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Equilibrium
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EquilibriumThe state where the concentrations of
all reactants and products remain constant with time.
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Reactions are reversibleA + B C + D ( forward)C + D A + B (reverse)Forward and Reverse Rxns can be
shown by double arrow
A + B C + D
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A + B C + D Initially there is only A and B so only the
forward reaction is possibleAs C and D build up, the reverse reaction
speeds up while the forward reaction slows down.
Eventually the rates are equal So concentrations of the reactants and
products no longer change with time
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Rea
ctio
n R
ate
Time
Forward Reaction
Reverse reaction
Equilibrium
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Static or Dynamic? At equilibrium, forward and reverse reaction rates
are equal May seem like no changes are occurring but
there are changes No NET changes On the molecular level, there is frantic activity.
Equilibrium is not static, but is a highly dynamic situation.
chemical reactions take place ,but concentrations of reactants and products remain unchanged
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Analogies and Metaphors to think about
1. In a football game, the number of players on the field is constant although exchange of players (substitution) changes actual persons.
2. Connected fish bowl analogy . Two fish tanks are connected by a tube large enough to allow passage of fish. A number of fish are placed in one of the tanks. At equilibrium, the number of fish in each tank will eventually become unchanged.
3. Two jugglers analogy.
4. Drinking fountain line: (a) Ten students waiting in line to get a drink of water on a hot day. As each
gets a drink, the same student reenters the line (equilibrium in a closed system).
(b) (b) Same situation as "a," except as each student gets a drink and leaves, a new student enters the line (steady-state in an open system).
5. Picture a number of horses and wranglers in a corral. As each wrangler mounts a horse, the wrangler is bucked off. The equilibrium is:
Horse + Wrangler Mounted wrangler
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Molecular Simulation In this simulation two gaseous reactants
collide to produce a more dense solid.
A + B C
gaseous R dense P http://www.absorblearning.com/media/atta
chment.action?quick=w8&att=2310
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Homo vs HeteroHomogeneous Equilibria all reacting
species are in the same phasegas phase
• equilibrium constant can be expressed in terms of pressure or concentration, Kp or Kc
Solution (aqueous) phase• concentration term for the pure liquid does not appear in the
expression for the equilibrium constant but aqueous substance concentrations do appear
Heterogeneous Equilibria all reacting species are not in the same phase
• concentration term for solid or liquid does not appear in the expression for the equilibrium constant
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Equilibrium SummarizedForward and Reverse rates are equalConcentrations are not.Rates are determined by
concentrations and activation energy.Molecular Motion is frantic and
constantly changingMacroscopically no net change is
occurring (We can’t observe any changes)
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Can you identify when the system reached equilbrium?
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Distinguishing between Physical and Chemical Equilibrium
As with physical and chemical changes, physical and chemical equilibrium follow the same rules:Physical no changes to the chemical
properties of the substances involved• Ex. equilibrium of water vapor with liquid water in a partly
filled sealed bottle
Chemical involve changes in the chemical composition of substances. Bond breaking and bond formation is involved.
• Ex. dissociation of acetic acid water into acetate and hydronium ion
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ActivityModel Dynamic Equilibrium with
CoinsNow lets plot the data using excel
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Law of Mass Action For a reaction: aA + bB cC + dD⇄
equilibrium constant: K
Pure liquids and pure solids have concentrations of 1.
c
c
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Playing with KIf we write the reaction in reverse.
cC + dD aA + bB⇄Then the new equilibrium constant is
c
c
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They are simply the inverse of one another.
Forward Reaction
aA + bB cC + dD⇄
So we call this K1
And K1= 1 = K2-1
K2
Reverse Reaction
cC + dD aA + bB⇄
So we call this K2
And K2= 1 = K1-1
K1
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The units for KAre determined by the various powers
and units of concentrations. They depend on the reaction.will always have the same value at a
certain temperature (Why will the T effect it?) no matter what amounts are initially
added ratio at equilibrium will always be same
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K Has no units Is constant at any given temperature. Is affected by temperature.
Equilibrium constants are reaction, phase, temperature and pressure dependent
There is a K for each temperature. Equilibrium constant values are thus established
for a specific reaction in a specific system and will be unchanging (constant) in that system, providing the temperatures does not change.
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What does the size of my K mean?
Large K > 1 products are "favored“Ex. 1 x 1034
K = 1 neither reactants nor products are
favoredSmall K < 1
reactants are "favored“Ex. 4 x 10-41
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Now Let’s Calculate Your KUsing your data from the simulation
calculate K
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Different Equilibrium Constants (All of these are known as Keq) Kc is the most used general form with molar
concentrations. Kp can be used with partial pressures when
working with a gas phase reaction. Ka is used for the dissociation of weak acids in
water. Kb is used for the dissociation of weak bases in
water. Kw is the equilibrium expression for the dissociation
of water into its ions. Ksp is used for the dissociation into ions of sparingly
soluble solids in water.
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Practice Writing the Equilibrium Expression
4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(g)
First write the equilibrium expression using no concentration values.
What is the value for K if the concentrations are as follows
NH3 1.0 MO2 1.0 MNO2 1.4 MH2O 1.8 M
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Equilibrium with GasesEquilibria involving only gases can be
described using pressures or concentrations If using pressures,
use pA not [A]KP not KC
be sure all pressure are in the same units
N2(g) + 3H2(g) 2NH3(g)
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Calculating Kc from Kp
whereΔn is the difference in moles of gas on either side
of the equation (np – nr)R is the gas law constant: 0.08206T is Kelvin temperatureFor: N2(g) + 3H2(g) 2NH3(g) Δ n = (2) – (1+3) = -2
h
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PracticeSetup the expression for KP in terms
of KC, R and T
2NO(g) + Cl2(g) 2NOCl(g)
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What the equilibrium constant tells us…
if we know the value of K, we can predict: tendency of a reaction to occur if a set of concentrations could be at
equilibriumequilibrium position, given initial
concentrations If you start a reaction with only reactants:
concentration of reactants will decrease by a certain amount
concentration of products will increase by a same amount
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Using this we can make ICE charts
The following reaction has a K of 16. You are starting reaction with 9 O3 molecules and 12 CO molecules.
Find the amount of each species at equilibrium.
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Consider the following reaction at 600ºC 2SO2(g) + O2(g) 2SO3(g)
In a certain experiment 2.00 mol of SO2,
1.50 mol of O2 and 3.00 mol of SO3 were
placed in a 1.00 L flask. At equilibrium 3.50 mol were found to be present. Calculate
The equilibrium concentrations of O2 and
SO2, K and KP
Practice 1 ICE Charts
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Consider the same reaction at 600ºCIn a different experiment .500 mol SO2
and .350 mol SO3 were placed in a 1.000 L container. When the system reaches equilibrium 0.045 mol of O2 are present.
Calculate the final concentrations of SO2 and SO3 and K
Practice Problem 2 ICE Charts
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The Reaction Quotient (Q)Tells you the directing the reaction
will go to reach equilibriumCalculated the same as the
equilibrium constant, but for a system not at equilibrium
Q = [Products]coefficient
[Reactants] coefficient
Compare value to equilibrium constant
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What Q tells us IF THEN
Q = K reaction is at equilibrium
Q > K too much products,
left shift
Q < K too much reactants,
right shift
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Example 1 Reaction QuotientFor the synthesis of ammonia at 500°C,
the equilibrium constant is 6.0 x 10-2. Predict the direction the system will shift to reach equilibrium in the following case:
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Ex 1 Cont.
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Example 2In the gas phase, dinitrogen tetroxidedecomposes to gaseous nitrogen dioxide:
Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a T where KP = 0.133. At equilibrium, the pressure of N2O4 was found to be 2.71 atm. Calculate the equilibrium pressure of NO2.
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Example 3
At a certain temperature a 1.00 L flask initially contained 0.298 mol PCl3(g) and 8.70x10-3 mol PCl5(g). After the system had reached equilibrium, 2.00x10-3 mol Cl2(g) was found in the flask.
PCl5(g) PCl3(g) + Cl2(g)
Calculate the equilibrium concentrations of all the species and the value of K.
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Ex 3 Cont
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Approximations If K is very small, we can assume that the
change (x) is going to be negligible compared to the initial concentration of the substances
can be used to cancel out when adding or subtracting from a “normal” sized number to simplify algebra
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Example 4
At 35°C, K=1.6x10-5 for the reaction 2NOCl(g) 2NO(g) + Cl⇄ 2(g)
Calculate the concentration of all species at equilibrium for the following mixtures
2.0 mol NOCl in 2.0 L flask
1.0 mol NOCl and 1.0 mol NO in 1.0 L flask
2.0 mol NO and 1.0 mol Cl2 in 1.0 L flask
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Ex 4 Cont
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Ex 4 Cont
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Ex 4 Cont
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Le Chatelier’s Principle can predict how certain changes or
stresses put on a reaction will affect the position of equilibrium
helps us determine which direction the reaction will progress in to achieve equilibrium again
system will shift away from the added component or towards a removed component
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Change Concentration equilibrium position can change but not K system will shift away from the added
component or towards a removed component
Ex: N2 + 3H2 2NH3
if more N2 is added, then equilibrium position shifts to right (creates more products)
if some NH3 is removed, then equilibrium position shifts to right (creates more products)
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Adding Gasadding or removing gaseous reactant
or product is same as changing concentration
adding inert or uninvolved gas increase the total pressure doesn’t effect the equilibrium position
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Change the Pressure by changing the Volume
only important in gaseous reactions decrease V
requires a decrease in # gas moleculesshifts towards the side of the reaction with
less gas molecules increase V
requires an increase in # of gas moleculesshifts towards the side of the reaction with
more gas molecules
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Change in Temperature all other changes alter the concentrations
at equilibrium but don’t actually change value of K
value of K does change with temperature if energy is added, the reaction will shift in
direction that consumes energy treat energy as a
reactant: for endothermic reactionsproduct: for exothermic reactions
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CatalystThe use of a catalyst may speed up
a reaction, but it speeds it up in both the forward and reverse direction, therefore a catalyst has no effect on the equilibrium state of the system.
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energy + N2(g) + O2(g) 2NO(g)⇄ endo or exo?
endothermic increase temp
to right remove O2
to left increase volume
no shift add N2
to right
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Le Chatlier’s Simulationhttp://www.learnerstv.com/
animation/animation.php?ani=120&cat=chemistry
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Solution Equilibria
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Another Dynamic EquilibriumEquilibrium occurs when the solution is
saturated
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Ksp
The solubility product constant (Ksp) is similar to the Keq.
When a solid is added to water, some dissolves (and splits into ions) while some remains a solid.
Ex: NaCl(s) Na+(aq) + Cl-(aq)
The mass action expression for this is:
Ksp = [Na+][Cl-]
(remember… solids are 1)
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Common Ion EffectThe solubility of a solid is lowered if
the solution already contains ions common to the solidDissolving silver chloride in a
solution containing silver ionsDissolving silver chloride in a
solution containing chloride ions
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PrecipitationOpposite of dissolution Can predict whether precipitation or
dissolution will occurUse Q: ion product
Equals Ksp expression but doesn’t have to be at equilibrium
Q > K: more reactant will form, precipitation until equilibrium reached
Q < K: more product will form, dissolution until equilibrium reached
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Example 1
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Example 1 Cont.
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Example 2
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Example 2 Cont
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Example 2 Cont
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Example 2 Cont
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Qualitative AnalysisProcess used to separate a solution
containing different ions using solubilities
Example Problem
A solution of 1.0x10-4 M Cu+ and 2.0x10-3 M Pb2+. If I- is gradually added, which will precipitate out first, CuI or PbI2?
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Solution to Example Problem
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ExampleThe molar solubility for MgCl2 is
0.0056M. Calculate Ksp
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Salt SimulatorIn groups of two use the simulator to
discover the Ksp and Le Chatlier’s Principle
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Acid-base Equilibria
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Solutions of Acids or Bases Containing a Common Ion
Common Ion Ion provided in solution by an aqueous acid (or base) as
well as a salta. HF(aq) and NaF (F- in common)
b. HF(aq) H+(aq) + F-(aq) Excess F- added by NaF
Equilibrium shifts away from added component. Fewer H+ ions present.
pH is higher than expected.
a. NH4OH and NH4Cl (NH4+ in common)
b. NH3(aq) + H2O(l) NH4 +(aq) + OH-(aq) Equilibrium shifts to the left. pH of the solution decreases due to a decrease in OH- concentration
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Equilibrium CalculationsConsider initial concentration of ion
from salt when calculating values for H+ and OH-
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Acid + Base Salt + Water
Acid + Base Conjugate Base + Conjugate Acid
Some solvents are amphiproticWater can act as an acid and a base!Methanol can act as an acid and a base!
AutoprotolysisSome solvents can react with themselves to
produce an acid and a base• Water is a classical example
Weak acids dissociate partially, weak bases undergo partial hydrolysis. Strong acids and bases are strong electrolytes.
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Kw (Dissociation of Water) Water is amphiprotic it also undergoes autoprotolysis
Kw = 1.0E-14 at about 25 ˚C This is where the pH scale we commonly use originates
from!
What is the concentration of hydronium and hydroxide ions in neutral solution? What is the pH? What is the pOH?
H2OBase
+ H2OAcid
Kw
H3OConjugate Acid
+ + OH-
Conjugate Base
Kw [H3O] x [OH- ]
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Weak Acid & Weak Base Equilibria
Weak acids produce weak conjugate bases, and weak bases produce weak conjugate acids
Ka is a “special” equilibrium constant for the dissociation of a weak acid (found in standard tables)
Kb is a “special” equilibrium constant for the hydrolysis (or dissociation” of a weak base.
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HA(aq) + H2O(l) H3O(aq)+ + A (aq)
-
Ka = [H3O(aq)
+ ] x [A(aq)- ]
[HA(aq) ]
NH3(aq) + H2O(l) NH4(aq)+ + OH(aq)
-
Kb = [NH4(aq)
+ ] x [OH(aq)- ]
[NH3(aq) ]
FOR ANY Conjugate Acid -Base Pair in Aqueous Solution
Ka x Kb = Kw
Only Ka values appear in most standardized tables.
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Calculations…….. What is the pH of a 1.0 M solution of acetic acid
(HAc)?
What assumption can you make? If [acid] is about 1000 times the Ka value, it’s
concentration in solution won’t change much!Use an “I-C-E” table to look at this.The text goes into a more elaborate discussion
of approximations. I will allow approximations if the concentrations or pH values do not change in the hundred’s decimal place.
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What is the pH of a 4.0 M solution of phosphate ion?Write reactionCalculate KbSetup “I-C-E” tableMake assumptionsSolve algebraically.
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BuffersBuffers resist the change in pH
because they have acid to neutralize bases and bases to neutralize acids.
Made from a weak acid (HA) and the salt of its conjugate base (A-, where the counter ion is gone for example), or a weak base and the salt of its conjugate acid.
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Features of BuffersBuffers work best at maintaining pH
near the Ka of the acid component, usually about +/- 1 pH unit. This is their buffer capacity .
Buffers resist pH changes due to dilution.
All seen when we use the “Buffer Equation”
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Henderson-Hasselbalch (Buffer) Equation A modification of the equation for the dissociation of a
weak acid.
The pH is the pH of the buffered solution, pKa is the pKa of the weak acid.
What is the pH of a buffer solution made from 1.0 M acetic acid and 0.9 M sodium acetate?
You add .10 moles of sodium hydroxide to the above solution? What is the new pH?
pH = pKa + log A - HA
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H-H Equation & Buffers….
If [A-] = [HA] pH = pKa!This is what we see at half-way to
the equivalence point in the titration of a weak acid with a strong base!
Dilution does not change the ratio of A- to HA, and thus the pH does not change significantly in most cases
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How do you prepare buffers? Select a buffer ‘system’ based on the pH you want to maintain. Use the H-H equation to calculate how much acid and
conjugate base you need Take one of three approaches:
Mix acid and base forms, measure pH and adjust with strong acid or strong base
Start with a solution of weak acid, and add strong base until you reach the desired pH
Start with a solution weak base and add strong acid until you reach the desired pH
REMEMBER, STRONG Acids or Bases react completely! Dilute as necessary, adjust pH further if needed with strong
acid or base.
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You want 1L of a buffer system that has a pH of 3.90?What acid/conjugate base pair
would you use?How would you go about figuring
out how much of each reagent you might need?
How would you prepare and adjust the pH of this solution?