chemistry 120 stoichiometry: chemical calculations chemistry is concerned with the properties and...
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Chemistry 120
Stoichiometry: Chemical Calculations
Chemistry is concerned with the properties and the
interchange of matter by reaction – structure and
change.
In order to do this, we need to be able to talk about
numbers of atoms
Chemistry 120
Stoichiometry: Chemical Calculations
The mole and atomic mass
The mole is defined as
the number of elementary entities as are present in 12.00 g of 12C.
Numerically, this is equal to Avogadro’s Number
6.022 x 1023
Therefore, in 12.00 g of 12C there are 6.022 x 1023 ‘elementary entities’, in this case atoms.
Chemistry 120
Stoichiometry: Chemical Calculations
The mole and atomic mass
Atomic masses, in atomic units, u, are defined relative to 12C.
Therefore,
The formula mass of an element or compound contains 1 mole, 6.022 x 1023, of particles
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
How many particles are there in 5 g of Na?
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
How many particles are there in 5 g of Na?
The particles are atoms – how many atoms are there in 5 g of Na?
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
How many particles are there in 5 g of Na?
The particles are atoms – how many atoms are there in 5 g of Na?
Atomic mass of Na = 22.9898 u
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
How many particles are there in 5 g of Na?
The particles are atoms – how many atoms are there in 5 g of Na
Atomic mass of Na = 22.9898 uAs
1 u = 1/12 x mass (12C) And
1 mole = 6.022 x 1023 particles = number of particles in 12 g 12C
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
How many particles are there in 5 g of Na?
The particles are atoms – how many atoms are there in 5 g of Na
Atomic mass of Na = 22.9898 u
Mass of 1 mole of Na = 22.9898 g
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
How many particles are there in 5.0000 g of Na?
22.9898 g Na = 1 mole Na
Then 1 g Na = 1 mol Na
22.9898
5 x 1 g Na = 5 x 1 mol Na
22.9898
5 g Na = 0.2175 mol Na
5 g Na = 0.2175 x (6.022 x 1023) particles Na
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
How many particles are there in 5.0000 g of Na?
1.310 x 1023 atoms
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
What is the mass of 0.23 mol of butane?
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
What is the mass of 0.23 mol of butane?
Molecular formula of butane: C4H10
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
What is the mass of 0.23 mol of butane?
Molecular formula of butane: C4H10
Atomic mass of C = 12.011g
Atomic mass of H = 1.0079g
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
What is the mass of 0.23 mol of butane?
Molecular formula of butane: C4H10
Atomic mass of C = 12.011g
Atomic mass of H = 1.0079g
Molecular mass of C4H10 = (4x12.011)+(10x1.0079)u
= 58.123 u
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
What is the mass of 0.23 mol of butane?
Molecular formula of butane: C4H10
Atomic mass of C = 12.011g
Atomic mass of H = 1.0079g
Molecular mass of C4H10 = (4x12.011)+(10x1.0079)u
= 58.123 u
Relative Molecular Mass of Butane = 58.123 g
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
What is the mass of 0.23 mol of butane?
Relative Molecular Mass of Butane = 58.123 g
1 mole of butane = 58.123 g
0.23 x 1 mole of butane = 0.23 x 58.123 g
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
What is the mass of 0.23 mol of butane?
Relative Molecular Mass of Butane = 58.123 g
1 mole of butane = 58.123 g
0.23 x 1 mole of butane = 0.23 x 58.123 g
0.23 mole of butane = 13.368 g
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
These are formulæ which show the chemical change taking place in a reaction.
Sr(s) + Cl2(g) SrCl2(s)
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
These are formulæ which show the chemical change taking place in a reaction.
Physical state
Sr(s) + Cl2(g) SrCl2(s)
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
These are formulæ which show the chemical change taking place in a reaction.
Physical state
Sr(s) + Cl2(g) SrCl2(s)
Reactants Product
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
As matter cannot be created or destroyed in a chemical reaction, the total number of atoms on one side must be equal to the total number of atoms on the other.
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
Example
Cyclohexane burns in oxygen to give carbon dioxide and water
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
Example
Cyclohexane burns in oxygen to give carbon dioxide and water
Reactants: Cyclohexane, C6H12
Oxygen, O2
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
Example
Cyclohexane burns in oxygen to give carbon dioxide and water
Reactants: Cyclohexane, C6H12
Oxygen, O2
Products: Carbon Dioxide, CO2
Water, H2O
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
Example
Initially, we can write the reaction as
C6H12 + O2 CO2 + H2O
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
Example
Initially, we can write the reaction as
C6H12 + O2 CO2 + H2O
This is NOT a correct equation – there are unequal numbers of atoms on both sides
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
Example
Initially, we can write the reaction as
C6H12 + O2 CO2 + H2O
This is NOT a correct equation – there are unequal numbers of atoms on both sides
Reactants: 6 C, 12 H, 2 O
Products: 1 C, 2 H, 3 O
Chemistry 120
Stoichiometry: Chemical Calculations
Balancing the equation
C6H12 + O2 CO2 + H2O
Chemistry 120
Stoichiometry: Chemical CalculationsBalancing the equation
6 C, 12 H, 2 O 1 C, 2 H, 3 O6 C on LHS means there must be 6 C on the RHS
C6H12 + O2 CO2 + H2O
C6H12 + O2 6CO2 + H2O
6 C, 12 H, 2 O 6 C, 2 H, 13 O
13 O on RHS means there must be 13 O on LHSC6H12 + 13/2 O2 6CO2 + H2O
6 C, 12 H, 13 O 6 C, 2 H, 13 O
Chemistry 120
Stoichiometry: Chemical CalculationsBalancing the equation
C6H12 + 13/2 O2 6CO2 + H2O
6 C, 12 H, 13 O 6 C, 2 H, 13 O
12 H on RHS means there must be 12 H on LHS
C6H12 + 13/2 O2 6CO2 + 6H2O
6 C, 12 H, 13 O 6 C, 12 H, 18 O
18 O on RHS means there must be 18 H on LHS
C6H12 +9O2 6CO2 + 6H2O
6 C, 12 H, 18 O 6 C, 12 H, 18 O
Chemistry 120
Stoichiometry: Chemical CalculationsThe final balanced equation is
and the coefficients are known as the
stoichiometric coefficients.
These coefficients give the molar ratios for reactants and products
This is a stoichiometric reaction – one where exactly the correct number of atoms is present in the reaction
C6H12 +9O2 6CO2 + 6H2O
Chemistry 120
Stoichiometry: Chemical Calculations
If cyclohexane were burnt in an excess of oxygen,
the quantity of oxygen used would be the same
although O2 would be left over.
Chemistry 120
Stoichiometry: Chemical CalculationsThe final balanced equation is
and the coefficients are known as the
stoichiometric coefficients.
These coefficients give the molar ratios for reactants and products
This is a stoichiometric reaction – one where exactly the correct number of atoms is present in the reaction
Chemistry 120
The Exam
Chemistry 120
Solutions
A solution is a homogenous mixture which is composed of two or more components
the solvent
- the majority component
and
one or more solutes
- the minority components
Chemistry 120
Solutions
Most common solutions are liquids where a solid, liquid or gas (the solute) is dissolved in the liquid solvent.
Some are solids where both the solvent and the solute are solids. Brass is an example
Chemistry 120
Solutions
Most common solutions are liquids where a solid, liquid or gas (the solute) is dissolved in the liquid solvent.
Some are solids where both the solvent and the solute are solids. Brass is an example
Chemistry 120
Solutions
Most common solutions are liquids where a solid, liquid or gas (the solute) is dissolved in the liquid solvent.
Some are solids where both the solvent and the solute are solids. Brass is an example
Cu
ZnHere copper is the solvent, zinc the solute.
Chemistry 120
Solutions
Gas-Solid solution: Hydrogen in palladium
Steel
Chemistry 120
Solutions
Common laboratory solvents are usually organic liquids such as acetone, hexane, benzene or ether or water.
Solutions in water are termed aqueous solutions and species are written as E(aq).
Water is the most important solvent. The oceans cover ~ ¾ of the surface of the planet and every cell is mainly composed of water.
Chemistry 120
Solutions
Aqueous Solutions
Water is one of the best solvents as it can dissolve many molecular and ionic substances.
The properties of solutions which contain molecular and ionic solutes are very different and give insight into the nature of these substances and solutions.
Chemistry 120
Solutions
Ionic Solutions
An ionic substance, such as NaClO4, contain ions – in this case Na+ and ClO4
-.
The solid is held together through electrostatic forces between the ions.
In water, the solid dissolves and the particles move away from each other and diffuse through the solvent. This process is termed
Ionic Dissociation
Chemistry 120
Solutions
Ionic Solutions
In an ionic solution, there are therefore charged particles – the ions – and as the compound is electrically neutral, then the solution is neutral.
When a voltage is applied to the solution, the ions can move and a current flows through the solution.
The ions are called charge carriers and whenever electricity is conducted, charge carriers are present.
Chemistry 120
Solutions
Molecular Solutions
A molecular solution does not conduct electricity as there are no charge carriers present.
The bonding in a molecule is covalent and involves the sharing of atoms and there is no charge separation.
Chemistry 120
Solutions
Electrolytes
A solute that, when dissolved, produces a solution that conducts is termed an electrolyte, which may be strong or weak.
A strong electrolyte is one which is fully dissociated in solution into ions
A weak electrolyte is one which is only partially dissociated.
Chemistry 120
Solutions
Moles and solutions
When a substance is dissolved in a solvent, we relate the quantity of material dissolved to the volume of the solution through the concentration of the solution.
The concentration is simply the number of moles of the material per unit volume:
C = n V
n = number of moles; V = volume of solvent
Chemistry 120
Solutions
Moles and solutions
The units of concentration are:
C = n = moles V L3
and we define a molar solution as one which has 1 mole per liter.
Alternatively,
Concentration = Molarity = number of molesvolume of solution
Chemistry 120
Solutions
Example
4 g of Na2SO4(s) is dissolved in 500 ml of water.
What is the concentration of the solution?
Chemistry 120
Solutions
Example
4 g of Na2SO4(s) is dissolved in 500 ml of water.
What is the concentration of the solution?
Formula mass of Na2SO4(s):
Molar Atomic Mass of Na: 22.9898 gmol-1
Molar Atomic Mass of S: 32.064 gmol-1
Molar Atomic Mass of O: 15.9994 gmol-1
Chemistry 120
Solutions
Example
4 g of Na2SO4(s) is dissolved in 500 ml of water.
What is the concentration of the solution?
Formula mass of Na2SO4(s):
(2 x 22.9898)+ 32.064+(4x15.9994)=142.041gmol-1
1 mole of Na2SO4(s) = 142.041g
1/142.041 mole of Na2SO4(s) = 1 g
Chemistry 120
Solutions
Example
4 g of Na2SO4(s) is dissolved in 500 ml of water.
What is the concentration of the solution?
1/142.041 mole of Na2SO4(s) = 1 g
Therefore 4 g of Na2SO4(s) = 4/142.041 mole
= 2.82 x 10-2 mole
Chemistry 120
Solutions
Example
4 g of Na2SO4(s) is dissolved in 500 ml of water.
What is the concentration of the solution?
2.82 x 10-2 mole is therefore dissolved in 500 ml of
water;
So in 1 L, there are 2 x 2.82 x 10-2 mole
Molarity of solution = 5.64 x 10-2 molL-1
Chemistry 120
Solutions
Example
The equation for the dissolution of Na2SO4(s) is
So if we have 5.64 x 10-2 molL-1 Na2SO4(s), we must
have 1.13 x 10-1 moles Na+(aq)
and 5.64 x 10-2 mol SO42-
(aq) as there are 2 Na
cations for every sulfate ion
Na2SO4(s)H2O
2Na+(aq) + SO4
2-(aq)
Chemistry 120
Solutions
If we change the volume of the solution then we change the concentration:
If the Na2SO4 solution is diluted with 500ml of water, the concentration or molarity would be halved:
2.82 x 10-2 mole is therefore dissolved in 1000 ml of water
Molarity = 2.82 x 10-2 molL-1
Chemistry 120
Solutions
Dissolution on an atomic level.
Solids are held together by very strong forces.
NaCl(s) melts at 801oC and
boils at 1465 oC but it
dissolves in water at room
temperature.
Chemistry 120
Solutions
Dissolution on an atomic level.
When we dissolve NaCl(s) in water we break the bonds between ions but make bonds between the ions and the water
Chemistry 120
Solutions
Dissolution on an atomic level.
When we dissolve NaCl(s) in water we break the bonds between ions but make bonds between the ions and the water
The ions are hydrated or solvated in solution and these bonds between solvent and solute make the dissolution energetically possible
If something does not dissolve then the energetics are wrong for it do do so.
Chemistry 120
Solutions
Solubility rules
All ammonium and Group I salts are soluble.
All Halides are soluble except those of silver, lead and mercury (I)
All Sulfates are soluble except those of barium and lead.
All nitrates are soluble.
Everything else is insoluble
Chemistry 120
Solutions
• Solutions are homogenous mixtures in which the
majority component is the solvent
and the
minority component is the solute
• Solutions are normally liquid but solutions of gases in solids and solids in solids are known.
• Ionic compounds dissolve in water to give conducting solutions – they are electrolytes
Chemistry 120
Solutions
• Electrolytes are either strong or weak depending on the degree of dissociation in solution
• Molecular solutions do not conduct as molecules do not dissociate in solution
• The concentration or molarity of a solution is defined by
C = n = moles V L3
and the units are molL-1 or moldm-3
Chemistry 120
Solutions
• When ionic substances dissolve,
bonds between particles in the solid break
and
bonds between the solvent and the ions are made
• There are general rules for the solubilities of ionic compounds
Chemistry 120
Reactions in Solution
Reactions in solution include
• Acid – base reactions
• Precipitation reactions
• Oxidation- reduction reactions
Chemistry 120
Reactions in Solution
Reactions and equilibria
Reactions are often written as proceeding in one
direction only – with an arrow to show the direction
of the chemical change, reactants to products.
Not all reactions behave in this manner and not all
reactions proceed to completion.
Even those that do are dynamic.
Chemistry 120
Reactions in Solution: Acid - Base
NaI*(s)
NaI(aq)
A saturated solution of NaI is placed in contact
with Na131I(s), which is radioactive.
Chemistry 120
Reactions in Solution: Acid - Base
NaI*(s)
NaI(aq)
A saturated solution of NaI is placed in contact
with Na131I(s), which is radioactive.
After time, the activity
in the solution is
measured and ..........
Chemistry 120
I-
I-
I-
I-
I-
I-
I-
I-I-
I-
I-
I-I-I-
I-
I- I-
Na+
Na+
Na+
Na+
Na+
Na+Na+
Na+
Na+
Na+
Na+
Na+
Na+
Na+
Na+
Na+Na+
I-
I-
I-
I-
I-
I-
I-
I-I-
I-
I-
I-I-I-
I-
I- I-
Na+
Na+
Na+
Na+
Na+
Na+Na+
Na+
Na+
Na+
Na+
Na+
Na+
Na+
Na+
Na+Na+
Reactions in Solution: Acid - Base
Radioactivity is found in the solution, even though
the concentration of I-(aq) has not changed.
Chemistry 120
Reactions in Solution: Acid - Base
The equilibrium here is composed of two reactions:
So we write
Na131I(s)H2O
Na+(aq) + 131I-
(aq)
H2ONa+
(aq) + I-(aq) NaI(s)
H2ONa+
(aq) + I-(aq)NaI(s)
Chemistry 120
Reactions in Solution: Acid - Base
Such reactions are termed equilibria and all chemical reactions are equilibria.
The symbol for an equilibrium is a double-headed arrow
Chemistry 120
Reactions in Solution: Acid - Base
Such reactions are termed equilibria and all chemical reactions are equilibria.
The symbol for an equilibrium is a double-headed arrow
Chemistry 120
Reactions in Solution: Acid - Base
Such reactions are termed equilibria and all chemical reactions are equilibria.
The symbol for an equilibrium is a double-headed arrow
Chemistry 120
+ =
Forward reaction
Reverse reaction
Reactions in Solution: Acid - Base
Such reactions are termed equilibria and all chemical reactions are equilibria.
The symbol for an equilibrium is a double-headed arrow
Chemistry 120
Reactions in Solution: Acid - Base
Equilibria are important in the chemistry of acids and bases
Strong acids and bases are completely ionized
But.....
Weak acids and bases are not.
Chemistry 120
Reactions in Solution: Acid - Base
The Arrhenius definition of acid and bases are:
an acid is a compound which dissolves
in water or reacts with water to give
hydronium ions, H3O+(aq)
a base is a compound which dissolves
in water or reacts with water to give
hydroxide ions, OH- (aq)
Svante Arrhenius
(1859 – 1927)
Chemistry 120
Reactions in Solution: Acid - Base
A strong acid is a compound which dissolves and dissociates completely in water or reacts with water to give hydronium ions, H3O+
(aq)
- the double arrow implies that the reaction can go both ways – it is an equilibrium.
As a strong acid, the reaction is completely on the RHS:
HCl(g)H2O
H3O+(aq) + Cl-(aq)
HCl(g)H2O
H3O+(aq) + Cl-(aq)
Chemistry 120
Reactions in Solution: Acid - BaseA strong base is a compound which dissolves and dissociates completely in water or reacts with water to give hydroxide ions, OH- (aq)
Again, we could write this reaction as an equilibrium with a double headed arrow, but the base is strong and the reaction is completely over to the right hand side.
NaOH(s)H2O
Na+(aq) + OH-
(aq)
Chemistry 120
Reactions in Solution: Acid - BaseIn a reaction such as
we write the reaction as going from LHS to RHS.
Chemical reactions run both ways, so in this reaction, there are two reactions present:
Ionization
Recombination
H2OMeCO2H H3O+
(aq) + MeCO2-(aq)
H2OMeCO2H H3O+
(aq) + MeCO2-(aq)
H2OMeCO2HH3O+
(aq) + MeCO2-(aq)
Chemistry 120
Reactions in Solution: Acid - BaseWe write the reaction for acetic acid, MeCO2H, as
an equilibrium to include the ionization and recombination. Ionization
Recombination
As the amount of ionization and recombination are the same, the concentrations of the ions and the molecular form are constant
H2OMeCO2H H3O+
(aq) + MeCO2-(aq)
H2OMeCO2HH3O+
(aq) + MeCO2-(aq)
H2OMeCO2H H3O+
(aq) + MeCO2-(aq)
Chemistry 120
Reactions in Solution: Acid - Base
In solution, weak acids establish an equilibrium between the un-ionized or molecular form and the ionized form:
un-ionizedmolecular form
ionized
H2OMeCO2H H3O+
(aq) + MeCO2-(aq)
Chemistry 120
Reactions in Solution: Acid - Base
In solution, strong acids are completely ionized and even though there is an equilibrium, it lies entirely on the RHS and recombination is negligible:
un-ionizedmolecular form
ionized
H2OHBr(g) H3O+
(aq) + Br-(aq)
Chemistry 120
Reactions in Solution: Acid - Base
Acids with more than one ionizable hydrogen are termed
Polyprotic
The common polyprotic acids are
H3PO4 Phosphoric acid
H2SO4 Sulfuric acid
Chemistry 120
Reactions in Solution: Acid - Base
Polyprotic acids can ionize more than once
H3PO4
Each proton is ionizable and the anions, dihydrogen phosphate (H2PO4
-(aq))
and hydrogen phosphate (HPO42-
(aq)) both act as acids, though H3PO4 is a weak acid.
H2SO4(aq) H3O+(aq)
+ HSO4-(aq)H2O
HSO4-(aq) H3O+
(aq) + PO4
2-(aq)H2O
HPO42-
(aq) H3O+(aq)
+ PO42-
(aq)H2O
Chemistry 120
Reactions in Solution: Acid - Base
Polyprotic acids can ionize more than once
H3PO4
H2SO4
H3PO4(aq) H3O+(aq)
+ H2PO4-(aq)H2O
H2PO4-(aq) H3O+
(aq) + HPO4
2-(aq)H2O
HPO42-
(aq) H3O+(aq)
+ PO42-
(aq)H2O
H2SO4(aq) H3O+(aq)
+ HSO4-(aq)H2O
HSO4-(aq) H3O+
(aq) + PO4
2-(aq)H2O
Chemistry 120
Reactions in Solution: Acid - Base
In contrast, H2SO4 is a strong acid and hydrogen
sulfate (HSO4-(aq)) is also a strong acid.
H2SO4(aq) H3O+(aq)
+ HSO4-(aq)H2O
HSO4-(aq) H3O+
(aq) + PO4
2-(aq)H2O
Chemistry 120
Reactions in Solution: Acid - BaseStrong or weak?
All acids can be assumed to be weak except the following:
HCl(aq) hydrochloric acid
HBr(aq) hydrobromic acid
HI(aq) hydriodic acid
HClO4(aq) perchloric acid
HNO3(aq) nitric acid
H2SO4(aq) sulfuric acid
Chemistry 120
Reactions in Solution: Acid - BaseHydrogens attached to carbon are not ionizable in water
Acetic acid, MeCO2H (or CH3CO2H) has the
structure H
H
H
O
O H
Chemistry 120
Reactions in Solution: Acid - Base
Only the hydrogen attached to oxygen is ionized in aqueous solution
The methyl hydrogens are NOT ionizable in aqueous solution.
H
H
H
O
O HH2O
H
H
H
O
O
O
HHH+
Chemistry 120
Reactions in Solution: Acid - Base
Strong bases are those which ionize in solution of react to generate hydroxide ion. The common strong bases are those which already contain the OH- ion in the solid. 2 Li
3
3 Na11 Mg12
4 K19
Ca20
5 Rb37
Sr38
6 Cs55
Ba56
Strong bases are therefore the hydroxides of the group I and II metals
Chemistry 120
Reactions in Solution: Acid - Base
Weak bases are the majority and are usually amines and ammonia. These react with water and deprotonate it, forming hydroxide ion and an ammonium ion:
Trimethylamine
Trimethylammonium
N
H3C CH3
CH3H2O
N
H3C CH3
CH3
H
+ OH-
Chemistry 120
Reactions in Solution: Acid - BaseReactions in Solution: Acid - Base
Neutralization reactions and titrations Hydroxide and hydronium ions will react to form water.
From the stoichiometry of the balanced equation, the hydroxide and hydronium react in a 1:1 ratio.
We can therefore neutralize a known concentration of base or acid with the same quantity of acid or base. This is an example of a titration.
2H2O(l)H3O+(aq) + OH-
(aq)
Chemistry 120
Reactions in Solution: Acid - BaseReactions in Solution: Acid - Base
Neutralization reactions and titrations We use an indicator to determine the acidity or basicity of a solution:
An indicator is a compound which changes color strongly at a certain level of acidity.
Chemistry 120
Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseNeutralization reactions and titrations
We add acid or base – the titrant - to a solution of unknown concentration containing a few drops of the indicator solution.
When the solution is still acid, no color change occurs; when the indicator changes color, we know the equivalence point – the point where the acidity or basicity has been neutralized.
By knowing the concentration and the volume of the titrant, we can calculate the concentration of the of the unknown solution.
Chemistry 120
Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseDecompostion in acid
A solid base, such as Ca(OH)2(s), will dissolve with
reaction in an acid. The anion, hydroxide, reacts
with the acid to form the calcium salt of the acid:
Chemistry 120
Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseDecompostion in acid
A solid base, such as Ca(OH)2(s), will dissolve with
reaction in an acid. The anion, hydroxide, reacts
with the acid to form the calcium salt of the acid:
Is this balanced?
Chemistry 120
Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseDecompostion in acid
A solid base, such as Ca(OH)2(s), will dissolve with
reaction in an acid. The anion, hydroxide, reacts
with the acid to form the calcium salt of the acid:
Is this balanced? No
Chemistry 120
Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseDecompostion in acid
A solid base, such as Ca(OH)2(s), will dissolve with
reaction in an acid. The anion, hydroxide, reacts
with the acid to form the calcium salt of the acid:
Is this balanced? No
2Ca(OH)2(s) + H2SO4(aq) Ca2SO4(s) + 2H2O(l)
Chemistry 120
Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseDecompostion in acid
Some anions also decompose in acid. These are usually anions which are derived from gases which are not soluble in water:
CO32-
(aq) carbonate CO2(g)
HCO3-(aq) hydrogen carbonate CO2(g)
S2-(aq) sulfide H2S(g)
HS-(aq) hydrogen sulfide H2S(g)
SO32-
(aq) sulfite SO2(g)
HSO3-(aq) hydrogen sulfite SO2(g)
Chemistry 120
Gases
Properties of Gases
Kinetic Molecular Theory of Gases
Pressure
Boyle’s and Charles’ Law
The Ideal Gas Law
Gas reactions
Partial pressures
Chemistry 120
Gases
Properties of Gases
All elements will form a gas at some temperature
Most small molecular compounds and elements are either gases or have a significant vapor pressure.
1 H1 Room Temperature Gases He
2
2 Li3
Be4
B5
C6
N7
O8
F9
Ne10
3 Na11
Mg12
Al13
Si14
P15
S16
Cl17
Ar18
4 K19
Ca20
Sc21
Ti22
V23
Cr24
Mn25
Fe26
Co27
Ni28
Cu29
Zn30
Ga31
Ge32
As33
Se34
Br35
Kr36
5 Rb37
Sr38
Y39
Zr40
Nb41
Mo42
Tc43
Ru44
Rh45
Pd
46
Ag47
Cd48
In49
Sn50
Sb51
Te52
I53
Xe54
6 Cs55
Ba56
Lu71
Hf72
Ta73
W74
Re75
Os76
Ir77
Pt78
Au79
Hg80
Tl81
Pb82
Bi83
Po
84
At85
Rn86
Chemistry 120
Gases
Properties of Gases
As the temperature rises, all elements form a gas at some point.
In the following diagram,
Blue represents solids
Green represents liquids
Red represents gases
At O K, all elements are solids
At 6000 K, all are gases
Chemistry 120
Gases
Chemistry 120
Gases
Properties of Gases
Gases have no shape and no volume.
They take the volume and shape of the container
Their densities are low – usually measured in gL-1
The atoms or molecules of the gas are far further apart than in a solid or a liquid.
Chemistry 120
Gases
Gases as an ensemble of particles
The attractive forces between liquids and solids are very strong
LiF: M.p.: 848°C B. p.: 1676°C
In a gas, the forces between particles are negligible and as there are no attractive forces, a gas will occupy the volume of the container.
Solid Liquid Liquid Gas
Chemistry 120
GasesGases as an ensemble of particles
The structures of liquids and solids are well ordered on a microscopic level
CaCl2Ethanol, C2H5OH
Chemistry 120
GasesGases as an ensemble of particles
In a gas, there is no order and all the properties of the gas are isotropic – all the properties of the gas are the same in all directions.
Gas particles are distributed uniformly throughout the container.
They can move throughout the container in straight line trajectories.
Chemistry 120
GasesGases as an ensemble of particles
The directions of the motions of the gas particles are random
and
The velocities form a distribution – there is a range of possible velocities around an average value.
The trajectories of the gas particles are straight lines and there are two possible fates for a gas molecule......
Chemistry 120
GasesGases as an ensemble of particles
A gas particle can collide with
– the walls of the container
Or
– another gas molecule
When this happens, the gas particle changes direction.
Chemistry 120
GasesGases as an ensemble of particles
Kinetic energy can be transferred between the two colliding particles
– one can slow down and the other speed up –
but the net change in kinetic energy is zero.
These collisions are termed elastic, meaning that there is no overall change in kinetic energy.
Chemistry 120
Gases
The average kinetic energy for a given gas is determined by the temperature alone and the width and peak maximum is also determined by the temperature.
The Maxwell-Boltzmann distribution for He
Chemistry 120
GasesGases as an ensemble of particles
The force exerted by the gas particles on the walls of the container gives rise to the pressure of the gas.
We define pressure as the force exerted per unit area:
P = Force = F Area A
The unit of pressure is the Pascal (Pa)
1 Pa = 1 Nm-2
In practice, the Pascal is too small - kPa or GPa
Chemistry 120
GasesPressure measurement
Pressure is also measured in several other non – SI units:
In industry: Pounds per square in (p.s.i.)
In research: Pascal, atmosphere, bar, Torr
Chemistry 120
GasesPressure conversion factors
Atmospheric pressure = 101,325 Pa
1 Atmosphere = 101,325 Pa = 1 bar
1 Atmosphere = 101,325 Pa
= 1 bar
= 760 Torr
= 760 mmHg
= 14.7 p.s.i.
Chemistry 120
GasesPressure Measurement
Pressure is measured using a manometer or barometer
– either one containing Hg or an electronic gauge
A mercury manometer is a U–tube connected to the gas vessel, with the other end either evacuated or open to the atmosphere.
The measurement of the height difference between the mercury levels on both sides of the ‘U’ gives the pressure........
Chemistry 120
GasesPressure Measurement
Let the height difference between the two Hg levels be h
Then the gas pressure is given by
Pgas = P0 + h
As
P = Force = F = mg where g = 9.81 ms-2
Area A A
Chemistry 120
GasesPressure Measurement
How is the height difference related to the pressure?
As density, = m V
Then m = V
The volume of the column of mercury is
V = A.h
And so m = V = A.h
Chemistry 120
GasesPressure Measurement
The pressure above the baseline pressure P0 is therefore
Pgas = mg =gA.h = gh A A
Chemistry 120
GasesGases as an ensemble of particles
Kinetic energy can be transferred between the two colliding particles
– one can slow down and the other speed up –
but the net change in kinetic energy is zero.
These collisions are termed elastic, meaning that there is no overall change in kinetic energy.
Chemistry 120
The Gas Laws
The factors that control the behavior of a gas are
• The nature of the gas
Chemistry 120
The Gas Laws
The factors that control the behavior of a gas are
• The nature of the gas
• The quantity of the gas - n
Chemistry 120
The Gas Laws
The factors that control the behavior of a gas are
• The nature of the gas
• The quantity of the gas - n
• The pressure - P
Chemistry 120
The Gas Laws
The factors that control the behavior of a gas are
• The nature of the gas
• The quantity of the gas - n
• The pressure - P
• The temperature - T
Chemistry 120
The Gas Laws
The factors that control the behavior of a gas are
• The nature of the gas
• The quantity of the gas - n
• The pressure - P
• The temperature - T
• The volume of a gas -V
Chemistry 120
The Gas Laws
These laws apply to a perfect gas or and ideal gas. All gases behave as ideal gases at ordinary temperatures and pressures.
The qualities of an ideal gas are:
• Zero size to the gas particles
We assume that the volume of the container is very much larger than the total volume of the gas molecules
• No attractive forces between atoms
Chemistry 120
The Gas Laws
These laws apply to a perfect gas or and ideal gas. All gases behave as ideal gases at ordinary temperatures and pressures.
At low temperatures and high pressures gases deviate from ideality.
The ideal gas laws are based on three interdependent laws – Boyle’s Law, Charles’ Law and Avogadro’s Law.
Chemistry 120
The Gas Laws
Boyle’s Law
Robert Boyle experimented with gases in Oxford in 1660.
He discovered that the product of the volume and the pressure of a gas is a constant, so long as the quantity of gas and the temperature are constant.
Chemistry 120
The Gas Laws
Boyle’s Law
Mathematically,
PV = a constant
as long as n and T are constant
Chemistry 120
The Gas Laws
Boyle’s Law
Mathematically,
PV = a constant, k
or
P = kV
as long as n and T are constant.
Chemistry 120
The Gas Laws
Boyle’s Law
A graph of Boyle’s data shows this relationship:
PV = k
Chemistry 120
The Gas Laws
Boyle’s Law
A graph of 1/P as the abscissa and V as the ordinate.
V = k P
The graph shows a straight line of slope k
Chemistry 120
The Gas Laws
Boyle’s Law
As the pressure rises, 1/P becomes smaller and the graph passes through the origin.
This implies that at infinitely large pressure, the volume of a gas is zero.We know that molecules and and atoms have a definite volume, so Boyle’s law must fail at very high pressures.
Chemistry 120
The Gas Laws
Charles’ Law
Jacques Charles was a Feench scientist and aeronaut who discovered (1787) that all gases expand by the same amount when the temperature of the gas rises by the same amount.
Chemistry 120
The Gas Laws
Charles’ Law
Mathematically, we express this as
V = k’T
And a graph of Charles’ Law
is a straight line:
Chemistry 120
The Gas Laws
The Combined Gas Law for a Perfect Gas
Combining Boyle’s Law, Charles’ Law and Avogadro’s Law,
V = k and V = k’T and V = k”n P
we can say thatV nT P
Chemistry 120
The Gas Laws
The Combined Gas Law for a Perfect Gas
V nT P
OrV =K nT P
Rearranging we find
PV = a constantnT
Chemistry 120
The Gas Laws
The Combined Gas Law for a Perfect Gas
The constant is termed the Universal Gas Constant, R, and takes the value
R = 8.314 Jmol-1K-1
So the Universal Gas Law is written as
PV = nRT
This Law applies to all gases as long as they fulfill the conditions for near ideal behavior – not at high pressure and not at low temperature
Chemistry 120
The Gas Laws
Using the Combined Gas Law
If the quantity of gas is the same, then changes in pressure, temperature or volume can be calculated easily as
P1V1 = n = P2V2 RT1 RT2
Or
P1V1 = P2V2 T1 T2
Chemistry 120
The Gas Laws
Using the Combined Gas Law
The advantage of this expression is that the units do not matter; the units used for P1 ,V1, and T1 will be returned in the calculation for P2 ,V2, and T2.
However, if you have to use PV = nRT, you must use the correct units which are consistent with R.
The easiest way is to convert all temperatures to K, all pressures to Pa and all volumes to m3; the value for R is then 8.314 Jmol-1K-1
Chemistry 120
The Gas Laws
The absolute temperature scale
From Charles’ Law, the decrease in volume per unit temperature is always the same and therefore there must be a minimum temperature that can be reached. This is absolute zero O K, and is the zero point for the absolute temperature scale.
The temperature in K is related to the temperature in oC through
T/K = T/oC + 273.16
Chemistry 120
The Gas Laws
Example: Molecular Mass determinations
If we know the mass of gas in a sample of known volume, pressure and temperature, then we can calculate the relative molecular mass as we can calculate n.
As n = m then, PV = mRT , so RMM = mRT RMM RMM PV
RMM = mRT PV
Chemistry 120
The Gas Laws
Example: Molar volumes
From Avogadro’s Law, equal quantities of gas occupy equal volumes.
The volume of one mole of gas is therefore independent of the nature of the gas, as long as the gas behaves as ideal.
One mole of a perfect gas at 0oC and 1 atm pressure occupies
22.4 L
Chemistry 120
The Gas Laws
Example: Volumes and moles
When we react solids or liquids, the easiest way is to measure the mass of the sample and then convert to moles by dividing by the relative molecular mass.
For gases, the easiest way is to measure the pressure or the volume, as the densities of gases are so low.
For these calculations, you must use the same temperatures and pressures for each gas.
Chemistry 120
The Gas Laws
Partial pressures
In a mixture of gases, we can measure the total pressure of the mixture – PTotal and therefore we can use PV = nRT to determine the total number of moles of gas present.
As the mixture contains more than one gas, we can write the contribution of the pressure of each gas to the total pressure
Chemistry 120
The Gas Laws
Partial pressures
So the total pressure Ptotal is written as the sum of all the individual pressures of the components of the gas mixture:
Chemistry 120
The Gas Laws
Partial pressures
So the total pressure Ptotal is written as the sum of all the individual pressures of the components of the gas mixture:
PTotal = P1 + P2 + P3 + P4 + ...........
Chemistry 120
The Gas Laws
Partial pressures
So the total pressure Ptotal is written as the sum of all the individual pressures of the components of the gas mixture:
PTotal = P1 + P2 + P3 + P4 + ...........
As PV = nRT then
Chemistry 120
The Gas Laws
Partial pressures
So the total pressure Ptotal is written as the sum of all the individual pressures of the components of the gas mixture:
PTotal = P1 + P2 + P3 + P4 + ...........
As PV = nRT then
nTotalRT = n1RT + n2RT + n3RT + n4RT + ...........
Chemistry 120
The Gas Laws
Partial pressures
So
PTotal = P1 + P2 + P3 + P4 + ...........
nTotalRT = n1RT + n2RT + n3RT + n4RT + ...........
Chemistry 120
The Gas Laws
Partial pressures
So
PTotal = P1 + P2 + P3 + P4 + ...........
nTotalRT = n1RT + n2RT + n3RT + n4RT + ...........
nTotal = n1 + n2 + n3 + n4 + ...........
Chemistry 120
The Gas Laws
Partial pressures
So the pressures of each component of the gas mixture correlate with the number of moles of the gas component of the mixture – a simple extension of Avogadro’s Law.
Chemistry 120
The Gas Laws
Partial pressures
We can also write the fraction of the total pressure that is due to one of the component:
PTotal = P1 + P2 + P3 + P4 + ...........
Chemistry 120
The Gas Laws
Partial pressures
We can also write the fraction of the total pressure that is due to one of the component:
PTotal = P1 + P2 + P3 + P4 + ...........
nTotal = n1 + n2 + n3 + n4 + ..........
Chemistry 120
The Gas Laws
Partial pressures
We can also write the fraction of the total pressure that is due to one of the component:
PTotal = P1 + P2 + P3 + P4 + ...........
nTotal = n1 + n2 + n3 + n4 + ..........
P1 = n1RT
Chemistry 120
The Gas Laws
Partial pressures
We can also write the fraction of the total pressure that is due to one of the component:
PTotal = P1 + P2 + P3 + P4 + ...........
nTotal = n1 + n2 + n3 + n4 + ..........
P1 = n1RT
So, P1 = n1
PTotal n1 + n2 + n3 + n4 + ..........
Chemistry 120
The Gas Laws
Partial pressures
P1 = n1
PTotal n1 + n2 + n3 + n4 + ..........
The fraction on the RHS is called the mole fraction
and is written as x1 so we can write
P1 = n1 PTotal
n1 + n2 + n3 + n4 + ..........
Or P1 = x1 PTotal
Chemistry 120
Thermochemistry
Energy
Energy is defined as the ability to do work.
There are several forms of energy
Kinetic energy – energy due to motion
Chemistry 120
Thermochemistry
Energy
Energy is defined as the ability to do work.
There are several forms of energy
Kinetic energy – energy due to motion EK = 1/2mv2
Potential energy – the energy due to the position of a particle in a field
e.g. Gravitational, electrical, magnetic etc.
Chemistry 120
Thermochemistry
Energy
The unit of energy is the Joule (J) and
1 J = 1 kgm2s-2
Thermochemistry is the study of chemical energy and of the conversion of chemical energy into other forms of energy.
It is part of thermodynamics – the study of the flow of heat.
Chemistry 120
Thermochemistry
Thermochemically, we define the system as the part of the universe under study and the surroundings as everything else.
Systems come in three forms:
Open The system can exchange matter and energy with the surroundings
Closed The system can exchange energy only with the surroundings
Isolated There is no exchange of matter or of energy with the surroundings
Chemistry 120
Thermochemistry
Matter is continually in motion and has an internal energy that is composed of several different types
There is
Translation Rotation Vibration Potential
between molecules and inside molecules.
The internal energy is written as U
Chemistry 120
Thermochemistry
Matter is continually in motion and has an internal energy that is composed of several different types
There is
Translation Rotation Vibration Potential
between molecules and inside molecules.
The internal energy is written as U
The internal energy is directly connected to heat and the transfer of heat.
Chemistry 120
Thermochemistry
Heat is the transfer of energy between the surroundings and the system or between systems.
The direction of the heat flow is indicated by the temperature
– heat flows along a Temperature gradient
from high temperature to low temperature.
When the temperature of the system and that of the surroundings are equal, the system is said to be
in thermal equilibrium
Chemistry 120
Thermochemistry
Heat is the transfer of energy between the surroundings and the system or between systems.
The direction of the heat flow is indicated by the temperature
– heat flows along a Temperature gradient
from high temperature to low temperature.
When the temperature of the system and that of the surroundings are equal, the system is said to be
in thermal equilibrium
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