1. Chemical Equilibrium Chapter 14
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2. Physical equilibrium Chemical equilibrium
Equilibrium is a state in which there are no observable changes as time goes by.
Chemical equilibrium is achieved when:
the rates of the forward and reverse reactions are equal and
the concentrations of the reactants and products remain constant
Physical equilibrium
H2O (l)
H2O (g)
Chemical equilibrium
N2O4 (g)
2NO2 (g)
14.1

3. N2O4 (g) 2NO2 (g) equilibrium equilibrium equilibrium Start with NO2
Start with NO2 & N2O4
14.1

4. constant
14.1

5. Chemical reactions often seem to stop before they are complete.
Actually, such reactions are reversible. That is, the original reactants form products, but then the products react with themselves to give back the original reactants.
When these two reactions—forward and reverse—occur at the same rate, a chemical equilibrium exists.

6. CO(g) + 3H2(g) CH4(g) + H2O(g)
The graph shows how the amounts of reactants and products change as the reaction approaches equilibrium.

7. CO(g) + 3H2(g) CH4(g) + H2O(g)
This graph shows how the rates of the forward reaction and the reverse reaction change as the reaction approaches equilibrium.

8. Chemical equilibrium is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal.
We can apply stoichiometry to compute the content of the reaction mixture at equilibrium.

9. When heated PCl5, phosphorus pentachloride, forms PCl3 and Cl2 as follows:
PCl5(g) PCl3(g) + Cl2(g)
When 1.00 mol PCl5 in a 1.00-L container is allowed to come to equilibrium at a given temperature, the mixture is found to contain mol PCl3. What is the molar composition of the mixture?

10. We will organize this problem by using the chemical reaction to set up a table of initial, change, and equilibrium amounts.
Initially we had 1.00 mol PCl5 and no PCl3 or Cl2.
The change in each is stoichiometric: If x moles of PCl5 react, then x moles of PCl3 and x moles of Cl2 are produced.
For reactants, this amount is subtracted from the original amount; for products, it is added to the original amount.

11. We can now find the amounts of the other substances.
PCl5(g)
PCl3(g) +
Cl2(g)
Initial
1.00 mol
Change –x +x
Equilibrium
1.00 – x x
We were told that the equilibrium amount of PCl3 is mol. That means x = mol.
We can now find the amounts of the other substances.

12. (given with 3 significant figures)
Moles PCl5 = 1.00 – = 0.87 mol
(2 decimal places)
Moles PCl3 = mol
(given with 3 significant figures)
Moles Cl2 = mol

13. The Equilibrium Constant, Kc
The equilibrium constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration term to a power equal to its coefficient in the balanced chemical equation.
The equilibrium constant, Kc, is the value obtained for the Kc expression when equilibrium concentrations are substituted.

14. N2O4 (g) 2NO2 (g) K = [NO2]2 [N2O4] = 4.63 x 10-3 aA + bB cC + dD K =
14.1

15. Equilibrium Will K = [C]c[D]d [A]a[B]b aA + bB cC + dD K >> 1
Lie to the right
Favor products
K << 1
Lie to the left
Favor reactants
14.1

16. Methanol (also called wood alcohol) is made commercially by hydrogenation of carbon monoxide at elevated temperature and pressure in the presence of a catalyst:
2H2(g) + CO(g) CH3OH(g)
What is the Kc expression for this reaction?

17. When Kc is very large (>102), the equilibrium mixture is mostly products.
When Kc is very small (<10-2), the equilibrium mixture is mostly reactants.
When Kc approaches 1, the equilibrium mixture contains appreciable amounts of both reactants and products.

18. Carbon dioxide decomposes at elevated temperatures to carbon monoxide and oxygen:
2CO2(g) CO(g) + O2(g)
At 3000 K, 2.00 mol CO2 is placed into a 1.00-L container and allowed to come to equilibrium. At equilibrium, 0.90 mol CO2 remains.
What is the value of Kc at this temperature?

19. We can find the value of x. 2.00 – 2x = 0.90 1.10 = 2x x = 0.55 mol
2CO2(g)
2CO(g) +
O2(g)
Initial
2.00 mol
Change
–2x
+2x +x
Equilibrium
2.00 – 2x 2x x
0.90 mol
1.10 mol
0.55 mol
We can find the value of x.
2.00 – 2x = 0.90
1.10 = 2x
x = 0.55 mol
2x = 2(0.55) = 1.10 mol

20. 2CO2(g) CO(g) + O2(g)
In the third and fourth lines, two of the [CO] entries should be changed to become [CO2] and [O2]. Which ones?
Corrected.

21. In a heterogeneous equilibrium, in the Kc expression, the concentrations of solids and pure liquids are constant (due to these substances’ constant density).
As a result, we incorporate those constants into the value of Kc, thereby making a new constant, Kc. In other words, equilibrium is not affected by solids and pure liquids as long as some of each is present.
More simply, we write the Kc expression by replacing the concentration of a solid or pure liquid with 1.

22. H2O(g) + C(s) CO(g) + H2(g)
Write the Kc expression for the following reaction:
H2O(g) + C(s) CO(g) + H2(g)

23. Given:
aA + bB cC + dD; K1
When the reaction is reversed:
cC + dD aA + bB; K2
The equilibrium constant expression is inverted:
K2 =

24. Given:
aA + bB cC + dD; K1
When the reaction is doubled:
2aA + 2bB cC + 2dD; K2
The equilibrium constant expression, K2 , is the square of the equilibrium constant expression, K1:
K2 =

25. aA(g) + bB(g) cC(g) + dD(g)
For the reaction
aA(g) + bB(g) cC(g) + dD(g)
The equilibrium constant expressions are
Kc = and Kp =

26. How are these related?
We know
From the ideal gas law, we know that
So,

27. When you express an equilibrium constant for a gaseous reaction in terms of partial pressures, you call it the equilibrium constant, Kp.
In general, the value of Kp is different from that of Kc.
We will explore this relationship on the next slides. Recall the ideal gas law and the relationship between pressure and molarity of a gas:

28. Kp = Kc (RT)Dn

29. CO(g) + 3H2(g) CH4(g) + H2O(g)
For catalytic methanation,
CO(g) + 3H2(g) CH4(g) + H2O(g)
the equilibrium expression in terms of partial pressures becomes
and

30. The value of Kc at 227°C is 0.0952 for the following reaction:
CH3OH(g) CO(g) + 2H2(g)
What is Kp at this temperature?
Kp = (RT)Dn
where
T = = 500. K
R = L  atm/(mol  K)
Dn = 2
Kp = 1.60 × 102

31. Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.
N2O4 (g) NO2 (g)
Kp =
NO2 P2
N2O4 P
Kc =
[NO2]2
[N2O4]
In most cases
Kc  Kp
aA (g) + bB (g) cC (g) + dD (g)
Kp = Kc(RT)Dn
Dn = moles of gaseous products – moles of gaseous reactants
= (c + d) – (a + b)
14.2

32. Homogeneous Equilibrium
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)
[CH3COO-][H3O+]
[CH3COOH][H2O]
Kc = ‘
[H2O] = constant
[CH3COO-][H3O+]
[CH3COOH] =
Kc [H2O] ‘
Kc =
General practice not to include units for the equilibrium constant.
14.2

33. 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

34. 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

35. 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

36. does not depend on the amount of CaCO3 or CaO
CaCO3 (s) CaO (s) + CO2 (g)
PCO 2
= Kp
PCO 2
does not depend on the amount of CaCO3 or CaO
14.2

37. Consider the following equilibrium at 295 K:
The partial pressure of each gas is atm. Calculate Kp and Kc for the reaction?
NH4HS (s) NH3 (g) + H2S (g)
Kp = P
NH3
H2S P
= x =
Kp = Kc(RT)Dn
Kc = Kp(RT)-Dn
Dn = 2 – 0 = 2
T = 295 K
Kc = x ( x 295)-2 = 1.20 x 10-4
14.2

38. ‘ ‘ ‘ ‘ ‘ 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

39. ‘ N2O4 (g) 2NO2 (g) 2NO2 (g) N2O4 (g) = 4.63 x 10-3 K = [NO2]2 [N2O4]
= 216
When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.
14.2

40. Writing Equilibrium Constant Expressions
The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm.
The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions.
The equilibrium constant is a dimensionless quantity.
In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.
If a reaction can be expressed as a 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

41. Chemical Kinetics and Chemical Equilibrium
ratef = kf [A][B]2
A + 2B AB2 kf kr
rater = kr [AB2]
Equilibrium
ratef = rater
kf [A][B]2 = kr [AB2] kf kr
[AB2]
[A][B]2 =
Kc =
14.3

42. Qc > Kc system proceeds from right to left to reach equilibrium
The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression. IF
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

43. Given:
H2(g) + F2(g) HF(g); Kc = 1.15 × 102
3.000 mol of each species is put in a L vessel. What is the equilibrium concentration of each species?
First calculate the initial concentrations:

44. H2(g) + F2(g) 2HF(g) Initial 2.000 M Change +2x Equilibrium 2.000 – x

46. Now compute the equilibrium concentrations:
Double-check by substituting these equilibrium concentrations into the Kc expression and solving. The answer should be the value of Kc.
This is within round-off error.

47. Calculating Equilibrium Concentrations
Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.
Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.
Having solved for x, calculate the equilibrium concentrations of all species.
14.4

48. 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

49. 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

50. Le Châtelier’s Principle
If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.
Changes in Concentration
N2 (g) + 3H2 (g) NH3 (g)
Add
NH3
Equilibrium shifts left to offset stress
14.5

51. Le Châtelier’s Principle
When a system in chemical equilibrium is disturbed by a change in
temperature,
pressure, or
concentration,
the system shifts in equilibrium composition in a way that tends to counteract this change of variable.

52. When a substance that is part of the equilibrium is added to the mixture, the equilibrium shifts to use it (in a direction that makes the substance a reactant).
When a substance that is part of the equilibrium is removed from the mixture, the equilibrium shifts to produce it (in a direction that makes the substance a product).

53. 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

54. 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

55. Le Châtelier’s Principle
Changes in Temperature
Change
Exothermic Rx
Endothermic Rx
Increase temperature
K decreases
K increases
Decrease temperature
K increases
K decreases
colder
hotter
14.5

56. 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 lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift equilibrium.
14.5

57. Life at High Altitudes and Hemoglobin Production
Chemistry In Action
Life at High Altitudes and Hemoglobin Production
Hb (aq) + O2 (aq) HbO2 (aq)
Kc =
[HbO2]
[Hb][O2]

58. Chemistry In Action: The Haber Process
N2 (g) + 3H2 (g) NH3 (g) DH0 = kJ/mol

59. Le Châtelier’s Principle
Change Equilibrium
Constant
Change
Shift Equilibrium
Concentration
yes no
Pressure
yes no
Volume
yes no
Temperature
yes
yes
Catalyst no no
14.5