1. Chapter 15 Chemical Equilibrium
Lecture Presentation
Chapter 15 Chemical Equilibrium
© 2012 Pearson Education, Inc.

2. The Concept of Equilibrium
Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.
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3. The Concept of Equilibrium
As a system approaches equilibrium, both the forward and reverse reactions are occurring.
At equilibrium, the forward and reverse reactions are proceeding at the same rate.
Once equilibrium is achieved, the amount of each reactant and product remains constant
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4. The Equilibrium Constant
Consider the generalized reaction
aA + bB
cC + dD
The equilibrium expression for this reaction would be
Kc =
[C]c[D]d
[A]a[B]b
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5. The Equilibrium Constant
N2O4(g)
2NO2(g)
Keq =
[NO2]2
[N2O4]
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6. The Equilibrium Constant
Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written
Kp =
(PCc) (PDd)
(PAa) (PBb)
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7. The equilibrium constant Kp for the reaction
is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = atm and PNO = atm?
2NO2 (g) NO (g) + O2 (g) 2
Kp = 2
PNO PO
PNO PO 2
= Kp
PNO PO 2
= 158 x (0.400)2/(0.270)2
= 347 atm
14.2

8. Relationship Between Kc and Kp
Kp = Kc (RT)n
where
n = (moles of gaseous product)  (moles of gaseous reactant)
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9. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = M, [Cl2] = M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g) COCl2 (g)
[COCl2]
[CO][Cl2] =
0.14
0.012 x 0.054
Kc =
= 220
Kp = Kc(RT)Dn
Dn = 1 – 2 = -1
R =
T = = 347 K
Kp = 220 x ( x 347)-1 = 7.7
14.2

10. Equilibrium Can Be Reached from Either Direction
It doesn’t matter whether we start with N2 and H2 or whether we start with NH3, we will have the same proportions of all three substances at equilibrium.
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11. What Does the Value of K Mean?
If K>>1, the reaction is product-favored; product predominates at equilibrium.
If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium.
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12. Manipulating Equilibrium Constants
The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction:
Kc = = at 100 C
[NO2]2
[N2O4]
N2O4(g)
2NO2(g)
Kc = = 4.72 at 100 C
[N2O4]
[NO2]2
N2O4(g)
2NO2(g)
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13. Manipulating Equilibrium Constants
The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number:
Kc = = at 100 C
[NO2]2
[N2O4]
N2O4(g)
2NO2(g)
Kc = = (0.212)2 at 100 C
[NO2]4
[N2O4]2
4NO2(g)
2N2O4(g)
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14. ‘ ‘ ‘ ‘ ‘ Kc = [C][D] [A][B] Kc = [E][F] [C][D] Kc A + B C + D Kc
C + D E + F
[E][F]
[A][B]
Kc =
A + B E + F Kc
Kc = Kc ‘ x
If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.
14.2

15. Sample Exercise 15.5 Combining Equilibrium Expressions
Given the reactions
determine the value of Kc for the reaction
Solution
Analyze We are given two equilibrium equations
and the corresponding equilibrium constants and are
asked to determine the equilibrium constant for a
third equation, which is related to the first two.
Plan We cannot simply add the first two equations to get the third. Instead, we need to determine how to manipulate the equations to come up with the steps that will add to give us the desired equation. 15

16. Sample Exercise 15.5 Combining Equilibrium Expressions
Continued
Solve
If we multiply the first equation by 2 and make the corresponding change to its equilibrium constant (raising to the power 2), we get
Reversing the second equation and again making the corresponding change to its equilibrium constant (taking the reciprocal) gives
Now we have two equations that sum to give the net equation, and we can multiply the individual Kc values to get the desired equilibrium constant. 16

17. Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.
CaCO3 (s) CaO (s) + CO2 (g)
Kc = ‘
[CaO][CO2]
[CaCO3]
[CaCO3] = constant
[CaO] = constant
Kc x ‘
[CaCO3]
[CaO]
Kc = [CO2] =
Kp = PCO 2
The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.
14.2

18. © 2012 Pearson Education, Inc.
Example
PbCl2(s)
Pb2+(aq) + 2Cl(aq)
Kc = [Pb2+] [Cl]2
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19. ICE At 12800C the equilibrium constant (Kc) for the reaction
Is 1.1 x If the initial concentrations are [Br2] = M and [Br] = M, calculate the concentrations of these species at equilibrium.
Br2 (g) Br (g)
Let x be the change in concentration of Br2
Br2 (g) Br (g)
Initial (M)
0.063
0.012
ICE
Change (M) -x
+2x
Equilibrium (M) x x
[Br]2
[Br2]
Kc =
Kc =
( x)2 x
= 1.1 x 10-3
Solve for x
14.4

20. Kc =
( x)2 x
= 1.1 x 10-3
4x x = – x
4x x = 0
-b ± b2 – 4ac  2a
x =
ax2 + bx + c =0
x =
x =
Br2 (g) Br (g)
Initial (M)
Change (M)
Equilibrium (M)
0.063
0.012 -x
+2x x x
At equilibrium, [Br] = x = M
or M
At equilibrium, [Br2] = – x = M
14.4

21. Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations
A closed system initially containing  103 M H2 and  103 M I2 at 448 C is allowed to reach equilibrium, and at equilibrium the HI concentration is 1.87  103 M. Calculate Kc at 448 C for the reaction taking place, which is
Solution
Analyze We are given the initial concentrations of H2 and l2 and the equilibrium concentration of HI. We are asked to calculate the equilibrium constant Kc for
Solve First, we tabulate the initial and
equilibrium concentrations of as many
species as we can. We also provide space
in our table for listing the changes in
concentrations. As shown, it is convenient
to use the chemical equation as the heading
for the table.
Second, we calculate the change in HI
concentration, which is the difference
Between the equilibrium and initial values:
Plan We construct a table to find equilibrium concentrations of all species and then use the equilibrium concentrations to calculate the equilibrium constant.
Change in [HI] = 1.87  103 M  0 = 1.87  103 M 21

22. Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations
Continued
Third, we use the coefficients in the balanced equation to relate the change in [HI] to the changes in [H2] and [I2]:
Fourth, we calculate the equilibrium concentrations of H2 and I2, using initial concentrations and changes in concentration. The equilibrium concentration equals the initial concentration minus that consumed:
Our table now looks like this (with equilibrium concentrations in blue for emphasis):
[H2] =  103 M   103 M =  103 M
[I2] =  103 M   103 M =  103 M
Notice that the entries for the changes are negative when a reactant is consumed and
positive when a product is formed. 22

23. Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations
Continued
Finally, we use the equilibrium-constant
expression to calculate the equilibrium
constant: 23

24. The Reaction Quotient (Q)
Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.
To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.
aA + bB
cC + dD
Q =
[Cinitial]c[Dinitial]d
[Ainitial]a[Binitial]b
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25. Qc > Kc system proceeds from right to left to reach equilibrium
Qc = Kc the system is at equilibrium
Qc < Kc system proceeds from left to right to reach equilibrium
14.4

26. Sample Exercise 15.10 Predicting the Direction of Approach to Equilibrium
At 448 C the equilibrium constant Kc for the reaction
is Predict in which direction the reaction proceeds to reach equilibrium if we start with 2.0  102 mol of HI,
1.0  102 mol of H2, and 3.0  102 of I2 in a 2.00-L container.
Solution
Analyze We are given a volume and initial molar amounts of the species in a reaction and asked to determine in which direction the reaction must proceed to achieve equilibrium.
Plan We can determine the starting concentration of each species in the reaction mixture. We can then substitute the starting concentrations into the equilibrium-constant expression to calculate the reaction quotient, Qc. Comparing the magnitudes of the equilibrium constant, which is given, and the reaction quotient will tell us in which direction the reaction will proceed.
Solve
The initial concentrations are
The reaction quotient is therefore
Because Qc < Kc, the concentration of HI must increase and the concentrations of H2 and I2 must decrease to reach
equilibrium; the reaction as written proceeds left to right to attain equilibrium. 26

27. Le Châtelier’s Principle
“If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”
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28. Le Châtelier’s Principle
Changes in Concentration continued
Remove
Add
Add
Remove
aA + bB cC + dD
Change
Shifts the Equilibrium
Increase concentration of product(s)
left
Decrease concentration of product(s)
right
Increase concentration of reactant(s)
right
Decrease concentration of reactant(s)
left
14.5

29. Le Châtelier’s Principle
Changes in Volume and Pressure
A (g) + B (g) C (g)
Change
Shifts the Equilibrium
Increase pressure
Side with fewest moles of gas
Decrease pressure
Side with most moles of gas
Increase volume
Side with most moles of gas
Decrease volume
Side with fewest moles of gas
14.5

30. Le Châtelier’s Principle
Adding a Catalyst
does not change K
does not shift the position of an equilibrium system
system will reach equilibrium sooner
uncatalyzed
catalyzed
Catalyst does not change equilibrium constant or shift equilibrium.
14.5