1. Chemical Equilibrium Chapter 14
PowerPoint Lecture Presentation by J. David Robertson University of Missouri
Chemical Equilibrium
Chapter 14
Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

2. Chemical equilibrium is achieved when:
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. EQUILIBRIUM EXPRESSION: Law of Mass Action
The Law of Mass Action states:
“the values of the equilibrium constant , K are constant for a particular reaction at a given temperature whatever equilibrium concentrations are substituted.”
EQUILIBRIUM CONSTANT, K:
“the constant that relates the concentration of reactants and products at equilibrium, raised to a power equal to the stoichiometric coefficient in the balanced equation. It can be expressed in terms of concentration, Kc or pressure, Kp.”

6. Equilibrium Will N2O4 (g) 2NO2 (g) K = [NO2]2 [N2O4] = 4.63 x 10-3
aA + bB cC + dD
K =
[C]c[D]d
[A]a[B]b
Law of Mass Action
Equilibrium Will
K >> 1
Lie to the right
Favor products
K << 1
Lie to the left
Favor reactants
14.1

7. Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.
N2O4 (g) NO2 (g)
Kp =
NO2
P 2
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

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

9. QUESTION #1: 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. Answer: QUESTION #2: The equilibrium constant Kp for the reaction:
2NO2 (g)  2NO (g) + O2 (g)
is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO2 = atm and PNO = atm?
Answer:
14.2

11. Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.
CaCO3 (s) CaO (s) + CO2 (g)
[CaCO3] = constant
[CaO] = constant
Kc = ‘
[CaO][CO2]
[CaCO3]
Kc x ‘
[CaCO3]
[CaO]
Kc = [CO2] =
Kp = PCO 2
The concentration of pure solids and pure liquids are excluded from the expression for the equilibrium constant because their concentrations “do not change”.
14.2

12. 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
The equilibrium pressure of CO2 is the same at the same temperature despite the presence of different amount of CaCO3 and CaO.
14.2

13. QUESTION #3: Consider the following equilibrium at 295 K:
NH4HS (s)  NH3 (g) + H2S (g)
The partial pressure of each gas is atm. Calculate Kp and Kc for the reaction?
Dn = 2 – 0 = 2
T = 295 K
14.2

14. 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

15. Manipulating the Equilibrium Constant Expressions
Kc = ‘
[C][D]
[A][B] Kc ‘
A + B C + D Kc ‘
C + D E + F
Kc = ‘
[E][F]
[C][D] Kc
A + B E + F
[E][F]
[A][B]
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

16. Manipulating the Equilibrium Constant Expressions
N2O4 (g) NO2 (g)
2NO2 (g) N2O4 (g)
= 4.63 x 10-3
K =
[NO2]2
[N2O4]
K =
[N2O4]
[NO2]2 ‘ = 1 K
= 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

17. Manipulating the Equilibrium Constant Expressions
N2O4 (g) NO2 (g)
½ N2O4 (g) NO2 (g)
= 4.63 x 10-3
K1 =
[NO2]2
[N2O4]
When the equation for a reversible reaction is halved, the new equilibrium constant is square root of the original constant.
14.2

18. Manipulating the Equilibrium Constant Expressions
N2O4 (g) NO2 (g)
2N2O4 (g) NO2 (g)
= 4.63 x 10-3
K1 =
[NO2]2
[N2O4]
When the reaction is doubled, the new equilibrium constant is square of the original constant.
Take note, if equilibrium constant for
A ⇌ 2 B is K1 , then for n A ⇌ 2n B,
the equilibrium constant K2 = (K1)n
14.2

19. USES OF EQUILIBRIUM CONSTANT, K:
To qualitatively interpreting the equilibrium.
To predict the direction of reactions – use reaction quotient, Qc or Qp. Then compare the values of Qc with Kc, or compare Qp with Kp.
To calculate equilibrium concentrations (ICE calculations).

20. The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression. Qc helps predict the reaction direction.
Qc > Kc : system proceeds from right to left to reach equilibrium
Qc = Kc : system is at equilibrium
Qc < Kc : system proceeds from left to right to reach equilibrium

21. Chemical Kinetics and Chemical Equilibrium
A + 2B AB2 kf kr
ratef = kf [A][B]2
rater = kr [AB2]
At equilibrium:
ratef = rater
kf [A][B]2 = kr [AB2] kf kr
[AB2]
[A][B]2 =
Kc =
14.3

22. Calculating Equilibrium Concentrations
(ICE Methods)
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

23. Question #4: ICE Calculation
Value of Kc is given.
At 12800C the equilibrium constant (Kc) for the reaction:
Br2 (g) ⇄ 2Br (g)
is 1.1 x If the initial concentrations are [Br2] = M and [Br] = M, calculate the concentrations of these species at equilibrium.
Values of initial concentration are given.
14.4

24. ⇄ Let x be the change in concentration: Br2(g) 2 Br (g) Initial (M)
Change (M)
Equilibrium (M)
0.063
0.012
 x
+ 2x
0.063  x x
Solve for x
14.4

25. Use quadratic equation:
Kc =
( x)2 x
= 1.1 x 10-3
4x x = – x
4x x = 0
Use quadratic equation:
x = – or x = –
If x = – , at equilibrium:
[Br] = x = – M
[Br2] = – x = M
If x = – , at equilibrium:
[Br] = x = M
[Br2] = – x = M
Concentration cannot be negative, so the answers above are not acceptable.
14.4

26. Question #5: ICE Calculation
In the water-gas shift reaction, a sample containing M CO2 and M H2 is allowed to reach equilibrium at 700 K. At this temperature, Kc = Determine the concentrations of each substance at equilibrium.
H2(g) + CO2 (g) ⇄ H2O (g) + CO (g)

27. Let x be the change in concentration:
H2 (g) +
CO2 (g) ⇄
H2O (g)
CO (g)
Initial
(M)
0.570
0.632
Change
Equilibrium (M)
 x
 x
+ x
+ x
+ x
+ x x x
Solve for x

28. Use quadratic equation:
x2 = (0.360 – 1.202x + x2)
0.894x x – = 0
Use quadratic equation:
x cannot be a negative value because x = [H2O] = [CO] , and concentration cannot be negative.
x = or – 0.290
[H2O] = [CO] = M
[H2] = M – M = M
[CO2] = M – M = M

29. Question #6: ICE Calculation
The reaction to form HF from molecular hydrogen and fluorine has Kc = 115 at a certain temperature. If 3.0 mol of each component is added to a 1.5 L flask, calculate the equilibrium concentrations of each species.
H2 (g) + F2 (g) ⇄ 2 HF(g)

30. Let x be the change in concentration:
H2 (g) +
F2 (g) ⇄
2 HF (g)
Initial (M)
Change (M)
Equilibrium
(M)
 x
 x
+2x
2.0  x
2.0  x x
Solve for x

31. Therefore, equilibrium concentration:
Taking the square root of each side:
 x = 1.53
Therefore, equilibrium concentration:
[H2] = [F2] = 2.0 M  1.53 M = 0.47 M
[HF] = 2.0 M + 2(1.53 M) = 5.06 M

32. Question #7: ICE Calculation
A 1.00-mol sample of NOCl was placed in a 2.00-L reactor and heated to 227oC until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain mol of Cl2. Calculate Kc for the reaction at this temperature.
2NOCl (g) ⇄ 2 NO (g) + Cl2 (g)

33. Let x be the change in concentration:
2NOCl (g) ⇄
2 NO (g) +
Cl2 (g)
Initial
(M)
Change (M)
Equilibrium (M) +x
 2x
+2x
+2x
0.500  2x +x
Info from the question: at equilibrium the reactor contain mol Cl2.

34. Put in value of x = 0.028 in the table:
2NOCl (g) ⇄
2 NO (g) +
Cl2 (g)
Initial
(M)
Change (M)
Equilibrium (M)
 2(0.028)
+2(0.028)
+0.028
0.444
0.056
0.028

35. Question #8: ICE Calculation
Atmospheric nitrogen and oxygen react to form nitric oxide:
N2 (g) + O2 (g) ⇄ 2NO (g)
Kp = 2.0  at 25oC.
What is the partial pressure of NO at equilibrium with N2 and O2 in the atmosphere?
Given (at 1 atm), PN2 = 0.78 atm and PO2 = 0.21 atm

36. Let x be the change in pressure:+
O2 (g) ⇄
2 NO (g)
Initial
(P, atm)
Change
Equilibrium
0.21
0.78
 x
 x
+2x
0.78  x
0.21  x 2x
Solve for x

37. x = 9.1  10-17 PNO = 2(9.1  10-17) atm = 1.82  10-16 atm
Kp value is too small, therefore value of x is negligible compared to the initial concentrations.
Assume: (0.78 – x) ≈ & (0.21 – x) ≈ 0.21
x = 9.1  10-17
Therefore, at equilibrium:
PNO = 2(9.1  10-17) atm = 1.82  atm
PN2 = atm – 9.1  atm ≈ 0.78 atm
PO2 = atm – 9.1  atm ≈ 0.21 atm

38. IMPORTANT!!! Assumption MUST be followed by validity check!!!
Validity checking is done using x value.
As a general rule, if validity is less than 5%, then the assumption is justified.
Otherwise, you MUST use the quadratic formula or some other approach as shown in the previous examples.

39. EQUILIBRIUM & EXTERNAL EFFECTS
“A change to a system, which is initially at equilibrium, can cause a change in the equilibrium composition.”
The position of the equilibrium will shift in a direction that tends to reduce that change.

40. The outcome of the change is governed by LE CHATELIER’S PRINCIPLE.
“…if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”
If a stress is applied to a system at equilibrium, the system will change so as to reduce the stress.
Henri Le Chatelier
By changing the reaction conditions, the yield of product can be increased or reduced.

41. Changing the temperature.
There are three (3) ways to alter the equilibrium composition of a reaction:
Changing the concentrations by removing products or adding reactants to the reaction vessel. 1.
Changing the partial pressure of gaseous reactants and products by changing the volume. 2.
Changing the temperature. 3.

42. 1. CHANGING THE CONCENTRATIONS:
Addition or removal of either reactant or product shifts the equilibrium to reduce the excess compound.
The disturbance of an added reactant or product is relieved by shifting to the direction that consumes the added substance.
The disturbance of a removed reactant or product is relieved by shifting to the direction that replenished the removed substance.

43. Example: Consider the formation of NH3(g) below, lets add N2 so that its concentration increases 2x:
N2(g) H2 (g) ⇄ 2NH3 (g)
Equil.Conc.
0.50 M
3.00 M
1.98 M
Kc = 0.291
After adding N2
1.00 M
Qc = 0.145
Qc < Kc. Equilibrium is shifted to the right which means that more product will be formed.

44. Le Châtelier’s Principle : Summary
Changes in Concentration:
aA + bB cC + dD
Change
Shifts the Equilibrium
Increase concentration of product(s)
To the left
Decrease concentration of product(s)
To the right
To the right
Increase concentration of reactant(s)
Decrease concentration of reactant(s)
To the left
14.5

45. 2. CHANGING THE PRESSURE BY CHANGING THE VOLUME:
Equilibria containing gases are sensitive to changes in pressure (volume).
Since pressure and volume are inversely proportional, an increase in pressure will give same result as a decrease in volume.
If the pressure is increased by decreasing the volume of a reaction mixture, the reaction shifts in the direction of fewer moles of gas.
Addition of inert gas does not affect the equilibrium position.

46. Example:
Determine the effect of a 2X increase in pressure (volume decreases by factor of 2; concentrations of reactants and products increase by a factor of 2).
N2(g) H2 (g) ⇄ 2NH3 (g)
Equil. Conc.
0.50 M
3.00 M
1.98 M
Kc = 0.291
After increase pressure
1.00 M
6.00 M
3.96 M
Qc = 0.073
Since Qc < Kc, the reaction proceeds to the right to form more NH3(g).
total moles product = 2
total moles of reactant = 4
product has fewer moles than reactants.

47. Le Châtelier’s Principle : Summary
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
Side with most moles of gas
Increase volume
Decrease volume
Side with fewest moles of gas
14.5

48. 3. CHANGING THE TEMPERATURE:
Whether Kc increases or decreases depends upon two factors: T and Hrxn.
An increase in temperature causes the equilibrium to shift towards the endothermic side.
A decrease in temperature causes the equilibrium to shift towards the exothermic side.

49. A + B ⇄ C + D + heat A + B + heat ⇄ C + D
For an exothermic reaction (H –ve), the amounts of products are decreased at equilibrium by an increase in temperature.
A + B ⇄ C + D + heat
HOT!!!
For an endothermic reaction (H +ve), the amounts of products are increased by an increase in temperature.
A + B + heat ⇄ C + D
HOT!!!

50. Question #9: Will an increase in temperature increase the amount of CO produced from the given reaction?
2CO2(g) ⇄ 2CO(g) + O2(g) H = +566 kJ
Answer:
Reaction forward is endothermic (absorb heat).
When temperature is increased, the system will adjust to remove the added heat by shifting from left to right.
Therefore, more CO will be produced.

51. Le Châtelier’s Principle : Summary
Changes in Temperature:
Change
Exothermic Rxn
Endothermic Rxn
Increase temperature
K decreases
K increases
Decrease temperature
K increases
K decreases
14.5

52. The Effect of Catalyst on Equilibrium:
A catalyst is a substance that increases the rate of a reaction but is not consumed by it.
A catalyst has no effect on the equilibrium composition of a reaction mixture.
A catalyst merely speeds up the reaction to achieve equilibrium.

53. Le Châtelier’s Principle : Summary
Change Equilibrium
Constant?
Change
Shift Equilibrium?
Concentration
yes no
Pressure
yes no
Volume
yes no
Temperature
yes
yes
Catalyst no no
14.5