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LP# 9. EXAMPLE (using Approximation Method):
For the reaction, K = 1.0 x 10-5 at 1500 K.
N2(g) + O2(g)  2 NO(g)
If 4.0 mol N2 and 1.0 mol O2 are mixed in a 5.0 L container, what are all
equilibrium concentrations?
I
C
E
N2(g)
O2(g)

2 NO (g)
Kc =
 We didn’t end up with a perfect square this time.
 We could factor and use the quadratic equation, but that is lots of work.
 Sometimes we can make an approximation that will get us very close to the
correct answer with a lot less work!
Let’s assume that x is ______________ IN COMPARISON TO
the thing it is added to or subtracted from. ____________
 If this is true we can ignore it’s impact on that number. This would simplify the
equation to:
Kc =
Always check the validity of the approximation to see if X was really less than 5% of
smallest value it was added to or subtracted from.
Was x really small enough to ignore?
 Where would the biggest error come in?
 What percent of 0.20 is X?
When is the approximation likely to work for us?
 When Kc is very large (>103)
or
When Kc is very small (<10-3)
EQUILIBRIUM 14-13
Qualitatively Predicting the Direction of a Reaction Shift:
Le Châtelier's Principle:
(Note: This is a good time to talk about LeChatlier Pre-lab write-up.)
Definition: If a stress is applied to a reaction mixture at equilibrium, a shift in the position of
the equilibrium occurs in the direction that relieves the stress.
1. It predicts changes in the composition of an equilibrium mixture if a system at
equilibrium is stressed.
Stresses that alter the composition of an equilibrium mixture:
1. __________________ of reactants or products. (for aqueous)
2. __________________ and volume. (for gasses)
3. __________________ (for exothermic & endothermic reactions)
Le Châtelier's Principle: Changes in Concentration
1. Stress: added reactant or product is relieved by the reaction shifting in the
direction that consumes the added substance.
a. add reactant - reaction shifts _______________
b. add product - reaction shifts _______________
2. Stress: removed reactant or product is relieved by the reaction shifting in the
direction that replenishes the removed substance.
a. remove reactant - reaction shifts _______________
b. remove product - reaction shifts _______________
Question for class: How do you “remove” a substance?
Ans:
Summary
 If reactant or product is added to a system at equilibrium,
the reaction will shift in a direction away from that reactant or
product.
 If reactant or product is removed from a system at equilibrium,
the reaction will shift in a direction toward that reactant or
product.
Rationale: Changes occur due to change in value of ____
EQUILIBRIUM 14-14
1. Add reactant - denominator in Qc expression becomes __________.
a. Qc < Kc
b. to return to equilibrium, Qc must increase
i. numerator of Qc must ____ and the denominator must ____
ii. reactants convert to products
iii. reaction ________________
2. Remove reactant - denominator in Qc expression becomes ________.
a. Qc > Kc
b. to return to equilibrium, Qc must decrease
i. numerator of Qc must ____ and the denominator must ____
ii. products convert to reactants
iii. reaction _________________
3. Add product – numerator in Qc expression becomes bigger
Reaction shifts left to reduce Qc.
4. Remove product - numerator in Qc expression becomes smaller
Reaction shifts right to increase Qc.
Question: what would happen if we increased the concentration of CO in the
following reaction.
We started at equilibrium.
When CO was added we had too many reactants compared to products to keep the ratio
of K constant.
Some of the reactants had to shift to the right to reduce the amount of reactants and
increase the amount of products to re-establish our constant ratio.
When the CO conc. came down, the H2 concentration had to fall as well.
The final concentration on CO is still larger than it started out.
EQUILIBRIUM 14-15
Question: What would happen if we increased the concentration of CH3OH instead?
Le Châtelier's Principle: Changes in Pressure and Volume
A. #moles of gaseous reactants  # moles of gaseous products.
1. Change in pressure (due to changing volume) changes composition of
equilibrium mixture.

Increase in pressure (due to decrease in volume) results in reaction in the
direction of ____________number of moles of gas.

Decrease in pressure (due to increase in volume) results in reaction in the
direction of _____________ number of moles of gas.
Rationale: due to change in value of Qc. (Shifts to relieve stress[overcrowding])
1. Decrease volume - molarity (= n/V) increases (& pressure increases).
2. If reactant side has more moles of gas.
Qc = [products]/[reactants]
a. increase in denominator is greater than increase in numerator
b. Qc < Kc
c. to return to equilibrium, Qc must increase
i. numerator of Qc expression must  and the denominator must 
ii. implies net conversion of ______________
(shifts towards fewer moles of gas)
3. If product side has more moles of gas.
a. increase in numerator is greater than increase in denominator
b. Qc > Kc
c. to return to equilibrium, Qc must decrease
i. denominator of Qc expression must  and the numerator must .
ii. implies net conversion of ______________
(shifts towards fewer moles of gas)
EQUILIBRIUM 14-16
B Reaction involves no change in the number of moles of gas.
1.
____________ on composition of equilibrium mixture.
C. For heterogeneous equilibrium.
1. Effect of pressure changes on solids and liquids can be ________.
a. volume is nearly independent of pressure
b. base the decision on gas constituents
D. Change in pressure due to addition of an inert gas.
1. No change in the molar concentrations of reactants or products.
(i.e., partial pressures don’t change)
2.
__________ on composition of equilibrium mixture.
LeChatlier’s Principle: Changes in Temperature
A. Affects Exothermic or Endothermic Reactions
This is an intrinsic part of the nature of the reaction.
 An exothermic reaction is exothermic, no matter what the temperature is.
 An endothermic reaction is endothermic at all temperatures.
 You should think of heat as
 ________________for endothermic reactions and as
 ________________ for exothermic reactions.
B. Le Châtelier's principle - add heat to an equilibrium mixture, net reaction
occurs in the direction that relieves the stress of the added heat.
Changes in temperature should be considered the ___________.
Changes in temperature should change the ________________ heat.
Changes in temperature DO NOT change which side of the equation the word heat ends
up on. (You cannot change an endothermic reaction into an exothermic one by changing
the temperature!)
EQUILIBRIUM 14-17
LP# 10.The reaction
 2 NO(g)
N2(g) + O2(g)
is endothermic.
In what direction will the reaction shift if the temperature is increased?
If the reaction is endothermic, heat is a ____________.
Just like with any other reaction, if we increase a reactant, the equilibrium
shifts to the ____________.
Example:
N2(g) + O2(g) + heat
 2 NO(g)
LP# 11. The reaction N2(g) + 3 H2(g)  2 NH3(g) has a negative value of H.
In what direction will the reaction shift if the temperature is increased?
If the value of H is negative, the reaction is ______________.
If a reaction is exothermic, it will NOT be favored by an increase in temperature. It will
shift to the left.
Example:
N2(g) + 3 H2(g)

2 NH3(g) + heat
C. Change in temperature always changes equilibrium constant.
1. Exothermic reaction (-Ho) - equilibrium constant __________
as the temperature increases.


A + B  C + D + heat
K = [products] 
[reactants] 
Therefore K
2. Endothermic reaction (+Ho) - equilibrium constant _________
as the temperature increases.
 
A + B + heat  C + D
K = [products] 
[reactants] 
Therefore K
EQUILIBRIUM 14-18
LP# 12. EXAMPLE:
How will the following changes alter the equilibrium for the reaction:
Ho = -150 kJ
3 Fe(s) + 4 H2O(g)  Fe3O4(s) + 4 H2(g)
SHIFT
a.
b.
c.
d.
e.
H2O is removed from the system.
H2 is removed from the system.
The volume of the container is increased.
Fe3O4 is added to the system.
The temperature is raised.
The Effect of a Catalyst
A. Catalyst increases the rate of a chemical reaction.
1. Provides a new, lower energy pathway.
2. Rate for forward and reverse reactions increases by the same factor
3. Does not affect the composition of the equilibrium mixture or Kc.
4. Equilibrium is reached faster.
Quantitative Evaluation of Equilibrium Shifts
LeChatlier’s Principle allows us to predict the direction that the equilibrium will shift, but
not the final concentrations. We will need ICE table calculations for that.
LP# 13. EXAMPLE:
For the following reaction at 500 K.
H2(g) + I2(g)  2 HI(g)
An equilibrium mixture in a 1.0L container is found to have the following
concentrations:
[H2]= 0.080M; [I2]=0.060M; [HI]=0.490M.
a) Calculate the value of Kc.
Kc =
EQUILIBRIUM 14-19
b) An additional 0.300 mol of HI is then added. (Ignore any volume change.)
What concentrations will be present when the new equilibrium is reached?
Step 1: What will the immediate [HI] be after addition but before the shift?

Step 2: Use LeChatlier’s Principle to determine the direction of the shift.

Since we added a ___________, we will shift to the _______
 If not sure of shift direction, calculate Q and compare to K.
Q=
Step 3: Set up an ICE table to help calculate new equilibrium concentrations.
PAY SPECIAL ATTENTION TO THE SIGN OF X!!!
I
C
E
H2(g)
I2(g)
0.24 + 7.0x + 50x2 = 0.624 – 3.16x + 4x2
46x2 + 10.2x – 0.38 = 0
Using Quadratic equation, x = _______
Equilibrium Concentrations:
[H2] = (0.080 + x) =
[I] = (0.060 + x) =
[HI] = (0.790 – 2x) =
EQUILIBRIUM 14-20

2 HI (g)
(this shows
the algebra)
DON’T STOP HERE!
Equilibrium Constant Variation with Temperature
The value of the equilibrium constant changes with temperature.
We can usually only find the value of K for any reaction at 1 temperature in the appendix.
If we know the value of Hrxn for the reaction and any single value of K at one
temperature, we can calculate K at any other temperature using:
K
ln  2
 K1
  H   1 1
 
 
R  T2 T1




This is the _____________equation.
(Whitten has (+)H and T values reversed.)
Note the similarity to the relationship between the rate constant k and the activation
energy, Ea, that we learned in the Kinetics chapter:
k
ln  2
 k1
  Ea  1
1
  
 
R  T2 T1 

This is the ______________ equation.
The Link Between Chemical Equilibrium and Chemical Kinetics
A. For the reaction: A + B  C + D.
1. Rate of forward reaction = kf[A][B].
2. Rate of reverse reaction = kr[C][D].
3. At equilibrium: Rate of forward Rxn = __________ .
k f C D
4. At equilibrium: kf[A][B] = kr[C][D] or k  A B   K c .
r
5. The equilibrium constant is equal to the ratio of the rate constant for the
forward reaction to the rate constant for the reverse reaction.
B. Relative values of kf and kr determine the composition of the equilibrium mixture.
1. kf >> kr; Kc is very large; reaction goes to completion.
2. kf  kr; Kc  1; reactants and products are present at equilibrium.
(Note we are comparing rate constants, not rates!)
EQUILIBRIUM 14-21
LP# 14. EXAMPLE:
Write the equilibrium constant expression and solve for the value of the constant Kc
for the reaction:
H2O (l)  H+ (aq) + OH- (aq)
-5
-1
9
-1 -1
given that kf = 2.4  10 s and kr = 2.4  10 M s
Kc =
The Relationship Between Equilibrium and Thermodynamics (Spontaneity)
 At Equilibrium, Q=____ and Grxn= ____.
Rxn is neither spontaneous nor non-spontaneous
 For the Rxn to be spontaneous in the forward direction:
Q<K and Grxn<0.
 For the Rxn to be spontaneous in the reverse direction:
Q>K and Grxn>0.
In Chapter 17, we noted that Grxn and not ______ truly predicts spontaneity.
(Grxn assumes that ALL reactants are completely converted to products.)
The formula that relates the Gibbs free energy for the reaction (Grxn) to the
standard Gibbs free energy (Grxn) is:
Grxn = Grxn + RT ln Q
(where R = 8.314 J/mol K)
At equilibrium:
Grxn = 0 and
Q=K
so; at equilibrium this becomes:
0 = Grxn + RT ln K
Grxn = -RT ln K
EQUILIBRIUM 14-22
LP# 15. EXAMPLE:
Determine the value of K at 100.C for the reaction
I2(g) + Cl2(g)  2 ICl(g)
Given H = -26.9 kJ and S = 11.3 J/K.
(Assume H and S are approximately constant with temp.)
G = H - TS
G = -RT ln K
ln K = G =
-RT
Calculating K from Hrxn and Srxn
The previous equation: Grxn = -RT ln K
Can be rearranged to solve for K and
substitution using the Gibbs Free Energy equation
to get:
0
o
H rxn
 1  S rxn
ln K  
 
R T 
R
This equation has already been linearized.
As well as being useful in solving for K,
This shows that if we make a plot of ln K on the y-axis and 1/T on the x-axis the slope is
_________and the intercept is__________.
EQUILIBRIUM 14-23