Chapter 18 Equilibrium

A + B AB• We may think that all reactions change all

reactants to products, or the reaction has gone to completion

• But in reality, products may start to change back into reactants; the reaction is reversible.

A + B ↔ AB

CO2 + H2O C6H12O6 + O2

O2 + C6H12O6 CO2 + H2O

Equilibrium

• Equilibrium is the state where the forward and reverse reactions balance because they are occurring at equal rates.

• NO OVERALL CHANGE!

• The Law of Chemical Equilibrium– the point in the reaction where a ratio of reactant

and product concentration has a constant value, Keq

Keq < 1 Reactants Favored

Keq = 1 Neither is favored

Keq > 1 Products favored

• Homogeneous equilibrium– States of all compounds are the ____________

H2 (g) + I2 (g) ↔ 2HI (g)

• Heterogeneous equilibrium– States of compounds are ________________

CaCO3 (g) ↔ CaO (s) + CO2 (g)

Remember we only pay attention to the GASES when we calculate K

eq!

What is the correct the Keq?N2 (g) + 3H2 (g) ↔ 2NH3 (g)

3][][

2][)

22

3

xHxN

xNHKA eq

23

322

][

][][)

NH

HxNKB eq

][][

][)

22

3

HN

NHKC eq

322

23

][][

][)

HxN

NHKD eq

What is the correct the Keq?2 NbCl4 (g) ↔ NbCl3 (g) + NbCl5 (g)

24

53

][

][][)

NbCl

NbClxNbClKA eq

][][

][)

53

4

NbClxNbCl

NbClKB eq

55

33

44

][][

][)

NbClxNbCl

NbClKC eq

44

55

33

][

][][)

NbCl

NbClxNbClKD eq

What is the correct Keq?

H2O (l) ↔ H2O(g)

][

][)

2

2

OH

OHKA eq

22 ][) OHKB eq

2][) 2 xOHKC eq

][) 2OHKD eq

If Keq = 2.4, which is more favored?

A. products

B. reactants

C. neither is favored

D. not enough information

Chemical Equilibrium Problems I

• Write the equilibrium constant expression (Keq) for these homogeneous equations.

1. N2O4 (g) ↔ 2 NO2 (g)

2. CO (g) + 3H2(g) ↔ CH4 (g)+ H2O (g)

3. 2 H2S (g) ↔ 2 H2 (g) + S2 (g)

• Write the equilibrium constant expression (Keq) for these heterogeneous equations.

1. C10H8 (s) ↔ C10H8 (g)

2. CaCO3 (s) ↔ CaO (s) + CO2 (g)

3. C (s) + H2O(g) ↔ CO(g) + H2 (g)

4. FeO(s) + CO(g) ↔ Fe(s) + CO2(g)

1. Calculate the Keq for the following equation using the data:

[N2O4] = 0.0613 mol/L

[NO2] = 0.0627 mol/L

N2O4 (g) ↔ 2NO2 (g)

A. 0.98

B. 15.59

C. 0.064

D. 1.02

E. 2.05

2. [CO] = 0.0613 mol/L

[H2] = 0.1839 mol/L

[CH4]=0.0387 mol/L

[H2O]=0.0387 mol/L

CO (g) + H2 (g) ↔ CH4 (g) + H2O (g)

A. 0.25

B. 0.72

C. 0.04

D. 3.93

E. 0.02

3.

[H2]=1.5 mol/L [N2]=2.0 mol/L [NH3]=1.8 mol/L

3H2 (g) + N2(g) ↔ 2NH3 (g)

A. 0.48

B. 0.24

C. 2.5

D. 0.40

3. If the Keq = 0.48, we know that the equilibrium favors…

a. reactants

b. products

c. neither

4.

[Mg]=2 mol/L [HCl]=3 mol/L

[MgCl]=6 mol/L [H2]=3 mol/L

2 Mg (s) + 2 HCl (g) ↔ 2 MgCl (g) + H2 (g)

A. 3

B. 12

4. Since Keq = 3, we know that the equilibrium favors

a. reactants

b. products

c. neither

5. [H2]=0.52 mol/L [I2]=0.23 mol/L [HI]=1.7 mol/L

H2 (g) + I2 (g) ↔ 2HI (g)

A. 202

B. 0.04

C. 24.16

5. Since Keq= 24.16, we know that the equilibrium favors

a. reactants

b. products

c. neither