Chapter 17.1, Problem 3PP

Interpretation Introduction

(a)

Interpretation:

- The equilibrium constant expression for the heterogeneous chemical equilibria needs to be determined.

$C_{10}{H}_8(s)\rightleftharpoons C_{10}{H}_8(g)$

Concept introduction:

- Ratio of concentration of product and concentration of reactant is equilibrium constant. The equilibrium constant expresses the relationship product and reactant of a reaction at equilibrium.

Consider a reversible chemical reaction

$\mathrm{aA}+bB\rightleftharpoons cC+dD$

${\mathrm{K}}_C=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}$

Answer to Problem 3PP

$K_C=[C_{10}{H}_8(g)]$

Explanation of Solution

For Heterogeneous equilibrium, certain conventions are followed.

1. In heterogeneous equilibria, where solids and gases are involved solids are omitted from the equilibrium constant expression.
2. Liquids are taken as “unity” i.e., liquids $=1$

Therefore, for the reaction:

$C_{10}{H}_8(s)\rightleftharpoons C_{10}{H}_8(g)$

$K_C=[C_{10}{H}_8(g)]$

Interpretation Introduction

(b)

Interpretation:

- The equilibrium constant expression for these heterogenous chemical equilibria needs to be determined.

$H_{2}O(l)\rightleftharpoons H_{2}O(g)$

Concept introduction:

- Ratio of concentration of product and concentration of reactant is equilibrium constant. The equilibrium constant expresses the relationship product and reactant of a reaction at equilibrium.

Consider a reversible chemical reaction

$\mathrm{aA}+bB\rightleftharpoons cC+dD$

${\mathrm{K}}_C=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}$

Answer to Problem 3PP

${\mathrm{K}}_C=[H_{2}O(g)]$

Explanation of Solution

For Heterogeneous equilibrium, certain conventions are followed.

1. In heterogeneous equilibria, where solids and gases are involved solids are omitted from the equilibrium constant expression.
2. Liquids are taken as “unity” i.e., liquids $=1$

Thus, for the reaction:

$H_{2}O(l)\rightleftharpoons H_{2}O(g)$

${\mathrm{K}}_C=[H_{2}O(g)]$

Interpretation Introduction

(c)

Interpretation:

- The equilibrium constant expression for these heterogenous chemical equilibria needs to be determined.

$CaC{O}_3(s)\rightleftharpoons CaO(s)+C{O}_2(g)$

Concept introduction:

- Ratio of concentration of product and concentration of reactant is equilibrium constant. The equilibrium constant expresses the relationship product and reactant of a reaction at equilibrium.

Consider a reversible chemical reaction

$\mathrm{aA}+bB\rightleftharpoons cC+dD$

${\mathrm{K}}_C=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}$

Answer to Problem 3PP

${\mathrm{K}}_C=[C{O}_2(g)]$

Explanation of Solution

For Heterogeneous equilibrium, certain conventions are followed.

1. In heterogeneous equilibrium, where solids and gases are involved solids are omitted from the equilibrium constant expression.
2. Liquids are taken as “unity” i.e., liquids $=1$

Thus, for the reaction:

$CaC{O}_3(s)\rightleftharpoons CaO(s)+C{O}_2(g)$

${\mathrm{K}}_C=[C{O}_2(g)]$

Interpretation Introduction

(d)

Interpretation:

- The equilibrium constant expression for these heterogenous chemical equilibria needs to be determined.

$C(s)+H_{2}O(g)\rightleftharpoons CO(g)+H_2(g)$

Concept introduction:

- Ratio of concentration of product and concentration of reactant is equilibrium constant. The equilibrium constant expresses the relationship product and reactant of a reaction at equilibrium.

Consider a reversible chemical reaction

$\mathrm{aA}+bB\rightleftharpoons cC+dD$

${\mathrm{K}}_C=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}$

Answer to Problem 3PP

${\mathrm{K}}_C=\frac{[CO(g)][H_2(g)]}{[H_{2}O(g)]}$

Explanation of Solution

For Heterogeneous equilibrium, certain conventions are followed.

1. In heterogeneous equilibria, where solids and gases are involved solids are omitted from the equilibrium constant expression.
2. Liquids are taken as “unity” i.e., liquids $=1$

Thus, for the reaction:

$C(s)+H_{2}O(g)\rightleftharpoons CO(g)+H_2(g)$

${\mathrm{K}}_C=\frac{[CO(g)][H_2(g)]}{[H_{2}O(g)]}$

Interpretation Introduction

(e)

Interpretation:

- The equilibrium constant expression for these heterogenous chemical equilibria needs to be determined.

$FeO(s)+CO(g)\rightleftharpoons Fe(s)+C{O}_2(g)$

Concept introduction:

- Ratio of concentration of product and concentration of reactant is equilibrium constant. The equilibrium constant expresses the relationship product and reactant of a reaction at equilibrium.

Consider a reversible chemical reaction

$\mathrm{aA}+bB\rightleftharpoons cC+dD$

${\mathrm{K}}_C=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}$

Answer to Problem 3PP

${\mathrm{K}}_C=\frac{[C{O}_2(g)]}{[CO(g)]}$

Explanation of Solution

For Heterogeneous equilibrium, certain conventions are followed.

1. In heterogeneous equilibria, where solids and gases are involved solids are omitted from the equilibrium constant expression.
2. Liquids are taken as “unity” i.e., liquids $=1$

Thus, for the reaction:

$FeO(s)+CO(g)\rightleftharpoons Fe(s)+C{O}_2(g)$

${\mathrm{K}}_C=\frac{[C{O}_2(g)]}{[CO(g)]}$