Eupper Elower (4)
Energy Lost in Emission Ephoton h Eupper Elower (5)
These equations apply to all atoms and molecules.
In the case of the H atom Eq (3) gives the energies of Eupper and Elower once the quantum
numbers, nupper and nlower are specified. Using these values in Eq (5) allowed Bohr to predict the
correct frequencies for the emission lines of the H atom:
E photon h RH 1
nlower
2
1
nupper
2
(Applies only to the H atom!!) (6)
27
______________________________________________________________________________
1. A violet photon has a frequency of 7.100 x 1014 Hz. (a) What is the wavelength (in nm) of
the photon? (b) What is the wavelength in Å ? (c) What is the energy of the photon? (d) What is
the energy of 1 mole of these violet photons?
Answer: (a) 422.5 nm ; (b) 4225 Å ; (c) 4.704 x 10-19 J ; (d) 283.3 kJ
______________________________________________________________________________
2. The lines in the visible region of the emission spectrum of the H atom are called the Balmer
lines. All of them involve transitions in which nlower = 2. Calculate (a) the frequency and (b) the
wavelength of the lowest energy Balmer emission line. (c) What color is this line?
Hint: Use Eq (6) with nupper = 3 (why?); RH = Rydberg constant = 2.18 x 10-18 J.
Answer: (a) 4.57 x 1014 s-1; (b) 656 nm; (c) It‟s the Balmer red line! (see Chang, pg 267).
______________________________________________________________________________
3. Derive Eq (6) from Eq (5) for the H atom.
Hint: First use Eq (3) to calculate Eupper and Elower in terms of nupper and nlower
respectively. Then substitute these expressions into Eq (5) and rearrange.
Quantum Numbers, Orbitals, and Electron Configurations
Quantum Mechanics and the H Atom
Bohr theory (1914) ran into problems when it was applied to atoms other than H, and was
soon replaced by quantum mechanics (1926). When quantum mechanics was applied to the H
atom, the same energies were calculated for the H atom energy levels as in the Bohr model,
En RH
1
n2
for n = 1, 2, 3, ... (Applies only to H atom!!) (1)
but the method by which they were calculated was entirely different.
Quantum mechanics assumes the electron is wave-like, and the planetary orbits of Bohr
theory (which views the electron as a particle) are discarded. Orbitals, which give us a picture of
the most probable locations for the electron in a particular energy state, replace the Bohr orbits.
Since the electron is viewed as a wave, it is impossible to describe its precise location: an
averaged picture is the best we can do! Another change is that while the quantum number n still
plays the prominent role, it is augmented by the quantum numbers l, ml, and ms. The allowed
quantum numbers for the H atom are given in the tables below.
28
Table 1.
Quantum
Number
Allowed Values Name and Meaning
n n = 1, 2, 3, ...... Principal quantum number: orbital energy
and size.
l l = (n-1), (n-2), ...., 0 Azimuthal (or orbital) quantum number:
orbital shape (and energy in a multi-electron
atom), letter name for subshell (s, p, d, f)
ml ml = l, (l-1), ..., 0, ..., (-l+1), -l Magnetic quantum number: orbital
orientation
ms ms = 1/2, -1/2 Electron spin quantum number: spin up ( )
or spin down ( ).
Table 2
l Value
Letter Equivalent
to l Value
No. of
Orbitals
in Set
Approximate Shape of Orbitals
with Specific l Values
0 s 1 spherical
1 p 3 px, py, pz are dumbbells along x, y, and z
axes
2 d 5 mostly cloverleaf shapes
3 f 7 very complicated shapes!
Table 3
Shell
(n)
Subshell
(l)
Orbital
Name (nl)
Orientations
(ml)
No. of
Orbitals
Maximum
Occupancy
n = 1 l = 0 1s ml = 0 1 2 e-
n = 2 l = 0 2s ml = 0 1 2 e-
l = 1 2p ml = 1, 0 -1
(or px, py, pz)
3 6 e-
n = 3 l = 0 3s ml = 0 1 2 e-
l = 1 3p ml = 1, 0, -1
(or px, py, pz)
3 6 e-
l = 2 3d ml = 2, 1, 0, -1, -2
(or dxy, dyz, dxz, dx2 y 2
, dz2
)
5 10 e-
Orbital Energies and Electron Configurations of Multi-Electron Atoms
For the H atom the orbital energy depends only on n, so all orbitals with the same value of n
have the same energy. This is not true, however, for any other atom!
The H atom orbitals may be used to approximate the orbitals for multi-electron atoms. But
since these atoms have more than one electron, electrons in the outer orbitals are shielded
somewhat from the nucleus: they do not feel the full nuclear charge. Orbitals with a lower l
value penetrate closer to the nucleus and are less shielded and have a lower energy than those
with a higher l value. The result is that for a given value of n the energy order is s < p < d < f.
29
Orbitals are filled from lowest energy to highest energy. Each orbital holds 2 electrons (Pauli
Exclusion Principle), one with spin up ( ) and one with spin down ( ). If more than one
orbital has the same energy (e.g., px, py, pz ), electrons first enter different orbitals with spins
parallel (Hund’s Rule); once each orbital in the set contains one electron, additional electrons
form pairs.
The order of orbital filling is easily remembered if correlated with the periodic chart. The
order is 1s (first row of chart); 2s, 2p (second row); 3s, 3p (third row); 4s, 3d, 4p (fourth row); 5s,
4d, 5p (fifth row); 6s, 4f, 5d, 6p (sixth row); 7s, 5f, 6d. There are a few exceptions to these rules.
______________________________________________________________________________
1. (a) List all the orbitals for which n = 4. (b) How many orbitals are there in all?
Answer: (a) When n = 4, allowed values of l are: l = 0=s, l = 1=p, l = 2=d, and l = 3=f. Thus we
have one 4s, three 4p, five 4d, and seven 4f orbitals. See Table 1 on p. 1 for the allowed ml
values for each value of l. For example, when l = 3, ml = 3, 2, 1, 0, -1, -2, -3. (b) 1 + 3 + 5 + 7
= 16 orbitals (each holds 2 e-).
______________________________________________________________________________
2. Which of the following subshells cannot exist: (a) 1p; (b) 4f; (c) 2d; (d) 5p; (e) 3f? Why not?
Answer: (a), (c), (e) [See Table 1 for the relation between the allowed values of l and value of n]
______________________________________________________________________________
3. List all possible values of ml for each of the indicated subshells. What role does the principal
quantum number n play in determining your answer?
Subshell Values of ml
(a) 4s
(b) 2p
(c) 3d
(d) 5f
Answer: (a) 0; (b) 1, 0, -1; (c) 2, 1, 0, -1, -2; (d) 3, 2, 1, 0, -1, -2, -3; Principal quantum number
plays no role: answers do not depend on n
______________________________________________________________________________
4. Give the formula that relates the number of possible values of ml to the value of l.
Answer: 2l +1 (why?)
30
Periodic Trends
Here we summarize trends for the main group elements (Columns 1A - 8A). Trends for the
transition metals, the lanthanides, and the actinides may differ.
Sizes of Atoms and Ions
Neutral Atoms (or Ions with the Same Charge).
• Size increases as you go down a column. Why? As you go down a column, electrons are
filling orbitals farther and farther out from the nucleus. Each row adds a new shell. Outer
electrons are shielded from the nucleus by electrons in inner shells; thus they are less tightly
held (in spite of the much increased nuclear charge).
• Size decreases as you go across a row. In this case electrons are being added to the same
shell. Thus they experience little additional shielding. On the other hand, the nuclear charge
of the atom increases with the atomic number. Thus as you go across a row, the electrons are
held more tightly and the size decreases.
Isoelectronic Series. These are series of atoms and ions in which the number of electrons stays
constant, but the number of protons increases with the atomic number. In this type of series, the
size of the atom decreases as the number of protons increases. The reason for the size decrease
is that more protons are pulling in the same number of electrons. Examples include the series
below in which the largest member of the series is listed first:
10 electron series: Ne > Na+ > Mg2+ > Al3+
18 electron series: P3- > S2- > Cl – > Ar
Cation Size as Compared to Parent Atom. The size decreases when cations form. The effect is
particularly pronounced when all the valence electrons are lost and only the noble gas core of
electrons remains. For example, the Mg2+ ion (65 pm radius) is considerably smaller than the
Mg atom (160 pm radius).
Anion Size as Compared to Parent Atom. The size increases when anions form. The added
electrons are going into the same shell. They repel each other and so the size increases. Thus the
Cl– ion (181 pm radius) is considerably larger than the Cl atom (99 pm radius).
Ionization Energies
The ionization energy I is the minimum energy needed to remove an electron from the
ground state of a gaseous atom, A(g).
A(g) A+(g) + e–(g) ∆E = I = I1
More precisely, this is the first ionization energy I1. Additional electrons may be removed with
ionization energies I2, I3, etc., for the removal of the second, third, etc., electrons. Ionization is
always an endothermic process: it requires energy to remove an electron from an atom or ion.
The overall trends in ionization energy are opposite to those for atomic and ionic radii. The
more tightly electrons are held, the higher the ionization energy, and the smaller the atom or ion
size. Some generalities are as follows:
• Noble gases have the highest ionization energies of the atoms in each row.
• Alkali metals have the lowest ionization energies of the atoms in each row.
31
• In general, ionization energies increase as you go across a row, but there are a few local ups
and downs. Dips occur with the loss of the first and the fourth p electron: Thus in the second
row, there are dips for boron and for oxygen.
• The ionization energy decreases for atoms as you go down a column.
• Higher ionization energies are always larger than lower ionization energies: I1< I2< I3 , etc.
A huge jump in ionization energy occurs when you first pull an electron out of the noble gas core.
Electron Affinity
The electron affinity EA is the energy released when an electron is added to a gas-phase
atom (or ion) of the element. The sign convention is opposite to that for ∆H. If the process is
exothermic, ∆H is (-) and EA is (+); if it is endothermic, ∆H is (+) and EA is (-).
A(g) + e–(g) A–(g) ∆H = - EA
While ionization energies are always positive numbers, electron affinities can be either positive
or negative. A high positive EA (and thus a (-) value of ∆H) indicates that gaining an electron is
a very favorable process. The halogens have the most positive electron affinities of all the
elements.
Electronegativity
The electronegativity (Greek letter chi) is a measure of the ability of an atom to attract and
hold electrons. Elements that readily form negative ions have high electronegativities, while a
low electronegativity correlates with the tendency to lose electrons and form positive ions.
Values of range from a high of = 4.0 for F to a low of = 0.7 for Cs. In general
electronegativities increase diagonally from the lower left (Cs) to the upper right (F) of the
periodic chart.
In practice, chemists use electronegativities far more than ionization energies or electron
affinities.
______________________________________________________________________________
Exercises:
1. In each of the following pairs, circle the species with the higher first ionization energy:
(a) Li or Cs (b) Cl- or Ar (c) Ca or Br (d) Na+ or Ne (e) B or Be
______________________________________________________________________________
2. In each of the following pairs, circle the species with the larger atomic radius:
(a) Mg or Ba (b) S or S2- (c) Cu+2 or Cu (d) He or H- (e) Na or Cl
______________________________________________________________________________
3. Circle the best choice in each list:
(a) highest first ionization energy: C, N, Si
(b) largest radius: S2–, Cl–, Cl
(c) highest electronegativity: As, Sn, S
32
(d) smallest atom: Na, Li, Be
(e) most paramagnetic: Fe, Co, Ni
(f) lowest first ionization energy: K, Na, Ca
(g) highest second ionization energy: Na, Mg, Al
(h) lowest second ionization energy: Ar, K, Ca
______________________________________________________________________________
Answers (be sure you can explain the reason for each answer!):
1. (a) Li; (b) Ar (isoelectronic pair); (c) Br; (d) Na+ (isoelectronic pair); (e) Be (common
exception: what is the rule here?).
2. (a) Ba; (b) S2-; (c) Cu; (d) H- (isoelectronic pair); (e) Na.
3. (a) N; (b) S2- (S2- and Cl- are isoelectronic); (c) S; (d) Be; (e) Fe (hint: determine no. of
unpaired spins for each element); (f) K; (g) Na; (h) Ca.
33
sp
C Atom
H
C
HH
H
2s 2p
Promotion Step
C Atom
2s 2p
Hybridization step( leads to energy release after bonding, since more bonds can be formed)
sp3
For 4 groups: tetrahedral
electron pair geometry
C C
Hybridization of Carbon
For 3 groups: trigonal planar
electron pair geometry
H
H
H
H
For 2 groups: linear
electron pair geometry
sp2 pz
C
py
leftover
H
H
sp py, pzTwo hybridized
AO'sleftover
sp
Example: CH 4
Use to form
3 single ()
bonds
Use pz to
form 1
pi () bond
H C
Example: C 2H4
bond
C
Each C forms three bonds and
one bond. Bond angles are 120 o.
H
xy plane
Use to form
2 single ()
bonds
Use py
and pz to
form 2 pi()
bonds
Example: C 2H2
Each C forms two bonds
and two bonds (which are
perpendicular to each other).
Bond angles are 180 o.
( requires energy)
Three hybridized AO's
Four hybridized AO's
pzxy plane
C
H
pz
H
C CH H
Use to form 4 single
(sigma = ) bonds
sp2
sp2sp2
sp3 sp3
sp3
sp3
C forms four bonds.
Bond angles are 109.5 o.
34
Total #
of
Groups of
e-
Electron Pair
Geometry
(Hybridization)
Approximate
Bond Angle
# of
Bonding
Directions
(# of X)
# of
Lone Pairs
(# of E)
Geometry Name
(VSEPR class)
Shape
Examples
2 linear
(sp)
180o 2 0 linear
(AX2)
BeH2,
CO2
3
trigonal planar
(sp2)
120o
3
0
trigonal planar
(AX3)
BF3, NO3–
2 1 bent
(AX2E)
SO2
4
tetrahedral
(sp3)
109.5o
4
0
tetrahedral
(AX4)
CH4
3 1 trigonal
pyramidal
(AX3E)
NH3
2 2 bent
(AX2E2)
H2O
5
trigonal
bipyramidal
(sp3d)
120o (in plane) &
90o (above &
below)
5
0
trigonal
bipyramidal
(AX5)
PCl5
4 1
seesaw
(AX4E)
SF4
3 2
T-shaped
(AX3E2)
ClF3
2 3
linear
(AX2E3)
XeF2
6
octahedral
(sp3d2)
90o
6
0
octahedral
(AX6)
SF6
5 1 square
pyramidal
(AX5E)
BrF5
35
4 2 square planar
(AX4E2)
XeF4
3 3 T-shaped
(AX3E3)
2 4 linear
(AX2E4)
Predicting Molecular Geometry and Hybridization
1. In each case, predict (a) the approximate bond angle(s), (b) the hybridization around the
underlined atom. (Note: It is helpful to first sketch the Lewis structure!)
Molecule or Ion (1) OF2 (2) H2CO (3) NO2+ (4) BF3 (5) SbF5
(a) No. of valence e -
„s
(b) Lewis structure
(c) Approximate
bond angle(s)
(d) Hybridization
(e) Polar or non-
polar molecule?
Ion: Not
applicable
(f) Geometry name
______________________________________________________________________________
2. For each of the molecules below fill in the indicated items in the chart. The central atoms are
underlined.
Molecule (1) SO2 (2) HBF2 (3) XeF4 (4) CH2Cl2 (5) NF3
(a) No. of valence e
- „s
(b) Lewis structure
(c) Approximate
bond angle(s)
36
(d) Hybridization
(e) Polar or non-
polar molecule?
(f) Geometry name
____________________________________________________________________________
3. Predict (a) the approximate bond angle, (b) the hybridization around the indicated atoms (the
atoms to which the arrows are drawn in the structures below). Write your answers near the
corresponding labels (1 to 5) in the drawings. (Note: the lone pairs on the F atoms are omitted.)
N
C
N
C
C
C
N
C
N
N
O
H
H H
H
H
3
4
5F S
F F
F
1
F Br
F F
F
F 2
37
Answers:
1. In each case, predict (a) the approximate bond angle(s), (b) the hybridization around the
underlined atom. (Note: It is helpful to first sketch the Lewis structure!)
Molecule or Ion (1) OF2 (2) H2CO (3) NO2+ (4) BF3 (5) SbF5
(a) No. of valence e -
„s
20 12 16 24 40
(b) Lewis structure
(c) Approximate
bond angle(s)
109.5o 120
o 180
o 120
o 90
o, 120
o
(d) Hybridization
sp3 sp
2 sp sp
2 sp
3d
(e) Polar or non-
polar molecule?
polar polar Ion: Not
applicable
non-polar non-polar
(f) Geometry name
bent trigonal planar linear trigonal
planar
trigonal
bypyramidal
______________________________________________________________________________
2. For each of the molecules below fill in the indicated items in the chart. The central atoms are
underlined.
Molecule (1) SO2 (2) HBF2 (3) XeF4 (4) CH2Cl2 (5) NF3
(a) No. of valence e
- „s
18 18 36 20 26
(b) Lewis structure
(c) Approximate
bond angle(s)
120o 120
o 90
o 109.5
o 109.5
o
(d) Hybridization
sp2 sp
2 sp
3d
2 sp
3 sp
3
(e) Polar or non-
polar molecule?
polar polar non-polar polar polar
(f) Geometry name
bent trigonal
planar
square
planar
tetrahedral trigonal
pyramidal
____________________________________________________________________________
3. Predict (a) the approximate bond angle, (b) the hybridization around the indicated atoms (the
atoms to which the arrows are drawn in the structures below). Write your answers near the
corresponding labels (1 to 5) in the drawings. (Note: the lone pairs on the F atoms are omitted.)
38
(1) 90
o, 120
o; sp
3d (2) 90
o; sp
3d
2 (3) 109.5
o; sp
3 (4) 120
o; sp
2 (5) 109.5
o; sp
3
N
C
N
C
C
C
N
C
N
N
O
H
H H
H
H
3
4
5F S
F F
F
1
F Br
F F
F
F 2
39
Note: The MO order for 2nd
row diatomics illustrated above is correct for Li2 through N2;
however, for O2 and F2 2p 2p MO so the order of these
MOs in the diagram above should be switched. This does not affect the bond order calculation.
MO Energy Level Diagrams
1st Row Homonuclear Diatomics
Energy
1sA 1sB
1s
*1s
Atom A Atom B
Molecule A-B
90% Contours of Orbitals
+
A B
1sA 1sB
node
*1s
anti-bonding MO
+
1sA 1sB
A B A B
1s bonding MO
2nd Row Homonuclear Diatomics
Energy
2pA2pB
*2p
*2p
2p
2p
2sA 2sB
*2s
2satom B
+
A B A B
*2p
anti-bonding MO
A
node
A B B
BA
2p
A B
*2p
bonding
anti-bonding
ABA
B
2p 2p bonding
Also: *2px, 2px formed from 2px + 2px
coordinate system
Bond Order = 1/2 (# of bonding electrons) ? 1/2 (# of anti-bonding electrons)
ZA ZB
A B
YA YB
XA XB 2p+z zA B
2p 2pz zA B
–
2pyA
2p yB
2py
2py
BA
z
+
+
–
+
+
A
molecule A-B
atom A
Electrons filled in for oxygen
+
–
node
xy plane
z
B
40
Intermolecular Forces
Types of Solids* Intermolecular Force(s) Between Particles
1. Metallic Crystals (Metals)
Examples: Na, Cu, Fe, Mn
Metallic bonding: Valence electrons form mobile sea of
electrons which comprise the metallic bond.
2. Ionic Crystals (Ionic Solids)
Examples: NaCl, MgCl2, MgO
Ionic Bonding: Attraction of charged ions for one another.
Lattice energy is a measure of ionic bond strength.
3. Covalent Crystals (Network
Solids)
Examples (small class!): C(diamond),
SiC(s), SiO2 (quartz)
Network covalent bonding. Network solids are extremely
hard compounds with very high melting and boiling points
due to their endless 3-dimensional network of covalent
bonds.
4. Molecular Crystals
Examples:
One or more of the following:
(a) Need H bonded to O, N or F: H2O,
HF, NH3.
(a) Hydrogen bonding: Hydrogen bonds are weaker than
covalent bonds, but stronger than (b) or (c) below.
(b) C6H6 (benzene), polyethylene, I2, F2,
and all the compounds from (a) above.
(b) Dispersion forces (induced dipole – induced dipole or
London dispersion forces): universal force of attraction
between instantaneous dipoles. These forces are weak for
small, low-molecular weight molecules, but large for
heavy, long, and/or highly polarizable molecules. They
usually dominate over (c) below.
(c) CHF3, CH3COCH3 (acetone) and
H2O, HF, NH3.
(c) Dipole-dipole forces: these forces act between polar
molecules. They are much weaker than hydrogen bonding.
Note: Van der Waals Forces is a category which includes both categories (b) and (c) above.
5. Atomic Crystals
Examples: He, Ne, Ar, Kr, Xe
Dispersion forces: See Section 4(b) above.
*Note: Many of the compounds given as examples are not solids at room temperature. But if
you cool them down to a low enough temperature, eventually they will become solids.
Physical properties depend on these forces. The stronger the forces between the particles,
(a) the higher the melting point.
(b) the higher the boiling point.
(c) the lower the vapor pressure (partial pressure of vapor in equilibrium with liquid or solid in a
closed container at a fixed temperature).
(d) the higher the viscosity (resistance to flow).
(e) the greater the surface tension (resistance to an increase in surface area).
41
______________________________________________________________________________
1. What type of crystal will each of the following substances form in its solid state? Choices to
consider are metallic, ionic, covalent, or molecular crystals.
(a) C2H6 __________ (b) Na2O ____________ (c) SiO2 ______________
(d) CO2 ______________ (e) N2O5 __________ (f) NaNO3 ______________
(g) Al ________________ (h) C(diamond) ______ (I) SO2 ________________
______________________________________________________________________________
2. Circle all the compounds in the following list which would be expected to form intermolecular
hydrogen bonds in the liquid state:
(a) CH3OCH3 (b) CH4 (c) HF (d) CH3CO2H (e) Br2 (f) CH3OH
(dimethyl ether) (acetic acid) (methanol)
______________________________________________________________________________
3. Specify the predominant intermolecular force involved for each substance in the space
immediately following the substance. Then in the last column, indicate which member of the
pair you would expect to have the higher boiling point.
Substance #1
Predominant
Intermolecular
Force
Substance #2
Predominant
Intermolecular
Force
Substance with
Higher Boiling
Point
(a) HCl(g) I2
(b) CH3F CH3OH
(c) H2O H2S
(d) SiO2 SO2
(e) Fe Kr
(f) CH3OH CuO
(g) NH3 CH4
(h) HCl(g) NaCl
(i) SiC Cu
Answers:
1. (a) molecular; (b) ionic; (c) covalent (network solid); (d) molecular; (e) molecular; (f) ionic;
(g) metallic; (h) covalent (network solid); (i) molecular.
2. Hint: Molecule must contain H bonded to O, N, or F, since only H bonded to O, N, or F can
form a hydrogen bond with an O, N, or F on another molecule. Thus (c), (d), and (f) should be
circled.
3. Hint: Choices for the predominant intermolecular force are metallic bonding, ionic bonding,
network covalent bonding, hydrogen bonding, and dispersion forces (induced dipole – induced
dipole forces). Dipole-dipole forces are generally dominated by dispersion forces and are rarely
predominant.
(a) dispersion forces; dispersion forces; I2.
(b) dispersion forces; hydrogen bonding; CH3OH.
42
(c) hydrogen bonding; dispersion forces; H2O.
(d) network covalent bonding; dispersion forces; SiO2.
(e) metallic bonding; dispersion forces; Fe.
(f) hydrogen bonding; ionic bonding; CuO.
(g) hydrogen bonding dispersion forces; NH3.
(h) dispersion forces; ionic bonding; NaCl.
(i) network covalent bonding; metallic bonding; SiC.
Chemical Kinetics: Introductory Concepts
Chemical kinetics is the study of the rates of chemical reactions and the mechanisms by
which reactions occur. A rate is the change of a property (in this case, concentration) per unit
time. The rate of a chemical reaction is found by following the rate of disappearance (or
decomposition) of one of the reactants or the rate of appearance (or formation) of one of the
products.
Suppose we consider the reaction
N2(g) + 3 H2(g) 2 NH3(g) (1)
Since three H2 molecules react with one N2 molecule to produce two NH3 molecules, the rate of
disappearance of H2 will be three times the rate of disappearance of N2, and the rate of
appearance of NH3 will be twice the rate of disappearance of H2. Thus,
rate
=
rate of
disappearance of N2
=
1/3 the rate of
disappearance of H2
=
1/2 times the rate of
appearance of NH3
rate
= [N2 ]
t
=
1
3
[H 2]
t
=
1
2 [NH3]
t
rate
=
vN 2
=
1
3 vH2
=
1
2 vNH 3
(2)
Note that rates are positive numbers (time doesn‟t go backwards!). That is the reason for the
negative sign in the expressions for the rate of disappearance of N2 and H2 in the Eq (2); for
example, since [N2 ] is a negative number, we need to multiply [N2 ] t by (-1) to obtain a
positive rate.
In general, for the chemical equation
a A + b B c C + d D
the rate is given by
43
rate 1
a
A
t
1
b
B
t
1
c C
t
1
d D
t
rate v 1
avA
1
bvB
1
cvC
1
dvD
(3)
Since the rates of appearance and disappearance of all reactants and products are related by
the equation stoichiometry, it doesn‟t matter which rate we actually measure – experimental
convenience governs our choice. However, since the rates differ by stoichiometric ratios, we
must specify the substance for which our rate is defined.
Rate Laws
The rate of most reactions changes with time. Initially, when concentrations of reagents are
highest, the rate is fastest. As reactants are consumed and products form, the forward reaction
slows down and the backward reaction speeds up. Eventually, either the reaction reaches
equilibrium, or it goes to completion and the rate goes to zero. At equilibrium, the forward
reaction rate equals the backwards reaction rate, and there is no further net change in
concentration.
In many cases the rate of reaction obeys a simple equation known as the rate law. A rate law
is an equation expressing the reaction rate as the product of a rate constant k and the
concentrations of species involved in the reaction raised to various powers:
rate = v = k [A]m [B]n .... (4)
A fast reaction is characterized by a large value for k. As a reaction proceeds, its rate changes
with concentration according to the rate law, but k remains constant. The rate constant k depends
on temperature, but is independent of concentration.
The powers to which the concentrations are raised (m, n, ...) are typically positive integers
(but may be fractions or negative numbers). They may also be zero, but in that case the factor is
usually omitted from the rate law, since, for example, [A]o = 1.
Reaction Order
The overall order of a reaction is the sum of the exponents in the rate law (Eq (4)). The
order of reaction in A is the power to which [A] is raised in the rate law, and so on for B, C, etc.
If the concentration of a substance does not appear in the rate law, the reaction is zeroth order in
that species. For example, if the reaction
2 A + B + C D + 2E
has the rate law
rate = v = k [A] [B]2
44
the reaction is first order in [A], second order in [B], zeroth order in [C], and third order overall.
The rate law must be determined experimentally. The reaction orders have no necessary
relation to the stoichiometric coefficients in the equation for the reaction.
Units
The units of a rate are always M s-1
= mol L-1
s-1
. Since in the rate law equation (Eq (4)), the
units on the left must equal the units on the right, the units of k depend on the overall order. For
example, if a reaction is first-order overall, k will have units of s-1
; if it is second-order overall, k
will have units of L mol-1
s-1
; and so on.
______________________________________________________________________________
1. For the reaction, 2 N2O5(g) 4 NO2(g) + O2(g), the rate of formation of NO2(g) is
4.0 x 10-3
mol L-1
s-1
.
(a) Calculate the rate of disappearance of N2O5(g)
(b) Calculate the rate of appearance of O2(g).
Answers: (a) 2.0 x 10-3 mol L-1 s-1; (b) 1.0 x 10-3 mol L-1 s-1
Hint: Eq(3) gives
rate v 12 vN2O5
14 vNO2
vO2 so here
rate v 14 vNO2
= (1/4)( 4.0 x 10
-3 mol L
-1s
-1) = 1.0 x 10
-3 mol L
-1s
-1.
______________________________________________________________________________
2. The reaction 2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g) is found experimentally to be second
order in NO(g) and first-order in H2(g).
(a) Write the rate law for the reaction.
(b) What is the overall order of the reaction?
(c) What are the units for the rate constant k?
(d) If [NO] is doubled (while keeping [H2] constant), by what factor will the reaction rate
increase?
(e) If [H2] is doubled (while keeping [NO] constant), by what factor will the reaction rate
increase?
Answers: (a) rate = v = k [NO]2 [H2] ; (b) third order overall; (c) L
2 mol
-2 s
-1; (d) 4-fold; (e) rate
will double.
45
How to Determine the Rate Law from Experimental Data
There are two general methods to determine the rate law from experimental data. The first is the
method of initial rates which is appropriate when the reaction is relatively slow so that the initial
rate v0 may be determined as a function of initial concentrations. The second method, which
makes use of integrated rate equations, is applicable when the reaction is sufficiently fast so that
concentration versus time data may be collected over several half-lives. The half-life of a
substance is the time needed for its concentration to fall to one-half its initial concentration.
Method I: Method of Initial Rates
The reaction is assumed to have a rate law in the form nm BAk ][][v rate so that the initial rate
is given by
nm BAk 000 ][][v rate initial (1)
We want to find the order of the reaction m in
[A]0 and the order n in
[B]0 . Data is gathered for
experiments in which
[B]0 (the initial concentration of B) is kept constant and the initial rate v0 is
determined for different values of
[A]0 (the initial concentration of A). If
[B]0 is held constant
and
[A]0 is increased by the factor f, the rates for the two runs will be
nm BAk 001run ,0 ][][v (2)
nm BAfk 002run ,0 ][)][ (v (3)
Dividing Eq(3) by Eq(2) gives the ratio of the initial rates:
m
nm
nm
fBAk
BAfk
00
00
1run ,0
2run ,0
][][
][)][ (
v
v (4)
By comparing the factor f by which
[A]0 was increased with the ratio of the initial rates
fm
from
Eq(4), the reaction order in [A] can be determined as shown in the table below:
Table 1. Reaction Order Using the Method of Initial Rates
Reaction Order in [A] Eq(4) Ratio Initial Rate Result
zeroth order, m = 0
fm f
01 1run ,02run ,0 vv
first order, m = 1
fm f
1 f 1run ,02run ,0 vv f
second order, m = 2
fm f
2
1run ,0
2
2run ,0 vv f
Suppose that in going from run 1 to run 2,
[A]0 is doubled so f = 2. Then, for example, if the
initial rate is constant ( 1run ,02run ,0 vv ) the reaction is zeroth order in [A] (i.e., m = 0 in the rate
law). If 1run ,02run ,0 v 2v when
[A]0 is doubled, the reaction is first order in [A] (m = 1 in the
rate law). Finally, if 1run ,01run ,0
2
2run ,0 v 4 v2v when
[A]0 is doubled, the reaction is second
order in [A] (m = 2 in the rate law).
46
An analogous process may be used to find n, the order of the reaction in [B] using runs where
[A]0 is held constant but
[B]0 is varied.
In general, if the initial concentration of a substance is increased by a factor f (while all other
initial concentrations are held constant), the initial rate v0 will increase by f n if the reaction has
an order of n for the substance.
Experimentally, to find each initial rate v0 we need to make up a solution with the desired initial
concentrations,
[A]0 and [B]o. The concentration of a reactant (or product) is then measured as a
function of the time t. A graph of concentration versus time is made with the data, and the initial
rate v0 is found from the slope of the graph (which should be linear at this initial stage of the
reaction).
______________________________________________________________________________
Problem Using the Method of Initial Rates
1. The following data were obtained for the reaction A + B + C products:
Experiment
[A]0
[B]0
[C]0
Initial rate, v0
(mol L-1
s-1
)
1
1.25 x 10-3 M
1.25 x 10-3 M
1.25 x 10-3 M
0.0087
2 2.50 x 10-3 M 1.25 x 10-3 M 1.25 x 10-3 M 0.0174
3 1.25 x 10-3 M 3.02 x 10-3 M 1.25 x 10-3 M 0.0508
4 1.25 x 10-3 M 3.02 x 10-3 M 3.75 x 10-3 M 0.457
5 3.01 x 10-3 M 1.00 x 10-3 M 1.15 x 10-3 M ?
(a) Write the rate law for the reaction. Explain your reasoning in arriving at your rate law. [Hint:
Table 1 is useful here.]
(b) What is the overall order of the reaction?
(c) Determine the value of the rate constant. [Hint: Use your rate law from (a) and appropriate
data.]
(d) Use the data to predict the reaction rate for experiment 5.
Answers to Problem 1.
(a) 2
0
2
000 ][][][v CBAkrate .
In runs 1 and 2, [B]0 and [C]0 are constant; when [A]0 doubles (f = 2), r0 increases by
fn
= 2 = 21: thus 1
st order in [A]0.
In runs 1 and 3, [A]0 and [C]0 are constant; when [B]0 increases by the factor f = 2.416, v0
increases by a factor of
fn = 5.839 = (2.416)
2: thus 2
nd order in [B]0.
In runs 3 and 4, [A]0 and [B]0 are constant; when [C]0 increases by a factor of f = 3, v0
increases by a factor of
fn = 9 = 3
2: thus 2
nd order in [C]0.
(b) Fifth order overall.
(c) k = 2.85 x 1012 L4 mol-4 s-1. (Use the rate law from (a) above and the data from any run to
calculate k.
47
(d) rate = v0 = 0.0113 mol L-1 s-1.
______________________________________________________________________________
Method II: Use of the Integrated Rate Equations Rate equations are differential equations that in simple cases can be easily integrated using
standard methods of calculus. For example, if the rate law has the form
rate d[A]
dt k [A]
n (5)
the equation may be integrated (from t = 0 to t, and from [A] =
[A]0 to [A]t) for different reaction
orders in [A] (i.e., different values of n). The equations given in the table below are obtained for
n = 0, 1, and 2. The useful feature of these equations is that they may be written in straight-line
form, y = mx + b (where m = slope and b = y-intercept are constants).
Table 2. Reaction Order Using the Integrated Rate Equations
Order in
[A]
Rate
Law
Integrated Form,
y = mx + b
Straight
Line Plot
Half-Life
t1/2
zeroth
order
(n = 0)
rate = k [A]o=
k
[A]t = - k t +[A]o
[A]t vs. t
(slope = - k)
t1/ 2 [A]0
2k
first
order
(n = 1)
rate = k [A]1
ln[A]t = - k t +
ln[A]o
ln[A]t vs. t
(slope = - k)
t1/ 2 ln2
k
0.693
k
second
order
(n = 2)
rate = k [A]2
1
[ A]t
k t 1
[A]0
1
[A]t
vs. t
(slope = k)
t1/ 2 1
k[A]0
Experimental data is collected for a reaction over time to give the concentration of A ([A]t) at a
number of times t. These data points are then tested graphically against the various rate
equations. The reaction order is determined by deciding which of the plots (see Table 2) gives
the best straight line.
Note that this method requires data from a single experiment. However the data must be
collected over several half-lives. If too little data is collected, all the plots will give a straight
line, and it will not be possible to determine the reaction order. Recall that the half-life t1/2 is the
time it takes for the concentration to fall to one-half its initial value [A]0. If the reaction is slow,
several half-lives will be a great deal of time, and the Method of Initial Rates is more convenient.
Other Forms of the First-Order Integrated Rate Equation The first-order rate equation is frequently re-arranged to give
48
ln[A]t
[A]0
k t or ln
[A]0
[A]t
k t (6)
These forms are obtained easily from the integrated form of the first-order rate equation in the
table above by using the property of logarithms,
lna
b
ln(a) ln(b) (7)
Another property of logarithms is that if
ln(y) = x (8)
then
y = ex (9)
These relations lead to additional forms of the first order integrated rate equation:
[A]t
[A]0
e
k t or [A]0
[A]t
e
k t (10)
The forms given in Eqs (6) and (10) are useful in solving problems in which the fraction of
material left [A]t/[A]o is related to the decay time t of the material. Many chemical decay
processes and all nuclear decay reactions follow first-order kinetics.
Note that for first-order reactions, the half-life t1/2 is easily determined from the rate constant k,
and vice versa (see Table 2 above).
______________________________________________________________________________
Problems Using the Integrated Rate Equations
2. A pesticide decomposes following first-order kinetics.
(a) If the half-life of the pesticide is 12 years, what is the rate constant k for the decomposition
reaction?
(b) What fraction of the pesticide will be left after 36 years?
(c) What fraction of the pesticide will be left after 100 years?
(d) How many years will it take for 99.9% of the pesticide to decompose?
Answers to Problem 2.
(a) k = 0.05775 yr-1 ≈ 0.058 yr-1; (b) 36 years = 3 half-lives. Thus 0.125 = 1/8 left; (c) 0.0031
(or 0.31%); (d) When 99.9% is gone, 0.1% is left, so the fraction left = 0.001. To reach this
point will take119.6 yr ≈ 120 yr.
______________________________________________________________________________
49
3. For a chemical reaction A+ 2 B C, a plot of 1/[A]t versus time t is found to give a straight
line with a positive slope.
(a) What is the order of the reaction? [Hint: See Table 2.]
(b) How could you determine the rate constant k for the reaction from your graph?
Answers to Problem 3:
(a) Second order; (b) Draw the best straight line through the points of the plot of 1/[A]t versus
time t . The slope of the line is equal to the rate constant k.
______________________________________________________________________________
4. The following data were obtained on the reaction 2 A B:
Time, s 0 5 10 15 20
[A], mol L-1 0.100 0.0141 0.0078 0.0053 0.004
(a) Plot the data and determine the order of the reaction.
[Hint: See Table 2. Fill in the table below and make three graphs on graph paper (or using a
computer graphing program): (1) Zeroth-order graph: plot [A]t versus t; (2) First-order graph:
plot ln[A]t versus t; Second-order graph: plot 1/[A]t versus t .]
Time, s [A], mol L-1 ln[A] 1/[A], L mol-1
0 0.1000
5 0.0141
10 0.0078
15 0.0053
20 0.0040
(b) Determine the rate constant.
[Hint: See Table 2. How does the slope m relate to the rate constant k for the reaction order you
determined in part (a)?]
Answers to Problem 4.
(a) You should find that the plot of 1/[A]t versus t gives the best straight line, so the reaction is
second order;
(b) k = 12 L mol-1 s-1.
50
Data Table and Plots for Problem 4
Time, s [A], mol L-1 ln[A] 1/[A], L mol-1
0 0.1000 -2.303 10
5 0.0141 -4.262 71
10 0.0078 -4.854 128
15 0.0053 -5.240 189
20 0.0040 -5.521 250
51
How to Work from a Mechanism to a Rate Law
Once you have experimentally determined the rate law, the next task to to find a mechanism
consistent with the rate law. Few reactions occur in a single step. Usually the mechanism is a
series of unimolecular and bimolecular steps that add to give the net equation for the reaction.
Mechanism 1 is a possible mechanism for the reaction
2 A + B C :
52
A + B
k1 k1
AB (fast to pre-equilibrium)
A + AB
k2 C (slow)
_______________________
Net equation:
2 A + B C
Here AB is an intermediate in the reaction, since it forms in an early step but is consumed in a
later step.
The rate of any step in the mechanism is found using the equation
(rate of step) vstep kstep (product of conc. of reactants in the step) (1)
Eq(1) gives the rates for all the steps of the mechanism above:
Rate of first step forward (bimolecular step)
v1 k1 [A][B]
Rate of first step backward (unimolecular step)
v-1 k1 [AB]
Rate of second step forward (bimolecular step)
v2 k2 [A][AB]
The rate law consistent with a the mechanism is found by assuming that the reaction rate equals
the rate of the slow step:
rate = v = vslow step (2)
Since in this example, the second step is slow
rate = v = vslow step = v2 k2 [A][AB] (3)
AB is an intermediate, so we are not finished yet. We need to re-express [AB] in terms of
concentrations of species that appear in the net equation. This is easy to do since the first step is
fast to pre-equilibrium. This tells us that v1 = v-1, so
k1 [A][B] k1 [AB]
Solving this equation for [AB] gives
[AB]k1 [A][B]
k1 (4)
Substituting Eq(4) into Eq(3) gives the rate law consistent with Mechanism 1:
rate = v = v2 k2 [A] k1 [A][B]
k1
or, equivalently,
53
rate = v =k2 k1 [A]2[B]
k1 k [A]
2[B] (5)
where k = k2k1/k-1. Thus Mechanism 1 predicts that the reaction is second order in A, first order
in B, and third order overall.
Mechanism 2 is another possible mechanism for the reaction:
A + B
k1 AB (slow)
A + AB
k2 C (fast)
_______________________
Net equation:
2 A + B C
In this case the first step is slow, and Eq(2) gives
rate = v = vslow step = v1 k1 [A][B] (6)
There are no intermediates in Eq(6), so we are finished. Thus Mechanism 2 predicts that the
reaction is first order in A, first order in B, and second order overall.
If either Eq (5) or Eq(6) is consistent with the experimental rate law, it is a possible mechanism
for the reaction. However, there may be other mechanisms that are also consistent with the
experimental rate law.
Enzyme Catalysis Catalysts are substances that react in an early step in a reaction mechanism, and are regenerated
in a later step (how do they differ from intermediates?). Enzymes are biological catalysts. Many
enzyme-catalyzed reactions have been found to follow the Michaelis-Menten Mechanism:
E + S
k1 k1
ES (fast to pre-equilibrium)
ES
k2 E + P (slow)
_______________________
Net equation: S
P
Here S is the substrate, E the enzyme, ES the enzyme-substrate complex, and P the product. The
problem below refers to this mechanism.
1. (a) What is the catalyst in the Michaelis-Menten Mechanism? ____________
(b) What intermediate (if any) is in the Michaelis-Menten Mechanism? _____________
(c) Derive the rate law for an enzyme reaction that follows the Michaelis-Menten
Mechanism. Express the rate law in a form involving no concentrations of intermediates.
(Hint: The method is analogous to that used for Mechanism 1 above.)
Answers: (a) E ; (b) ES ; (c) rate = v = k [E][S] where k = k2k1/k-1.
54
How to Solve Equilibrium Problems
Equilibrium problems are a common type of problem in Chemistry which involve the
equilibrium constant K. More specifically, Kc is used when the equilibrium constant is written in
terms of concentrations, and Kp is appropriate for a gas reaction if partial pressures of gases are
given.
For a reaction
a A + b B c C + d D
Kc is defined to be
Kc [C]
c[D]
d
[A]a[B]b
(at equilibrium)
where all concentrations are equilibrium concentrations. The double arrow in the equation
indicates the reaction goes in both directions (and never reaches completion). At equilibrium, the
rates of the forward and reverse reactions are equal, and there is no further change in
concentration. Kc is a constant for a given reaction at a given temperature.
Methods for solving equilibrium problems are best shown with examples.
______________________________________________________________________________
Example 1
Calculate the equilibrium constant Kc at 25 oC for the reaction
2 NOCl(g) 2 NO(g) + Cl2(g)
using the following information. In one experiment 2.00 mol of NOCl is placed in a 1.00 -L
flask, and the concentration of NO after equilibrium is achieved is 0.66 mol/L.
Method:
(a) Calculate the initial concentration of NOCl from the information given.
(b) Form an equilibrium table and fill in all known quantities. Represent unknown quantities
with variables. Use the information in the table to calculate the concentration of NOCl and Cl2 at
equilibrium.
(c) Form the Kc equation and fill in all known information. Calculate Kc.
Solution:
Step(a): The initial concentration of NOCl is
[NOCl]0 = 2.0 mol
1.00 L 2.00 M
Step(b): The equilibrium table is:
55
Balanced Equation 2 NO(g) + Cl2(g)
Initial Concentrations (M) 2.00 0 0
Change (M) - 2x 2x x
Equilibrium Concentrations (M) (2.00 - 2x) 2x = 0.66 x
From the table we have
[NO] = 2x = 0.66 M
x = 0.33 M
Thus
[NOCl] = (2.00 - 2x) M = (2.00 - 2(0.33)) M = 1.34 M
[Cl2] = x = 0.33 M
Step(c): We now have the information we need to calculate Kc:
Kc [NO]
2[Cl2 ]
[NOCl]2
at equilibrium
(0.66)
2(0.33)
(1.34)2 0.080
______________________________________________________________________________
Example 2
When solid ammonium carbamate sublimes, it dissociates completely into ammonia and carbon
dioxide according to the equation
N2H6CO2(s) 2 NH3(g) + CO2(g)
At 25 oC, experiment shows that the total pressure of the gases in equilibrium with the solid is
0.116 atm. What is the equilibrium constant Kp?
Method:
The steps are:
(a) Form an equilibrium table and fill in all known quantities. Represent unknown quantities
with variables.
(b) Use the table together with the rule that the sum of the partial pressures equals the total
pressure to find the partial pressures of NH3 and CO2 at equilibrium.
(c) Form the Kp equation and fill in all known information. Calculate Kp.
Solution:
Step (a): The equilibrium table is:
Balanced Equation N2H6CO2 2 NH3(g) + CO2(g)
Initial Partial Pressures (atm) (solid!) 0 0
Change (atm) not applicable 2x x
Equilibrium Partial Pressures (atm) not applicable 2x = p(NH3) x = p(CO2)
56
Step (b): The total pressure at equilibrium (given as 0.116 atm) equals the sum of the partial
pressures. The last line of the equilibrium table allows us to express p(NH3) and p(CO2) in terms
of x. Thus we have
pTotal = p(NH3) + p(CO2)
0.116 atm = 2x + x = 3x
x = 0.03867 atm
It follows that
p(NH3) = 2 x = 2 (0.0387 atm) = .07734 atm
p(CO2) = x = 0.03867 atm
Step (c): Now we have the information needed to calculate Kp:
Kp pNH3
2 pCO2
at equilibrium
(0.07734)2(0.03867) 2.3110
4
______________________________________________________________________________
Example 3
A sample of 0.0020 moles of F2 was sealed into a 2.0 L reaction vessel and heated to 1000 K to
study the dissociation into F atoms:
F2 2 F
At this temperature, Kc = 1.210-4
. What are [F2] and [F] at equilibrium? What is the percent
dissociation of F2?
Method:
The steps are:
(a) Calculate the initial concentration of F2 from mole and volume information given.
(b) Form an equilibrium table and fill in all known quantities.
(c) Form the Kc equation and fill in all known information. Solve for the unknown.
(d) Calculate results asked for in the problem.
Solution:
Step (a): The initial concentration of F2 is
[F2]0 = 0.0020 mol
2.0 L
0.0010 M
Step (b): The equilibrium table is given below. Note that as the reaction proceeds, reactants are
consumed and products are formed in proportion to their stoichiometric coefficients.
57
Balanced Equation F2 2 F
Initial Concentration (M) 0.0010 0
Change (M) - x 2x
Equilibrium Concentration (M) 0.0010 - x 2x
Step (c): Next we form the Kc equation and fill in equilibrium values. Note that [F] = (2x) and
that the entire quantity is squared:
Kc 1.2 104
[F]2
[F2 ]
(at eq)
2x 2
0.0010 x
4x2
0.0010 x
(0.0010 - x)(1.210-4
) = 4 x2
Next we rearrange the equation above to quadratic form, a x2 + b x + c = 0:
4 x2 + 1.210
-4 x - 1.210
-7 = 0
In our case,
a = 4, b = 1.210-4
, and c = - 1.210-7
and, using the quadratic formula, we have:
x b b
2 4ac
2a(1.210
4) (1.2 10
4)
2 4(4)(1.2 10
7)
2(4)
x 1.2 10
4 1.93410
6
8
This gives x+ = 1.5910-4
and x- = - 1.8910-4
. Clearly, only the positive root makes sense
physically; there is no such thing as a (-) concentration! Thus x = x+ = 1.5910-4
.
Step (d): Now we are ready to calculate the results asked for in the problem. For this we go
back to the last line of our equilibrium table, and use the value of x calculated above to obtain
[F2] = (0.0010 - x) = (0.0010 - 1.5910-4
) = 8.410-4
[F] = 2 x = 2 (1.5910-4
) = 3.210-4
The equation for percent dissociation is:
% dissociation = quantity dissociated
initial amount
100%
In this case,
58
% dissociation of F2 = x
[F2 ]0
100%=
1.5910-4
0.0010
100%
= 15.9 % = 16 %
______________________________________________________________________________
Approximate Methods
In many cases calculations may be simplified by making approximations. For example, if Kc
is very small (i.e., < 5.010-5
) and the initial reactant concentration c0 is fairly large (i.e., c0 >
1000Kc ), to a good approximation
(c0 x) c0
Use of this type of approximation eliminates the need to solve quadratic equations in many cases.
Making Use of Le Chatelier’s Principle
Le Chatelier’s Principle is very useful in predicting how a system at equilibrium will respond
to a change. It states that when a system at equilibrium is disturbed, the equilibrium shifts so as
to undo, in part, the effect of the disturbance.
There are three common ways an equilibrium may be disturbed:
Change in the concentration (or partial pressure) of one of the reactants or products.
Change in the temperature.
Change in the volume of the container.
Effect of Changes in Concentration (or Partial Pressure)
If a system at equilibrium is disturbed by the addition of a reactant (or the removal of a
product), Le Chatelier‟s principle predicts that the equilibrium will shift right. Shifting right will
use up some of the added reactant (or replace some of the removed product), and therefore “undo,
in part” the disturbance.
Similarly, if the disturbance is the removal of a reactant (or the addition of a product), Le
Chatelier‟s principle predicts that the equilibrium will shift left. Shifting left will replace some
of the removed reactant (or use up some of the added product), and therefore “undo, in part” the
disturbance.
The same conclusions may be reached by considering the values of Q and K. When the
equilibrium is disturbed, the reaction quotient Q changes so that it is no longer equal to K. If the
resulting Q is greater than K, the reaction will proceed backwards until once again Q equals K.
Conversely, if after the disturbance Q is less than K, the reaction will shift forwards until Q
equals K. In all cases, the shifts stop when Q = K.
Since concentrations of solids are constants and do not appear in expressions for Q or K,
removing or adding some solid does not cause shifts. However, shifts in the equilibrium do
change the amount of solid present!
59
Effect of Changes in Temperature
If a reaction is endothermic (∆H > 0), heat is absorbed in the forward reaction and released in
the backward reaction; thus in endothermic reactions, heat behaves like a reactant. Increasing
the temperature (adding heat) shifts the reaction right, since that is the direction which absorbs
heat and “undoes, in part” the disturbance. Similarly, decreasing the temperature of an
endothermic reaction shifts the reaction left.
Exothermic reactions (∆H < 0) release heat in the forward direction, and absorb heat in the
reverse direction; thus heat acts like a product in exothermic reactions. Increasing the
temperature (adding heat) shifts the reaction left since that is the direction that absorbs heat and
“undoes, in part” the disturbance. Conversely, lowering the temperature shifts it right.
The equilibrium constant K is temperature dependent and the shifts above change its value.
Reaction Type Role of heat
Endothermic (∆H > 0) reactants + heat products K K
Exothermic (∆H < 0) reactants products + heat K K
Effect of Changes in the Volume of the Container
When the volume of a reaction vessel is decreased, the partial pressures of all gases in the
container increase so the total pressure increases. Following Le Chatelier‟s principle, the
reaction shifts to reduce the total pressure since that “undoes, in part” the disturbance. This
means that the shift is in the direction which contains the fewest moles gas.
Similarly, if the volume of the reaction vessel is increased, the total pressure decreases, and
the shift is in the direction which contains the most moles gas.
In either case, if both sides of the equation have the same number of moles gas, the change in
the volume of the container has no effect on the equilibrium.
Effect of the Addition of a Catalyst
Catalysts speed up the rate at which equilibrium is obtained, but have no effect on the
magnitude of K or Q. They increase both the forward and backward rate of reaction. Since
catalysts do not appear in the net equation for a reaction, they are not involved in the expressions
for K or Q.
______________________________________________________________________________
Exercises
In the problems below, for each change given in the first column of the table, use Le Chatelier's
principle to predict
• the direction of shift of the equilibrium.
• the effect on the quantity in the third column.
______________________________________________________________________________
1. For the following reaction
5 CO(g) + I2O5(s) I2(g) + 5 CO2(g) ∆Ho = -1175 kJ
for each change listed, predict the equilibrium shift and the effect on the indicated quantity.
60
Change
Direction
of Shift
(; ; or no change)
Effect on
Quantity
Effect
(increase,
decrease,
or no change)
(a) decrease in volume Kc
(b) raise temperature amount of CO(g)
(c) addition of I2O5(s) amount of CO(g)
(d) addition of CO2(g) amount of I2O5(s)
(e) removal of I2(g) amount of CO2(g)
Answers: # 1(a) , no change; (b) , increase; (c) no change, no change; (d) , increase; (e) ,
increase.
______________________________________________________________________________
2. Consider the following equilibrium system in a closed container:
Ni(s) + 4 CO(g) Ni(CO)4(g) ∆Ho = - 161 kJ
In which direction will the equilibrium shift in response to each change, and what will be the
effect on the indicated quantity?
Change
Direction
of Shift
(; ; or no change)
Effect on
Quantity
Effect
(increase,
decrease,
or no change)
(a) add Ni(s) Ni(CO)4(g)
(b) raise temperature Kc
(c) add CO(g) amount of Ni(s)
(d) remove Ni(CO)4(g) CO(g)
(e) decrease in volume Ni(CO)4(g)
(f) lower temperature CO(g)
(g) remove CO(g) Kc
Answers: # 2(a) no change, no change; (b) , decrease; (c) , decrease; (d) , decrease; (e) ,
increase; (f) , decrease; (g) , no change.
61
Conjugate Acid-Base Pairs Arranged by Strength
The stronger the acid, the weaker the conjugate base. The stronger the base, the weaker the
conjugate acid.
Conjugate bases of diprotic acids are often atypical (see entries in italics for examples).
ACID BASE
Strength of Acid Name Formula Formula Name Strength of Base
STRONG
ACIDS
perchloric acid HClO4 ClO4– perchlorate ion Neutral Anion
sulfuric acid H2SO4 HSO4– hydrogensulfate ion
Moderately Strong
Acid!
hydroiodic acid HI I– iodide ion
Neutral
Anions
hydrobromic acid HBr Br– bromide ion
hydrochloric acid HCl Cl– chloride ion
nitric acid HNO3 NO3– nitrate ion
Strong Acid hydronium ion H3O+ H2O water Neutral
Moderately
Strong
Acid!
hydrogensulfate ion HSO4– SO4
2– sulfate ion Neutral Anion!
WEAK ACIDS
Acid strength
INCREASES as you go
UP the column.
hydrofluoric acid HF F– fluoride ion
WEAK BASES
Base strength
INCREASES as
you go DOWN the
column.
nitrous acid HNO2 NO2– nitrite ion
acetic acid HC2H3O2 C2H3O2– acetate ion
carbonic acid H2CO3 HCO3– hydrogencarbonate
ion
hydrosulfuric acid H2S HS– hydrogensulfide ion
ammonium ion NH4+ NH3 ammonia
hydrocyanic acid HCN CN– cyanide ion
Basic Anion! hydrogencarbonate
ion
HCO3– CO3
2– carbonate ion
Weak Acid methylammonium
ion
CH3NH3+ CH3NH2 methylamine
Neutral water H2O OH–
hydroxide ion Strong Base
Basic Molecule ammonia NH3 NH2
– amide ion
STRONG
BASES Neutral
Molecules
hydrogen H2 H– hydride ion
methane CH4 CH3– methide ion
Strong Base!
hydroxide ion OH– O2– oxide ion
62
Acid-Base Calculations
The Ion-Product Constant for Water, Kw
Water undergoes ionization to a small extent:
H20(l) H+(aq) + OH
–(aq)
The equilibrium constant for the reaction is the ion-product constant for water Kw:
Kw [H][OH
]1.010
14 (1)
This is a key equation in acid-base chemistry. Note that the product of [H+] and [OH
–] is a
constant at a given temperature (Eq(1) value is for 25oC). Thus as the hydrogen ion concentration
of a solution increases, the hydroxide ion concentration decreases (and vice versa).
The pH scale is widely used to report the molar concentration of hydrogen ion H+(aq) in aqueous
solution. The pH of a solution is defined as
pH log10[H] (2)
Similarly, pOH and pKw are defined as
pOH log10[OH] (3)
pKw log10(Kw)14.00 (4)
If you take the log10 of both sides of Eq(1), multiply the resulting equation by (-1), and use the
definitions of pH, pOH and pKw above, the result is the very useful equation
pH + pOH = pKw = 14.00 (5)
Equations (2) and (3) above may be solved for [H+] and [OH
–] respectively to give
[H]10
pH (6)
[OH]10
pOH (7)
(Here we use the well known rule that if
log10 y x , then
y 10x
.) In practice, the pH scale is
only used when [H+(aq)] is less than 1.0 M.
Acidic, basic, and neutral solutions can be distinguished as shown below:
63
Type of Solution pH [H+] Color of litmus
Acidic < 7.00 > 1.0 107 pink
Neutral = 7.00 = 1.0 107 in between
Basic > 7.00 < 1.0 107 blue
pH and [H+] Calculations for Strong Acids and Bases
By definition, strong acids and bases are 100% ionized in water solution. Ionization of a strong
acid gives rise to H+ ions, and ionization of a strong base produces OH
– ions. The equilibrium
constant for a strong acid or strong base is undefined, since the reaction the ionization is
complete. There is no equilibrium!
In nearly all cases of practical interest the [H+] for a strong acid (or the [OH
–] for a strong base)
is determined completely by the stoichiometry of the reaction. Once the [OH–] or pOH is known
for a base, the [H+] or the pH of the base may be calculated using Eq(1) and/or Eq(5).
Exercises
1. Complete the following table:
pH [H+] pOH [OH
–]
Acidic,
basic, or
neutral?
(a) 5.4 x 10–4
(b) 7.8 x 10-10
(c) 10.75
(d) 5.00
Answers:
(a) pH = 3.27; pOH = 10.73; [OH–] = 1.85 x 10
–11 = 1.9 x 10
–11, acidic (since pH < 7).
(b) pH = 4.89, [H+] = 1.3 x 10
–5, pOH = 9.11, acidic (since pH < 7).
(c) [H+] = 1.8 x 10
-11, pOH = 3.25, [OH
–] = 5.6 x 10
–4, basic (since pH > 7).
(d) pH = 9.00, [H+] = 1.0 x 10
–9, [OH
–] = 1.0 x 10
–5, basic (since pH > 7).
2. Calculate the pH of a 0.0430 M HNO3 solution.
64
Answer:
Since HNO3 is a strong acid, the nitric acid solution will be 100% ionized. Thus [H+] = [NO3
–] =
0.0430 M. The pH = 1.37 (use Eq(2)).
3. Calculate the pH of a 0.020 M Ba(OH)2(aq) solution.
Answer:
Since Ba(OH)2 is a strong base it is 100% ionized. Note that ionization gives 2 OH– ions for each
mole of Ba(OH)2. Thus [OH–] = 2 x 0.020 M = 0.040 M. Eq(3) gives pOH = 1.40. Then using
Eq(5), pH = 12.60.
pH and [H+] Calculations for Weak Acids and Bases
Weak acids and bases are usually less than 5% ionized. The equilibrium constant for a weak acid
equilibrium is the acid ionization constant Ka, and for a weak base equilibrium is the base
ionization constant Kb.
A typical monoprotic weak acid equilibrium can be written in two forms, the second of which
emphasizes the Brønsted acid-base nature of the reaction:
HA H+(aq) + A
–(aq)
HA + H2O H3O+(aq) + A
–(aq) (8)
In Eq(9) the Brønsted acid HA donates a proton H+ to the Brønsted base H2O to form H3O
+ and
the conjugate base A–. The acid ionization constant (using the second form) is
Ka [H3O
][A]
[HA] (9)
A typical weak base equilibrium is
B + H2O BH+(aq) + OH
–(aq) (10)
In Eq(10) the Brønsted base B accepts a proton H+ from the Brønsted base H2O to form the
conjugate acid BH+ and OH
–. The base ionization constant is
65
Kb [BH][OH]
[B] (11)
Exercises
4. Calculate (a) the pH and (b) the percent ionization of a 0.250 M HC2H3O2 solution.
Ka(HC2H3O2) = 1.8 x 10-5
. (The formula for acetic acid may also be written as CH3COOH.)
HINT: Begin by filling out the equilibrium table below.
Balanced Equation HC2H3O2 H+
+ C2H3O2–
Initial Concentration (M)
Change (M)
Equilibrium Concentration (M)
Answer:
Balanced Equation HC2H3O2 H+
+ C2H3O2–
Initial Concentration (M) 0.250 0 0
Change (M) - x x x
Equilibrium Concentration (M) 0.250 - x x x
(a)
Ka [H][C2H3O2
]
[HC2H3O2 ]1.810
5
x 2
0.250 - x
x2
0.250 . This approximation is OK if the %
ionization is < 5%; it is in this case -- see answer to (b) below. Thus x2 = 4.5 x 10
-6; x = 2.12 x 10
-
3 = [H
+]. pH = 2.67.
(b)
% ionization = x
0.250
100%
2.12103
0.250
100% 0.85%.
5. Calculate the pH of a 0.600 M solution of methylamine CH3NH2. Kb = 4.4 x 10–4
.
HINT: Methylamine is a weak base. First write the equation for the reaction following the pattern
of Eq(10). Then fill out the equilibrium table below.
Balanced Equation CH3NH2 CH3NH3+ + OH
–
Initial Concentration (M)
Change (M)
Equilibrium Concentration (M)
Answer:
Since CH3NH2 is a weak base, the balanced equation for the reaction is CH3NH2 + H2
CH3NH3+ + OH
–.
66
Balanced Equation CH3NH2
CH3NH3
+ + OH
–
Initial Concentration (M) 0.600 0 0
Change (M) - x x x
Equilibrium Concentration (M) 0.600 - x x x
Kb [BH][OH]
[B]
[CH3NH3
][OH]
[CH 3NH 2]
x2
0.600 x
x2
0.600 4.4 10
4. Thus x = 1.62 x 10
-2 =
[OH–], and pOH = 1.79. It follows from Eq(5) that pH = 12.21. NOTE: The approximation used
is OK since the % ionization is 2.7% (i.e., less than 5 %).
6. The pH of a 0.10 M solution of a weak base is 9.67. What is the Kb of the base?
Answer:
The balanced equation for a weak base B is given in Eq(10). The equilibrium table required is
given below.
Balanced Equation B BH+ + OH
–
Initial Concentration (M) 0.10 0 0
Change (M) - x x x
Equilibrium Concentration (M) 0.10 - x x x
At equilibrium, [OH–] = [BH
+] = x. Use the pH to calculate the [OH
–] at equilibrium (which is
the value of x). Here pOH = 14.00 – pH = 14.00 – 9.67 = 4.33. Thus
[OH]10
pOH10
4.33 4.6810
5 x .
82522
102.210.0
)1068.4(
10.010.0[B]
]][OH[BH
x
x
xKb . The approximation is OK
since the % ionization is well under 5%.
Relationship between Ka for a Weak Acid and Kb for its Conjugate Base
The relationship between Ka for a weak acid HA and Kb for its conjugate base A– is
Ka(HA)
Kb(A–) = Kw = 1.0 x 10
-14 (12)
If we define pKa = - log10(Ka) and pKb = - log10(Kb), the logarithmic form of Eq(12) is
pKa(HA) + pKb(A–) = pKw = 14.00 (13)
67
The stronger the acid, the larger the Ka and the smaller the pKa. Likewise the stronger the base,
the larger the Kb and the smaller the pKb. Eqs(12) and (3) show that as the Ka increases (and the
pKa decreases), the Kb decreases (and the pKb increases). These equations give quantitative
support to the statement “the stronger the acid, the weaker the conjugate base.”
The justification for Eq(12) follows from the equations below. Recall that if Eq(1) + Eq(2) =
Eq(3), then K1
K2 = K3.
Eq(1), Weak Acid: HA + H2O H3O+(aq) + A
–(aq)
Ka(HA) [H 3O
][A]
[HA]
Eq(2), Conjugate Base: A–(aq) + H2O HA(aq) + OH
–(aq)
Kb(A)
[HA][OH]
[A]
___________________________
Eq(3) = Eq(1) + Eq(2) 2 H20(l) H3O+(aq) + OH
–(aq)
Kw [H3O][OH
]
Relationship between Kb for a Weak Base and Ka for its Conjugate Acid
Analogous equations to Eqs(12) and (13) above can be written the relationship between Kb for a
weak base B and Ka for its conjugate acid HB+:
Kb(B)
Ka(BH+) = Kw = 1.0 x 10
-14 (14)
pKb(B) + pKa(BH+) = pKw = 14.00 (15)
The equations below provide justification for these results:
Eq(1), Weak Base: B + H2O BH+(aq) + OH
–(aq)
Kb(B)[BH][OH]
[B]
Eq(2), Conjugate Acid: BH+(aq) + H2O H3O
+(aq) + B(aq)
Ka(BH+)
[H3O][B]
[BH+]
___________________________
Eq(3) = Eq(1) + Eq(2) 2 H20(l) H3O+(aq) + OH
–(aq)
Kw [H3O][OH
]
___________________________________________________________________________
Exercise
7. Use the following acidity constants to help answer the questions below:
Ka(HC2H3O2) = 1.8 x 10 – 5
; Ka(HCN) = 4.9 x 10 – 10
; Ka(HCOOH) = 1.7 x 10 - 4
(a) Which of the three acids is the weakest? ________________
(b) Which of the following bases is the strongest: C2H3O2-, CN
- , or HCOO
- ? __________
68
(c) What is the pKa of HCN? _____________
(d) What is the Kb for CN- ?____________
Answers:
(a) HCN; (b) CN - ; (c) 9.31; (d) 2.04 x 10
-5
2.0 x 10-5
.
_____________________________________________________________________
Strong and Weak Acids and Bases
Definitions
An Arrhenius acid is a compound that contains hydrogen and releases hydrogen ions (H+)
in water. A Brønsted acid is a proton donor (where a proton is a hydrogen ion, H+). A
Lewis acid is an electron pair acceptor.
An Arrhenius base is a compound that produces hydroxide ions (OH–) in water. A
Brønsted base is a proton acceptor. A Lewis base is an electron pair donor.
Strong Acids
Strong acids are 100% ionized in aqueous solution to form the hydronium ion, H3O+ (also
written as H+(aq)) and an anion. For example, HCl in water ionizes completely:
HCl + H2O H3O+(aq) + Cl–(aq) [goes to completion]
(or, equivalently, HCl + water H+(aq) + Cl–(aq) [goes to completion])
There are very few strong acids, but they are extremely important in chemistry since they are
excellent sources of H+(aq), a highly reactive ion!
Strong Acid Examples: HCl (hydrochloric acid), HBr (hydrobromic acid), HI (hydroiodic acid),
HNO3 (nitric acid), H2SO4 (sulfuric acid), HClO4 (perchloric acid), and a small number of non-
metallic oxides which react with water to give a strong acid (eg., SO3(g) + H2O H2SO4).
Weak Acids
Most acids are weak. Weak acids are typically less than 5% ionized in water; thus the
predominant species is the un-ionized form. Since relatively small amounts of H+(aq) are
formed, weak acids are not very reactive. Typical weak acid ionizations in water are
HC2H3O2 + H2O H3O+(aq) + C2H3O2–(aq)
(or, equivalently, HC2H3O2 + water H+(aq) + C2H3O2–(aq))
SO2(g) + H2O H2SO3 H+(aq) + HSO3–(aq)
(or, equivalently, SO2(g) + 2 H2O H3O+(aq) + HSO3–(aq))
69
In each case above, reaction proceeds only to a very limited extent; typically over 95% of the
weak acid remains un-ionized! Since the predominant form is un-ionized, chemists do not split
up weak acids into ions when writing an ionic equation.
Weak Acid Examples:
Molecular compounds with an acidic hydrogen: HC2H3O2 = CH3COOH (acetic acid), HF
(hydrofluoric acid), HNO2 (nitrous acid), HCN (hydrocyanic acid), C6H5COOH (benzoic
acid).
Non-metallic oxides: SO2(g) (sulfur dioxide), CO2(g) (carbon dioxide), and NO(g) (nitrogen
oxide). A few non-metallic oxides such as SO3(g) (sulfur trioxide) and N2O5(g) (dinitrogen
pentoxide) give strong acids when dissolved in water. Acidic oxides of S and N are major
contributors to acid rain.
Most cations are acidic. Examples include H+, ammonium ion (NH4
+), and amine-type
cations such as CH3NH3+. Numerous metallic cations are acidic; the more acidic metal
cations are those such as Be+2
, Al+3
, and Fe3+
which have a high charge to radius ratio.
Anions of type HX- which are conjugate to strong or moderately strong acids H2X. An
example is HSO4- (conjugate base of H2SO4) which is a moderately strong acid. This is a
tiny class!!
Strong Bases
Strong bases are 100% ionized in aqueous solution to form the hydroxide ion, OH–, and a
cation. There are very few strong bases, but they are extremely important in chemistry since they
are excellent sources of OH–(aq), a highly reactive ion! Typical ionization reactions are
NaOH(s) + water Na+(aq) + OH–(aq) [goes to completion]
Na2O(s) + H2O 2 Na+(aq) + 2 OH–(aq) [goes to completion]
Strong Base Examples:
Alkali metal hydroxides and the more soluble alkaline earth hydroxides: NaOH(s) (sodium
hydroxide), KOH(s) (potassium hydroxide), Ba(OH)2(s) (barium hydroxide).
Alkali metal oxides and the more soluble alkaline earth oxides: Na2O(s) (sodium oxide),
K2O(s) (potassium oxide), BaO(s) (barium oxide).
The less soluble hydroxides and oxides of the alkaline earth cations are weak bases. Since
solubility increases for these compounds as you go down Column II, the hydroxides and
oxides of Ba2+ and Sr2+ are generally considered strong bases, while those for Ca2+ are on
the borderline between strong and weak due to their limited solubility in water.
Weak Bases
The vast majority of bases are weak. Much like weak acids, weak bases are typically less
than 5% ionized. Since their water solutions contain low concentrations of OH–(aq), they are not
very reactive. Examples of weak base ionization reactions include
NH3 + H2O NH4+(aq) + OH–(aq)
CH3NH2 + H2O CH3NH3+(aq) + OH–(aq)
70
Cu(OH)2(s) + water Cu2+(aq) + 2 OH–(aq)
CuO(s) + H2O Cu2+(aq) + 2 OH–(aq)
In each case above, reaction proceeds only to a very limited extent; typically over 95% of the
weak base remains un-ionized! Weak bases are therefore not split up into ions when writing
ionic equations.
Weak Base Examples
Metal hydroxides and metal oxides other than those listed as strong bases above. Examples
include Mg(OH)2(s) (magnesium hydroxide), MgO(s) (magnesium oxide), and Cu(OH)2(s)
(copper(II) hydroxide).
Ammonia (NH3) and amine-type bases such as CH3NH2. Typical amine-type bases have an
ammonia-type structure, but with one or more of the H atoms replaced by a hydrocarbon
group.
Most anions are basic. Common examples include HCO3–(aq) (hydrogen carbonate or
bicarbonate ion) which is present in sodium bicarbonate (NaHCO3(s)) and CO32–(aq)
(carbonate ion), a constituent of calcium carbonate (CaCO3(s)).
Neutral Compounds and Ions
So many compounds and ions are acidic or basic, you may wonder whether anything is
neutral! Examples of neutral substances include:
Water, H2O.
Hydrocarbons, alcohols, sugars, starch, and many other organic molecules.
The cations present in the strong hydroxide and oxide bases: Na+, K+, Ba2+, etc.
Most anions produced upon ionization of the strong acids: Cl–, Br–, I–, NO3–, SO42–.
Predicting the pH of Salt Solutions To determine the pH of a salt solution, you need to consider whether the salt cations or anions (or both) hydrolyze. Hydrolysis is a Brønsted acid-base reaction of an ion with water to give excess H
+ or OH
–. All acidic or basic ions hydrolyze.
CATIONS Most cations are acidic!! Hydrolysis of acidic cations gives rise to H
+ (also written as H3O
+). Examples include:
(a) Amine-type cations such as NH4
+, CH3NH3
+, etc.
NH4
+ + H2O NH3 + H3O
+
CH3NH3
+ + H2O CH3NH2 + H3O
+
(b) Metal ions with a high charge to radius ratio such as H
+, Be
2+, Al
+3, Cr
+3, and Fe
+3.
71
Here it is useful to think of a 2-step process: (1) hydration (universal for ions)
Al+3
+ 6 H2O Al(H2O)63+
(aq)
(2) hydrolysis Al(H2O)6
3+ + H2O Al(OH)(H2O)5
2+ + H3O
+
The exceptions are neutral (no cations are basic!!) Neutral ions do not hydrolyze. Examples of neutral cations are metal ions which are relatively large for their charge:
Li+, Na
+, K
+, Rb
+, Cs
+, Ba
+2, Sr
+2, Ca
+2 (cations associated with the most basic
hydroxides)
ANIONS Most anions are basic!! Hydrolysis of a basic anions gives rise to OH
–:
acetate: C2H3O2
– + H2O HC2H3O2 + OH
–
cyanide: CN
– + H2O HCN + OH
–
bicarbonate: HCO3
– + H2O H2CO3 + OH
– 2O + CO2(g) + OH
–
Most of the exceptions are neutral. Neutral ions do not hydrolyze. They are all conjugate bases of strong, or moderately strong, acids. The most common neutral anions are
Cl–, Br
–, I
–, NO3
–, ClO4
–, and SO4
2–.
A few anions are acidic. They typically contain hydrogen and are conjugate bases of strong, or moderately strong, acids. Examples of acidic anions are HSO4
– (conjugate to H2SO4), HC2O4
–
(conjugate to H2C2O4 = oxalic acid), and HC4H4O6– (conjugate to H2C4H4O6 = tartaric acid).
They hydrolyze in water:
HSO4– + H2O SO4
2– + H3O
+
Buffers
A buffer is a solution that resists changes in pH. The pH of a buffer changes very little when
small amounts of a strong acid or strong base are added to the buffer.
A buffer consists of approximately equal amounts of a conjugate weak acid/weak base pair in
equilibrium with each other. Strong acids and their conjugate bases don‟t produce a buffer since
strong acid ionization is complete: there is no equilibrium!
72
Acidic Buffers
In an acidic buffer both the weak acid and conjugate base are present initially in roughly equal
concentrations. The equilibrium is
HA H+ + A
–
weak acid conjugate base
For a weak acid,
Ka [H ][A– ]
[HA] so
pKa pH log[A]
[HA]
. Thus
pH pKa log[A]
[HA]
: this is
the Henderson-Hasselbalch equation.
These equations are useful for estimating the pH of a buffer. When [A-] = [HA], it follows that
Ka = [H+] and pH = pKa.
Acidic Buffer Example: Acetic acid (HC2H3O2), sodium acetate (NaC2H3O2)
NaC2H3O2 is a soluble salt that dissolves in water to give Na+ and C2H3O2
- ions. Na
+ ions are
neutral spectator ions, so can be ignored. The C2H3O2- ions provide the conjugate base for the
buffer. The buffer equilibrium is
HC2H3O2 H+ + C2H3O2
–
acetic acid acetate ion
For acetic acid, Ka(HC2H3O2) = 1.8 x 10-5
so pKa = -log10Ka = 4.74. Thus when [HC2H3O2] =
[C2H3O2–], pH = pKa = 4.74.
Basic Buffers
In a basic buffer both the weak base and conjugate acid are present initially in roughly equal
concentrations. The equilibrium is
B + H2O BH+ + OH
–
weak base conjugate acid
For a weak base,
Kb [BH][OH– ]
[B] so
pKb pOH log[BH]
[B]
=
pKb (14.00pH) log[BH]
[B]
. Thus
pH 14.00pKb log[BH]
[B]
. Since pKa = 14.00 – pKb,
pH pKa (BH) log
[B]
[BH]
. This is once again the Henderson-Hasselbalch equation.
73
These equations are useful for estimating the pH of a buffer. When [BH+] = [B], Kb = [OH
–].
Consequently, pOH = pKb for the buffer. It follows that pH = 14.00 – pKb = pKa(BH+) for the
buffer.
Basic Buffer Example: Ammonia (NH3), ammonium chloride (NH4Cl)
NH4Cl is a soluble salt that dissolves in water to give NH4+ and Cl
– ions. Cl
– ions are neutral
spectator ions, so can be ignored. The NH4+ ions provide the conjugate acid for the buffer. The
buffer equilibrium is
NH3 + H2O OH– + NH4
+
ammonia ammonium ion
For ammonia, Kb(NH3) = 1.8 x 10-5
so pKb = -log10Kb = 4.74. Thus when [NH3] = [NH4+], pOH
= 4.74 and pH = pKa(NH4+) = 14.00 – 4.74 = 9.26.
Neutral Buffers
Neutral buffers have a pH close to 7.00. A good example is a NaH2PO4/Na2HPO4 buffer. Since
Na+ ions are neutral spectator ions, this is a dihydrogen phosphate/hydrogen phosphate (H2PO4
–
/HPO42–
) buffer. The buffer equilibrium is
H2PO4– H
+ + HPO4
2–
dihydrogen phosphate ionhydrogen phosphate ion
Here
Ka [H ][HPO4
2–]
[H2PO4
]
= 6.2 x 10–8
. If [H2PO4-] = [HPO4
2-], Ka = [H
+] and pH = pKa = 7.21.
Response of a Buffer to the Addition of a Strong Acid or a Strong Base
Added Acid
When a small amount of a strong acid such as HCl is added to a buffer, the H+ from the acid
reacts with the basic part of the buffer to give more of the acidic part of the buffer. The reaction
is assumed to go 100%. The new concentration of the acidic part of the buffer (increased from
the initial value) and the new concentration of the basic part of the buffer (decreased from the
initial value) are then used to calculate the pH of the buffer.
Added Base
When a small amount of a strong base is added to a buffer, the OH– reacts with the acidic part of
the buffer to give more of the basic part of the buffer. The reaction is assumed to go 100%. The
new concentration of the acidic part of the buffer (decreased from the initial value) and the new
74
concentration of the basic part of the buffer (increased from the initial value) are then used to
calculate the pH of the buffer.
Exercises
1. Identify which of the following mixed systems could function as a buffer solution. For each
system that can function as a buffer, write the equilibrium equation for the conjugate acid/base
pair in the buffer system
(a) KF/HF (b) NH3/NH4Br (c)KNO3/HNO3 (d) Na2CO3/NaHCO3
Answer:
(a) Buffer: HF H+ + F
–
(b) Buffer: NH3 + H2O NH4+ + OH
– (or NH4
+ H
+ + NH3)
(c) Not a buffer since HNO3 is a strong acid. It is 100% ionized.
(d) Buffer: HCO3– H
+ + CO3
2–
2. What is the pH of a 1 L solution containing 0.240 mol HC2H3O2 and 0.180 mol NaC2H3O2?
Ka(HC2H3O2) = 1.8 x 10-5
. HINT: Begin by filling out the equilibrium table below.
Balanced Equation HC2H3O2 H+
+ C2H3O2–
Initial Concentration (M)
Change (M)
Equilibrium Concentration (M)
Answer:
Balanced Equation HC2H3O2 H+
+ C2H3O2–
Initial Concentration (M) 0.240 0 0.180
Change (M) - x x x
Equilibrium Concentration (M) 0.240 - x x 0.180 + x
Ka [H][C2H3O2
]
[HC2H3O2 ]1.810
5
x (0.180 x)
(0.240 - x)
x (0.180)
0.240. This approximation is OK if the %
ionization is < 5%; it is in this case. Thus x = 2.4 x 10-5
= [H+]. pH = 4.62.
3. What would be the pH of the buffer from #2 above if 0.010 mol of HCl were added per liter of
the buffer?
75
Answer:
The new initial concentrations are:
[C2H3O2-] = (0.180 - 0.010) M = 0.170 M (basic part decreases)
[HC2H3O2] = (0.240 + 0.010) M = 0.250 M (acidic part increases)
Thus
Ka [H][C2H3O2
]
[HC2H3O2 ]1.810
5
x (0.170 x)
(0.250 - x)
x (0.170)
0.250. Thus x = [H
+] = 2.647 x 10
-5;
pH = 4.58. Note that the pH has decreased very little!!
4. What would be the pH of the buffer from #2 above if 0.010 mol NaOH were added per liter of
the buffer?
Answer:
The new initial concentrations are:
[C2H3O2-] = (0.180 + 0.010) M = 0.190 M (basic part increases)
[HC2H3O2] = (0.240 - .010) M = 0.230 M (acidic part decreases)
Thus
Ka [H][C2H3O2
]
[HC2H3O2 ]1.810
5
x (0.190 x)
0.230 - x
x (0.190)
0.230. Thus x = [H
+] = 2.179 x 10
-5;
pH = 4.66. Note that the pH has increased very little!!
5. What would be the pH of 0.010 M HCl without the buffer?
Answer:
HCl is a strong acid and is therefore 100% ionized. Thus [H+] = 0.010M, so the pH = 2.00.
Note how much lower the pH is than it was with the buffer in #3 above (pH = 2.00 here versus
4.58 in #3 above)!
6. What would be the pH of 0.010 M NaOH without the buffer?
Answer:
NaOH is a strong base and is therefore 100% ionized. Thus [OH–] = 0.010M, so the pOH = 2.00.
Thus pH = 14.00 – pOH = 12.00.
Note how much higher the pH is than it was with the buffer in #4 above (pH = 12.00 here versus
4.66 in #4 above)!
76
Predicting Solubility
Solubility problems are equilibrium problems. The reactant in a solubility equilibrium is a
slightly soluble salt and the equilibrium constant for the reaction is the solubility product
constant, Ksp. Note that since the reactant is a solid, its concentration does not appear in the Ksp
expression. For example, the solubility equilibrium and Ksp for the salt SrF2(s) are
SrF2(s) Sr2+
(aq) + 2 F –
(aq) Ksp = [Sr2+
][F –]
2 = 2.0 x 10
-10
The molar solubility of a salt in water can be formed by setting up an equilibrium table and
solving for x. The solubility is the same quantity expressed in g/L (rather than M = mol/L).
______________________________________________________________________________
Example 1: Calculate (a) the molar solubility and (b) the solubility of SrF2(s) in water.
Solution: We first set up an equilibrium table:
Balanced Equation SrF2(s) Sr2+
(aq) + 2 F –
(aq)
Initial Concentrations (M) co 0 0
Change (M) - x x 2x
Equilibrium Concentrations (M) (co - x) x 2x
To determine the solubility we use the equilibrium concentrations from the table and the Ksp
value given above, and solve for x:
Ksp = [Sr2+
][F –]
2 = (x)( 2x)
2 = 4x
3 = 2.0 x 10
-10
x3 = 5.0 x 10
-11
x = 3.7 x 10-4
= molar solubility
To find the solubility we use the molar mass to convert the molar solubility to g/L:
? g/ L 3.7104 mol SrF2
1 L
108.62 g SrF 2
1 mol SrF2
4.010
2 SrF2 /L
______________________________________________________________________________
Predicting Precipitation
Away from equilibrium, the ion product Qsp may be defined. As always, Q has the same form as
K, but typically involves non-equilibrium concentrations. Precipitation can be predicted by
comparing Qsp with Ksp:
Relationship Solution Type Result
Qsp < Ksp Unsaturated More salt can dissolve without ppt forming
Qsp = Ksp Saturated No more salt can dissolve
Qsp > Ksp Supersaturated Salt will precipitate until Qsp = Ksp
77
Chemical Separations
If two precipitates are possible, the least soluble salt will precipitate first (the one with the
smaller Ksp). For example, if you add NaOH(aq) to a solution containing equal amounts of Ca2+
and Mg2+
, Mg(OH)2(s) will precipitate before Ca(OH)2(s) since Ksp(Mg(OH)2) = 1.2 x 10-11
is
less than Ksp(Ca(OH)2) = 8.0 x 10-6
. The reason is that for Mg(OH)2, Qsp > Ksp will occur at lower
[OH-] since the Ksp is smaller. If the Ksp values for two salts differ widely, the salts may be
separated by fractional precipitation.
Reduction in Solubility due to the Common Ion Effect
Addition of an ion “common” to a solubility equilibrium will reduce solubility. This can be
predicted in a qualitative way using Le Chatelier‟s Principle. For example, adding fluoride ion,
F–(aq), to the SrF2(s) equilibrium above will shift it left. The shift will increase the amount of
SrF2(s) in solid form, and thus decrease solubility.
The new solubility can be calculated as illustrated in the following example.
______________________________________________________________________________
Example 2. What would be the molar solubility of SrF2(s) in a 0.10 M NaF(aq) solution?
Again we set up an equilibrium table, but now we have an initial concentration of fluoride ion:
Balanced Equation SrF2(s) Sr2+
(aq) + 2 F –
(aq)
Initial Concentrations (M) co 0 0.10
Change (M) - x x 2x
Equilibrium Concentrations (M) (co - x) x (0.10 + 2x)
We next use the equilibrium concentrations in the table and the Ksp value given above, and solve
for x. Note that since x is small, (0.10 + 2x) 0.10.
Ksp = [Sr2+
][F –]
2 = (x)( 0.10 + 2x)
2 (x)( 0.10)
2 = 0.010 x = 2.0 x 10
-10
x = 2.0 x 10-8
= molar solubility
______________________________________________________________________________
Note that as predicted by the common ion effect, this solubility is much lower than what we
calculated in Example 1 for pure water!
Increase in Solubility due to addition of a Species that Reacts with an Ion
The solubility of a salt will increase if a species is added which reacts with one of its ions. Once
again this is an example of shifts predicted by Le Chatelier‟s Principle. For example if the anion
reacts with the added substance, the concentration of the anion will be reduced. Thus the
equilibrium will shift right to “undo, in part, the disturbance”. The shift right will reduce the
amount of salt in solid form, and thereby increase its solubility. There are two common
examples of this phenomena:
(1) Reaction of the basic anion of a salt with a strong acid. Salts such as Mg(OH)2(s), BaCO3(s),
and NiS(s) are much more soluble in a strong acid solution than in water. On the other hand,
the solubility of AgCl(s) does not increase when 6 M HNO3 is added.
78
(2) Reaction of the acidic cation (Lewis acid) of a salt with a Lewis base to form a soluble
complex ion. Salts such as AgCl(s) and Cu(OH)2(s) are much more soluble in 6 M NH3(aq)
solution than in pure water due to the formation of Ag(NH3)2+(aq) and Cu(NH3)4
2+(aq)
complex ions respectively. Likewise, amphoteric hydroxides such as Al(OH)3(s) and
Zn(OH)2(s) are soluble in 6 M NaOH(aq) due to the formation of Al(OH)4-(aq) and
Zn(OH)42-
(aq) ions.
Thermodynamics and Equilibrium: Important Equations
Second Law of Thermodynamics
Entropy S is a measure of the disorder of a system: the greater the disorder the greater the
entropy. The second law of thermodynamics tells us the direction of spontaneous (natural)
change in the universe. The law states that in any spontaneous change the entropy of the
universe increases, and in an equilibrium process it remains unchanged: Suniverse 0 (where the
equality applies at equilibrium). A process for which Suniverse 0 is impossible (much like
going backwards in time)!
Gibbs Free Energy
A much more convenient criteria for spontaneity which can be derived from the second law
involves changes in Gibbs free energy G, where GH – TS. A change is spontaneous (product-
favored) in the direction written at a constant temperature T and constant pressure p if ∆G for the
change is negative. Mathematically, for a spontaneous change we must have
G p,T 0 (1)
At equilibrium, G p,T 0 . If G p,T 0 , the change cannot occur in the direction written
(but will be spontaneous in the opposite direction).
There are a number of ways to calculate ∆G . Since GH – TS, when the temperature T is
constant
G H TS (2)
Go H
o TS
o (3)
In these equations (and in all which follow), the temperature is in Kelvin. At standard conditions
(superscript “o” means standard conditions):
Go nprod
products
G f
oprod nreact
reactants
G f
oreact (4)
Ho nprod
products
H f
oprod nreact
reactants
H f
oreact (5)
So nprod
products
So
prod nreact
reactants
So
react (6)
79
Another method of calculating ∆G follows from the connection between the criteria for
spontaneous change, G p,T 0 , and our earlier Q < K criteria. Recall that when Q < K,
reaction is predicted to proceed forward until Q = K. It can be shown that
G RT ln K RT ln Q (7)
Note that this equation predicts (as required) that when Q < K, ∆G < 0. Also, when Q = K, ∆G =
0.
By definition at standard conditions all reactants and products are in their standard states:
they are pure if solid or liquid, and at 1 M concentration if dissolved. It follows that at standard
conditions, Q = 1 and ln Q = 0. Thus, Eq(7) reduces to
Go RT lnK (8)
where, as always, the “o” superscript means standard conditions. Rearranging Eq(8) gives us a
convenient equation for calculating K:
ln K G
o
RT or K e
Go / RT (9)
Finally, Eq(7) and Eq(8) may be combined to give
G Go RT lnQ (10)
The table below summarizes our equations for G under various conditions. Note how the
more specialized equations are easily derived from those for the general case.
General Case
G RT ln K RT ln Q
G Go RT ln Q
G H TS
Equilibrium Conditions
Q K
G 0
H TS
Standard Conditions
Q 1
G Go; H H
o; S S
o
Go RT ln K
Go H
o TS
o
Third Law of Thermodynamics
You may have noticed the difference in notation in the So equation, Eq(6), as compared to
that in Eq(4) and Eq(5). The reason for the difference is that entropy values are absolute. They
are based on the third law of thermodynamics which says that the entropy of a perfect crystalline
substance is zero at the absolute zero of temperature (0 K). We therefore assign the value of zero
80
entropy to perfect crystals at absolute zero. All non-perfect crystals and all substances (including
pure elements) at temperatures above 0 K have positive entropy. Entropy increases with
temperature, since disorder increases when the temperature is raised.
In contrast the zero of enthalpy and zero of free energy are undefined. We follow the
convention of setting the heat of formation H f
o and free energy of formation Gf
o of a pure
element in its standard state equal to zero.
How to Balance Equations for Oxidation-Reduction Reactions
Oxidation-reduction (redox) reactions are reactions in which oxidation numbers change.
Oxidation numbers are either real charges or formal charges which help chemists keep track of
electron transfer. In practice, oxidation numbers are best viewed as a bookkeeping device.
Oxidation cannot occur without reduction. In a redox reaction the substance which is
oxidized contains atoms which increase in oxidation number. Oxidation is associated with
electron loss (helpful mnemonic: LEO = Loss of Electrons, Oxidation). Conversely, the
substance which is reduced contains atoms which decrease in oxidation number during the
reaction. Reduction is associated with electron gain (helpful mnemonic: GER = Gain of
Electrons, Reduction).
Chemists often talk about oxidizing and reducing agents. Be careful with these terms!
An oxidizing agent is a substance which oxidizes something else: it itself is reduced! Also, a
reducing agent is a substance that reduces another reactant: it itself is oxidized. A
disproportionation reaction is a reaction in which the same element is both oxidized and reduced.
How to Assign Oxidation Numbers: The Fundamental Rules
Rules for assigning oxidation numbers are as follows:
• The oxidation number of any pure element is zero. Thus the oxidation number of H in H2 is zero.
• The oxidation number of a monatomic ion is equal to its charge. Thus the oxidation number of
Cl in the Cl- ion is -1, that for Mg in the Mg+2 ion is +2, and that for oxygen in O2- ion is -2.
• The sum of the oxidation numbers in a compound is zero if neutral, or equal to the charge if an
ion.
• The oxidation number of alkali metals in compounds is +1, and that of alkaline earths in
compounds is +2. The oxidation number of F is -1 in all its compounds.
• The oxidation number of H is +1 in most compounds. Exceptions are H2 (where H = 0) and the
ionic hydrides, such as NaH (where H = -1).
• The oxidation number of oxygen (O) is -2 in most compounds. Exceptions are O2 (where O = 0)
and peroxides, such as H2O2 or Na2O2, where O = -1.
• For other elements, you can usually use rule (3) to solve for the unknown oxidation number.
Examples:
NO(g) has O = -2, so N = +2.
NO2(g) has O = -2, so N = +4.
SO42-
has O = -2. Thus S + 4(-2) = -2. Solving the equation gives S = -2 + 8 = +6.
K2Cr2O7 has K = +1 and O = -2. Thus 2(+1) + 2 Cr + 7(-2) = 0; 2 Cr = 12; Cr = +6.
81
How to Balance Redox Reactions Using the Method of Half-Reactions
Oxidation-reduction reactions are often tricky to balance without using a systematic
method. We shall use the method of half-reactions which is outlined in detail below.
Method in Acidic (or Neutral) Solution
Suppose you are asked to balance the equation below:
NO2– + MnO4
– NO3– + Mn+2 (in acid solution)
Begin by writing the unbalanced oxidation and reduction half-reactions (you do not need to
know which is which):
NO2– 3
–
MnO4– Mn+2
Next, balance for atoms. First do this for atoms other than O and H. (Both equations above are
already balanced for N and Mn, so no change is needed in this example.) Then balance for O
atoms by adding H2O to the reaction side deficient in O:
H2O + NO2– 3
–
MnO4– Mn+2 + 4 H2O
This leaves H atoms unbalanced. In acidic (or neutral) solution, balance for H atoms by adding
H+ to the side deficient in H:
H2O + NO2– 3
– + 2 H+
8 H+ + MnO4– Mn+2 + 4 H2O
The next step is to balance for charge. To do this, add electrons (e-) to the more positive side:
H2O + NO2- 3
- + 2 H+ + 2 e-
5 e- + 8 H+ + MnO4- Mn+2 + 4 H2O
Now you need to multiply the equations by appropriate factors so that the number of electrons
lost in the oxidation half-reaction (LEO) is equal to the number of electrons gained in the
reduction half-reaction (GER):
5 x [ H2O + NO2- 3
- + 2 H+ + 2 e- ]
2 x [ 5 e- + 8 H+ + MnO4- Mn+2 + 4 H2O ]
Then, sum the above equations to obtain
5H2O + 5NO2- + 10 e- + 16H+ + 2MnO4
- 3- + 10H+ + 10 e- + 2Mn+2 + 8H2O
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Finally, simplify by subtracting out species that are identical on both sides. Our final balanced
redox equation is
5 NO2- + 6 H+ + 2 MnO4
- 3- + 2 Mn+2 + 3 H2O
Check this equation to confirm that it is balanced for atoms and balanced for charge.
Method in Basic Solution
Suppose you are asked to balance the equation below:
I- + MnO4- 2
+ MnO2 (in basic solution)
Begin by writing the unbalanced oxidation and reduction half-reactions (you do not need to
know which is which):
I- 2
MnO4- MnO2
Next, balance for atoms. First do this for atoms other than O and H:
2 I- 2
MnO4- MnO2
Then balance for O atoms by adding H2O to the reaction side deficient in O:
2 I- 2
MnO4- MnO2 + 2 H2O
This leaves H atoms unbalanced. In basic solution (just as in acidic or neutral solution) first
balance for H atoms by adding H+ to the side deficient in H:
4 H+ + MnO4
- MnO2 + 2 H2O
In basic solution, follow this step by neutralizing the H+; do this by adding an equivalent amount
of OH- to both sides of the equation.
4 OH- + 4 H
+ + MnO4
- MnO2 + 2 H2O + 4 OH-
Then form water on the side which has both H+ and OH
- (recall that H
+ + OH
- H2O): in this
case we form 4 H2O on the left:
4 H2O + MnO4- MnO2 + 2 H2O + 4 OH-
Next simplify the water by subtracting 2 H20 from both sides. The half-reactions are now:
2 I– 2
83
2 H2O + MnO4– MnO2 + 4 OH
–
At this point the equations should be balanced for atoms. The next step is to balance for
charge. To do this, add electrons (e–) to the more positive side:
2 I– 2
+ 2 e–
3 e– + 2 H2O + MnO4
– MnO2 + 4 OH–
Now you need to multiply the equations by appropriate factors so that the number of electrons
lost in the oxidation half-reaction (LEO) is equal to the number of electrons gained in the
reduction half-reaction (GER):
3 x [ 2 I- I2 + 2 e- ]
2 x [ 3 e- + 2 H2O + MnO4- MnO2 + 4 OH- ]
Sum the equations to obtain
6 I- + 6 e- + 4 H2O + 2 MnO4- I2
+ 6 e- MnO2 + 8 OH-
Finally, simplify by subtracting out species that are identical on both sides:
6 I- + 4 H2O + 2 MnO4- I2
+ 2 MnO2 + 8 OH-
Check our final equation above to confirm that it is balanced for atoms and balanced for charge.
Exercises:
Balance the following redox reactions. In each case
• (a) give the balanced half-reactions; identify the oxidation half-reaction and the reduction
half-reaction.
• (b) give the balanced net reaction.
• (c) identify the oxidizing agent and the reducing agent.
______________________________________________________________________________
1. Cl2(g) + S2O32-(aq) Cl-(aq) + SO42-(aq) in acid solution.
Answers:
(a) S2O32-(aq) + 5 H20 2 SO42-(aq) + 10 H+(aq) + 8 e- (oxidation half-reaction – LEO);
Cl2(g) + 2 e- 2 Cl-(aq) (reduction half-reaction – GER).
(b) S2O32-(aq) + 5 H20 + 4 Cl2(g) 2 SO42-(aq) + 10 H+(aq) + 8 Cl-(aq)
(c) S2O32-(aq) is the reducing agent; Cl2(g) is the oxidizing agent.
______________________________________________________________________________
84
2. O3(g) + Br-(aq) O2(g) + BrO-(aq) in basic solution.
Answers:
(a) Br - (aq) + H20 + 2 OH-(aq) BrO-(aq) + 2H2O + 2 e- or, after simplifying,
Br - (aq) + 2 OH-(aq) BrO-(aq) + H2O + 2 e- (oxidation half-reaction – LEO);
O3(g) + 2 H2O + 2 e- O2(g) + H2O + 2 OH-(aq) or, after simplifying,
O3(g) + H2O + 2 e- O2(g) + 2 OH-(aq) (reduction half-reaction – GER).
(b) Br - (aq) + O3(g) BrO-(aq) + O2(g)
(c) Br - (aq) is the reducing agent; O3(g) is the oxidizing agent.
______________________________________________________________________________
3. Balance the reaction, Br2(l) Br-(aq) + BrO3-(aq) in basic solution. Hint: this is a
disproportionation reaction!
Answer: 6 Br2(l) + 12 OH-(aq) 10 Br-(aq) + 2 BrO3-(aq) + 6 H2O
Factors Affecting Corrosion
Much of the current concern in this country and throughout the world about the need to
replace the "infrastructure" is related to the rusting of iron. It has been estimated that 25% of the
annual steel production in the United States goes towards the repair and replacement of metals
lost by corrosion. Bridges are in need of being replaced, steel rods in the cement of older
buildings and parking ramps are deteriorating, and the rusting of many machine parts causes
concern for equipment reliability.
What is Corrosion?
Corrosion is a general term which refers to the undesired oxidation of metals to oxides
and other compounds. When the metal is iron, the corrosion process is called rusting. While
iron is structurally strong, the iron oxides formed as the products of rusting are not. Other metals
such as copper and silver become tarnished from surface oxidation of the metal. Billions of
dollars must be spent annually to replace materials that have corroded. The greatest damage is
from the rusting of iron and steel. Thus prevention of the corrosion of iron is a topic of immense
practical importance.
The surface of metals can provide a voltaic (galvanic) cell if there is a difference in
potential between two or more points on the metal. A galvanic cell can be established in the
presence of water vapor and impurities in the metal or stressed areas created in the process of
85
machining something as simple as a nail.
The metal which is oxidized (corroded) may go into solution as ions. For example, in the
case of iron, Fe2+
ions can result: Fe(s) Fe2+
(aq) + 2e -. Air can then further oxidize the Fe
2+
ions to rust or Fe2O3.n H2O. Corrosion is often localized and limited to small areas on the
surface of the metal.
Half-Reactions in the Corrosion of Iron
Since corrosion reactions are oxidation-reductions reactions, they may be resolved into two
half-reactions. The oxidation half-reaction represents the oxidation of the metal. In the case of
iron it is
Fe(s) Fe2+
(aq) + 2e - Eoox = 0.44 v (1)
The reduction half-reaction involved in corrosion in an aqueous environment depends on the pH
and on the availability of oxygen. Possibilities, with their standard half-cell reduction potentials,
are
2 H2O + 2 e- H2(g) + 2 OH-(aq) Eored = - 0.83 v (- 0.42 v at pH=7) (2)
2 H+ + 2 e- H2(g) Eored = 0.00 v (- 0.42 v at pH=7) (3)
O2(g) + 2 H2O + 4 e- 4 OH-(aq) Eored = 0.40 v (0.81 v at pH=7) (4)
O2(g) + 4 H+ + 4 e- 2 H2O Eored = 1.23 v (0.81 v at pH=7) (5)
Note that the reduction potential of each of these half-reactions gets more positive as the [H+]
increases and more negative as it decreases. (Why? Use Le Chatelier's Principle!)
Which one of these reduction half-reactions actually occurs may be deduced experimentally.
If Eq(2) or Eq(3) is the reduction half-reaction, bubbles of H2 gas will be visible in the reaction
mixture. Also note that Eq(2) and Eq(4) both have OH- as a product which will turn
phenolphthalein indicator pink.
Water in contact with air contains sufficient dissolved oxygen to provide the reactants for
Eq(4). Thus the net reaction for rusting under normal conditions is obtained by combining Eqs(1)
and (4) to give
2 Fe(s) + O2(g) + 2 H2O 2 Fe(OH)2(s) (6)
The iron(II) hydroxide formed is then further oxidized to iron(III) hydroxide:
4 Fe(OH)2(s) + O2(g) + 2 H2O 4 Fe(OH)3(s) (7)
Iron(III)hydroxide is an idealized formula, often written as Fe2O3.3H2O. Variable amounts of
water may be removed from it to form the material commonly called rust, Fe2O3(H2O)x.
Methods of Preventing Corrosion
86
So what can we do to prevent corrosion? Surely we have found ways to prevent it, or at least to
slow it down.
(3) We can plate a metal with an inert material, like chromium, provided there are no cracks in
the plating to allow corrosion to begin. The tin coating of iron on “tin” cans acts in a similar
way. However, once the surface of a tin can is scratched, rusting occurs rapidly with iron
acting as the anode and the tin as the cathode.
(4) We can coat with another metal, like zinc, which oxidizes more easily than iron-based metals
and so is "sacrificed" to prevent corrosion of the metal coated. Zn(s) acts as a “sacrificial
anode” and is oxidized preferentially to iron; thus no rusting occurs even if the Zn(s) coating
is scratched. “Galvanized” iron is iron coated with Zn(s).
(5) Some metals form their own protective coating which prevents or slows down corrosion.
Aluminum is very easily oxidized and when exposed to air forms a very stable coating of
Al2O3(s) which protects the Al(s) underneath from corrosion. Copper becomes coated with
CuCO3(s) which gives it a green “patina” and protects it somewhat from further corrosion.
(6) There are also additional protective coatings, like the paint used on titanium alloys in the
racing bicycles, which prevent rapid corrosion and slow ultimate deterioration.
(7) Annealing (heating of material followed by slow cooling) also slows down the rate of
corrosion.
What Factors Increase the Rate of Corrosion?
On the other hand, salt increases the rate of corrosion. Think of the bridges and cars
attacked by ocean spray, or by putting salt on highways in the areas of the country where snow
and ice need to be melted in the winter. As is well known, acid rain also greatly increases the
rate of corrosion. In addition, places where a metal has been worked or stressed corrode most
rapidly; examples include the head or tip of a nail, or where a nail is bent.
Use of Indicators to Determine Products of Corrosion
Two indicators may be used to give evidence of corrosion, or protection from corrosion.
The first is phenolphthalein, an acid/base indicator which is colorless in pHs below 8.8 and pink
in pHs above that reading. The second is hexacyanoferrate(III) ion in the form of
K3Fe(CN)6(aq), an indicator which turns a deep blue in the presence of Fe2+
due to the reaction
Fe2+(aq) + K+(aq) + Fe(CN)63- KFe[Fe(CN)6](s) (8)
(from iron oxidation) (yellow solution) (deep blue precipitate)
These indicators enable us to detect products of Eqs(1)-(5) above. You will find them useful in lab!
Electrolysis
Many nonspontaneous reactions may be forced to occur electrolytically. All that is
required are two electrodes, a battery, and a conducting solution. Graphite pencil leads work
well as electrodes since they are excellent conductors and are non-reactive. To produce an
87
electrolytic cell, graphite electrodes are hooked up to the terminals of a nine-volt battery and then
dipped in an electrolyte solution in a Petri dish.
A battery is a voltaic cell and thus has a positive cell potential (Ecell). The spontaneous
reaction occurring in the battery pumps electrons out of the negative battery terminal and pulls
electrons into the positive battery terminal. Thus reduction in the solution in the Petri dish
occurs at the electrode attached to the (-) terminal, and oxidation occurs at the electrode attached
to the (+) terminal.
The cell potential of the solution being electrolyzed is always negative since the reaction
is nonspontaneous. If the battery is to be successful in forcing this nonspontaneous reaction to
occur, the battery voltage must be large enough to overcome the negative electrolytic cell
potential so that the sum of the battery potential and electrolytic cell potential is greater than zero.
In practice, some "overvoltage" is also required. A nine volt battery gives sufficient voltage to
electrolyze most solutions of interest.
Predicting Electrolysis Products
To predict the products of electrolysis, begin by writing down all the possible reactants
that are present in significant amounts in the solution being electrolyzed (we assume here that the
electrodes themselves are inert). A possible reactant that it is easy to forget about is water!
Water can be either oxidized or reduced (or both!):
Reduction: 2 H2O + 2 e- H2(g) + 2 OH-(aq) Eored = - 0.83 v (1)
( - 0.42 v at pH = 7)
Oxidation: 2 H2O O2(g) + 4 H+(aq) + 4 e- Eoox = - 1.23 v (2)
( - 0.81 v at pH = 7)
Note that both of these reactions have a gaseous product, but the reduction of water produces
OH -
(aq) while the oxidation of water produces H+(aq).
Other possible reactants are the dissolved anions and cations in the solution. Metal ions
are good candidates for reduction while halide ions are good candidates for oxidation:
2 Cl-(aq) Cl2(g) + 2 e- Eoox = - 1.36 v (3)
2 Br-(aq) Br2(l) + 2 e- Eoox = - 1.07 v (4)
2 I-(aq) I2(s) + 2 e- Eoox = - 0.53 v (5)
All three halogens are easily detected. Chlorine, Cl2(g), is recognized by the observation of gas
bubbles and its distinctive odor. Bromine dissolved in water gives a yellowish color, while in
cyclohexane it turns orange. Iodine dissolved in water gives a reddish-brown color, while it turns
lavender in cyclohexane. If sufficient iodine is produced, crystals of I2(s) will form on the anode:
these look metallic.
If more than one reactant is available for reduction, the one with the highest Eored potential
88
is most likely to be reduced. Similarly, if more than one reactant is available for oxidation, the
one with the most positive (least negative) Eoox potential is most likely to be oxidized. However,
if two possible reduction reactions have similar Eored values, kinetic effects may become the
deciding factor; either reduction reaction, or both, may occur. The same applies if two possible
oxidation reactions have Eoox values which are close. Careful observation and testing of products
is essential to determine which oxidation and reduction reactions have occurred.
______________________________________________________________________________
Exercises
1. A student has run an electrolysis reaction of an aqueous solution of MgBr2(aq) in a petri dish
using inert electrodes and a 9 volt battery. Observations are made and the following data is
recorded:
At the electrode attached to the (-) terminal of the battery, bubbles are observed. A white
precipitate forms around the electrode.
At the electrode attached to the (+) terminal of the battery, a small amount of bubbles are also
observed. The solution near the electrode has turned yellowish. When a few drops of the
yellowish liquid are mixed with cyclohexane, an orange layer forms at the top of the mixture.
When a few drops of phenolphthalein are added to the petri dish, a magenta pink color is
observed near the electrode attached to the (-) terminal of the battery.
Account for all the observations above. Write equations for the redox reactions which occurred
and account for the colors, precipitate, and bubbles.
Answer:
At the cathode (the (-) terminal) the possible reductions are:
2 H2O + 2 e- H2(g) + 2 OH-(aq) Eored = - 0.83 v ( - 0.42 v at pH = 7)
Mg+2
+ 2 e - Mg(s) Eored = - 2.37 v
At the anode (the (+) terminal) the possible oxidations are:
2 Br – (aq) Br2(l) + 2 e- E
oox = - 1.07 v
2 H2O O2(g) + 4 H+(aq) + 4 e- Eoox = - 1.23 v ( - 0.81 v at pH = 7)
The reduction reaction which occurs is usually the one with the more (+) Eored, and the oxidation
reaction is usually the one with the more (+)Eoox. Thus here we expect that H2O is reduced at the
cathode and Br -
(aq) is oxidized at the anode. The data indicates, however, that some H2O is
also oxidized at the anode. This is not surprising since the Eoox values are close.
The bubbles at the (-) terminal are due to H2(g) and the white precipitate is Mg(OH)2(s). The
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magenta pink color at the (-) electrode is due to phenolphthalein which turns that color in the
presence of OH –
(aq), one of the products of the reduction of H2O.
The small amount of bubbles at the (+) electrode are O2(g). The other product at the (+)
electrode is Br2 which is yellowish in water but turns orange when dissolved in an organic
solvent such as cyclohexane. The cyclohexane layer is on top since cyclohexane is less dense
than water.
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2. Suppose that molten MgBr2(l) is electrolyzed. What reactions would take place, and what is
the minimum emf required for reaction?
Answer: No water is present in a molten salt, so only the salt ions are available to react. Thus,
Mg+2
would be reduced: Mg+2
+ 2 e - Mg(s) Eored = - 2.37 v
(e) Br – would be oxidized: 2 Br
– (aq) Br2(l) + 2 e
_ E
oox = - 1.07 v
The minimum emf required is equal to - Eocell = - (E
ored + E
oox) = - (- 2.37 v + - 1.07 v) = + 3.44
v
Sources: http://spiepho.sbc.edu