1. Chemical Equilibrium Chapter 16
General Chemistry - Principles and Modern Applications
Petrucci • Harwood • Herring, 8th Edition; Prentice-Hall © 2002
An Equilibrium is reached when two opposing processes take place at equal rates.

2. Dynamic Equilibria of “Physical” Processes
Equilibrium – two opposing processes taking place at equal rates.
Saturated aqueous solution of I2 in contact with CCl4
Phase Transfer (Chapter 14)
I2(H2O) I2(CCl4)
Evaporation (Chapter 14)
H2O(l) H2O(g)
Dissolution of a solid in a solvent (water) (Chapter 19)
H2O
H2O
CCl4
CCl4
NaCl(s) Na+ (aq) + Cl-(aq)
Time = 0
Time = Equilibrium
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

3. S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005

4. Dynamic Equilibrium of a Chemical Process
CO(g) + 2 H2(g) CH3OH(g) k1
Forward:
CO(g) + 2 H2(g) → CH3OH(g)
Rfwrd = k1[CO][H2]2
k-1
Reverse:
CH3OH(g) → CO(g) + 2 H2(g)
Rrvrs = k-1[CH3OH] k1
CO(g) + 2 H2(g) CH3OH(g)
k-1
At Equilibrium:
Rfwrd = Rrvrs 
k1[CO][H2]2 = k-1[CH3OH] k1
[CH3OH] =
= Kc (Equilibium Constant)
k-1
[CO][H2]2
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

5. S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005

6. S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005

7. 3 Trial & Error Approaches to the Equilibrium Constant of the Methanol Reaction (Data from Table 16.1) k1
CO(g) + 2 H2(g) CH3OH(g)
k-1
[CH3OH]
[CH3OH]
[CH3OH]
Kc =
[CO][H2]
[CO](2[H2])
[CO][H2]2
Kc(1) =
1.19 M-1
0.596 M-1
14.2 M-2
Kc(2) =
2.17 M-1
1.09 M-1
14.2 M-2
Kc(3) =
2.55 M-1
1.28 M-1
14.2 M-2
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

8. Relationship between Kc and the Balanced Chemical Equation
a A + b B … g G + h H ….
[G]g[H]h ….
Equilibrium Constant (concentrations) = Kc =
[A]m[B]n ….
A more accurate equilibrium constant is Keq, which will be discussed in detail in Physical Chemistry I (59-240).
(aG)g(aH)h ….
Thermodynamic Equilibrium constant = Keq =
(aA)a(aB)b ….
[B]
aB =
= B[B]
cB0 is a standard reference state
= 1 mol L-1 (ideal conditions)
cB0
aB is dimensionless and so is Keq!
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

9. Important Relationships
When we reverse a chemical equation, the value of Kc inverses;
When we multiply the coefficients in a balanced equation by a common factor, we raise the equilibrium constant to the corresponding power;
When we divide the coefficients in a balanced equation by a common factor, we take the corresponding root of the equilibrium constant;
[CH3OH]
CO(g) + 2 H2(g) CH3OH(g)
= Kc = 14.5
[CO][H2]2
[CO][H2]2
CH3OH(g) CO(g) + 2 H2(g)
= Kc-1 =
[CH3OH]
[CH3OH]
2 CO(g) + 4 H2(g) CH3OH(g)
= (Kc)2 = 2.10102
[CO][H2]2
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

10. Chemical Equilibrium Expressions Might Be Combined to Find an Unknown KC
N2O(g) + ½O NO(g) Kc= ?
Known:
[N2][O2]½
[N2O] =
N2(g) + ½O N2O(g) Kc(2)= 2.710-18
[N2][O2]
[NO]2 =
N2(g) + O NO(g) Kc(3)= 4.710-31
[NO]2
[NO]2
[N2][O2]½ 1
Kc= = =
Kc(3)
= 1.710-13
[N2O][O2]½
[N2][O2]
[N2O]
Kc(2)
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

11. Working with Gases – KP [SO3]2 2 SO2(g) + O2(g) 2 SO3(g) Kc =
Mixtures of gases are solutions as are mixtures with liquids.
With gases it is more convenient to use KP, which is based upon partial pressures of gases (Chapter 6), than the concentration based KC.
[SO3]2
2 SO2(g) + O2(g) SO3(g)
Kc =
[SO2]2[O2]
Use the ideal gas law (Chapter 6):
nSO2
PSO2
[SO2] = =
nSO3
PSO3 V RT
[SO3] = = V RT
[O2] = V
nO2 = RT
PO2
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

12.  Kc = KP(RT) and KP = Kc(RT)-1
PSO3 2
[SO3]2 RT
PSO3 2
Kc = = = RT
[SO2]2[O2]
PSO2
PO2
PSO3 2 2
PSO2 2
PO2
KP = RT RT
PSO2 2
PO2
 Kc = KP(RT) and KP = Kc(RT)-1
For a general balanced chemical equation
a A + b B … g G + h H ….
KP = Kc(RT)Δn(gas)
Dn = the sum of the stoichiometric coefficients of gaseous products minus the sum of the stoichiometric coefficients of gaseous reactants;
Dn = (g + h + …) – (a + b + …)
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

13. Do Examples 16-2 and 16-3

14. Pure Liquids and Solids
Concentration terms for solid or liquid phases of a single component (that is, pure solids or liquids) are, by definition, set to 1 and, therefore, do not appear in the equilibrium constant expressions!
C(s) + H2O(g) CO(g) + H2(g)
Kc =
[H2O]
[CO][H2] =
PH2O
PCOPH2
(RT)-1
CaCO3(s) CaO(s) + CO2(g)
CaCO3
CaO
Kc = [CO2]
KP = PCO2
KP = Kc(RT)
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

15. Wide Range of KP Values
Reversible Reactions (KC or P = > and < 1010);
Quasi Non-Reversible Reactions (KC or P = < and > 1010);
The large value of K does not imply that the reaction proceeds at a given temperature. Why?
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

16. The Reaction Quotient, Q: Predicting the Direction of Net Change
An equilibrium might be approached in various ways;
The calculation of Q allows us to qualitatively determine the change of an initial condition as the equilibrium is approached;
Qc =
[A]ta[B]tb
[G]tg[H]th
At equilibrium Qc = Kc
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

17. Example 16-5 CO(g) + 2 H2(g) CH3OH(g)k1
CO(g) + 2 H2(g) CH3OH(g)
k-1
Direction of Net Chemical Change in Established Equilibrium
KC = 1 at 1100 ºC; Starting amounts are 1 mol each of CO and H2O as well as 2 mols each of CO2 and H2.
Which substance will be present in greater amounts and which in lesser amounts once the equilibrium has been established?
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

18. Altering Equilibrium Conditions: Le Châtellier’s Principle
When a system in equilibrium is subjected to a change in temperature, pressure, or concentration of a reacting species, the system responds by attaining a new equilibrium that partially offsets the impact of the change.
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

19. Effect of Changing the Amounts of Reacting Species on Equilibriumk1
2 SO2(g) + O2(g) SO3(g)
Kc = 2.8102 at 1000K
k-1
Q > Kc
Q =
= Kc
[SO2]2[O2]
[SO3]2
new equilibrium concentrations
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

20. Effect of Changes in Pressure or Volume on Equilibrium
Addition or removal of a gaseous reactant or product changes Pgases at equilibrium as discussed before.
Addition of an inert gas changes the total pressure.
Relative partial pressures and, therefore K, remain unchanged!
Change of the volume of the system causes a change in the equilibrium position. V
nSO3 2
nSO2
nO2 =
Kc =
[SO2]2[O2]
[SO3]2
nSO3 2 =
 V = 2.8  102
nSO2 2
nO2
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

21. Do not confuse with Kc = KP(RT) and KP = Kc(RT)-1nG nH
[G]g[H]h g h
Kc = =
 V(a+b)-(g+h)
[A]a[B]b nA a nB b
[A]a = (nA / VA)a
= nAa / VAa nG g nH h =
 V-Δn nA a nB a
Dn = (g + h + …) – (a + b + …)
When the volume of an equilibrium mixture of gases is reduced, a net change occurs in the direction that produces fewer moles of gas. When the volume is increased, a net change occurs in the direction that produces more moles of gas.
Do not confuse with Kc = KP(RT) and KP = Kc(RT)-1
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

22. Effect of Temperature on Equilibrium
Raising the temperature of a reaction mixture at equilibrium shifts the equilibrium condition in the direction of the endothermic reaction.
Lowering the temperature causes a shift in the direction of the exothermic reaction.
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

23. Effect of a Catalyst on Equilibrium
A catalyst changes the mechanism of a reaction to one with a lower activation energy.
A catalyst has no effect on the equilibrium concentrations and constant.
But does affect the rate at which equilibrium is attained!
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

24. Synthesis of ammonia (NH3) from H2 and N2
The optimum conditions are only for the equilibrium position and do not take into account the rate at which equilibrium is attained.
low temp. and high pressure
high temp. and lower pressure
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005

25. Examples 16-10 to -12 and Feature Problem 91 (p. 664) Do it!
S. Holger Eichhorn; University of Windsor; General Chemistry II ; © 2005