Chapter 15, Problem 15.19QP

(a)

Interpretation Introduction

Interpretation: The expressions $K_{\text{c}}$ and $K_{\text{p}}$ for the given gas-phase reactions are to be written.

Concept introduction: In equilibrium the concentration of the reactant is equal to the concentration of the product.

In reversible reaction reactants give products and these attained products again react to give the reactants back in their form.

A constant that express the relation between the concentrations of the reactant with the product at equilibrium is termed as equilibrium constant. The equilibrium constant is represented by $K_{\text{c}}$ and in the gas phase this equilibrium constant is represented by $K_{\text{p}}$.

To determine: The expressions $K_{\text{c}}$ and $K_{\text{p}}$ for the given gas-phase reactions.

Answer to Problem 15.19QP

Solution

The equilibrium constant $\left(K_{\text{c}}\right)$ for the given reversible reaction is,

$K_{\text{c}}=\frac{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\right]}{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right]\left[{\text{H}}_{\text{2}}\right]}$

The equilibrium constant for the given reaction in gas phase that is $K_{\text{p}}$ is represented as,

$K_{\text{p}}=\frac{\left[p_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\right]}{\left[p_{{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}}\right]\left[p_{{\text{H}}_2}\right]}$

Explanation of Solution

Explanation

The given reversible equation is,

${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(g\right)+{\text{H}}_{\text{2}}\left(g\right)\rightleftharpoons {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)$

In the reaction at equilibrium the rate constants for forward and backward reaction are shown as,

${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(g\right)+{\text{H}}_{\text{2}}\left(g\right)\underset{k\text{'}}{\overset{k}{\rightleftharpoons }}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)$

Where,

• $k$ is the rate constant for forward reaction.
• $k\text{'}$ is the rate constant for backward reaction.

The expression for the rate of forward reaction is,

$r_{\text{f}}=k\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right]\left[{\text{H}}_{\text{2}}\right]$

The expression for the rate of backward reaction is,

$r_{\text{b}}=k\text{'}\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\right]$

The rate of forward reaction is equal to the rate of backward reaction at equilibrium.

$\begin{array}{c}{r}_{\text{f}}=r_{\text{b}}\\ k\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right]\left[{\text{H}}_{\text{2}}\right]=k\text{'}\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\right]\\ K_{\text{c}}=\frac{k}{k\text{'}}\\ K_{\text{c}}=\frac{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\right]}{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right]\left[{\text{H}}_{\text{2}}\right]}\end{array}$

Where,

• $K_{\text{c}}$ is the equilibrium constant.

Hence, the equilibrium constant $\left(K_{\text{c}}\right)$ for the given reversible reaction is,

$K_{\text{c}}=\frac{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\right]}{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right]\left[{\text{H}}_{\text{2}}\right]}$

The equilibrium constant in gas phase that is $K_{\text{p}}$ is represented as,

$K_{\text{p}}=\frac{\left[p_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\right]}{\left[p_{{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}}\right]\left[p_{{\text{H}}_2}\right]}$

Where,

• $p$ is the partial pressures of the corresponding components involved in the above reversible reaction occurring in gas phase.

(b)

Interpretation Introduction

To determine: The expressions $K_{\text{c}}$ and $K_{\text{p}}$ for the given gas-phase reactions.

Answer to Problem 15.19QP

Solution

The equilibrium constant $\left(K_{\text{c}}\right)$ for the given reversible reaction is,

$K_{\text{c}}=\frac{{\left[{\text{SO}}_3\right]}^2}{{\left[{\text{SO}}_2\right]}^2\left[{\text{O}}_{\text{2}}\right]}$

The equilibrium constant for the given reaction in gas phase that is $K_{\text{p}}$ is represented as,

$K_{\text{p}}=\frac{{\left[p_{{\text{SO}}_3}\right]}^2}{{\left[p_{{\text{SO}}_2}\right]}^2\left[p_{{\text{O}}_2}\right]}$

Explanation of Solution

Explanation

The given reversible equation is,

${\text{2SO}}_2\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\rightleftharpoons 2{\text{SO}}_3\left(g\right)$

In the reaction at equilibrium the rate constants for forward and backward reaction are shown as,

${\text{2SO}}_2\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\underset{k\text{'}}{\overset{k}{\rightleftharpoons }}{\text{2SO}}_3\left(g\right)$

Where,

• $k$ is the rate constant for forward reaction.
• $k\text{'}$ is the rate constant for backward reaction.

The expression for the rate of forward reaction is,

$r_{\text{f}}=k{\left[{\text{SO}}_2\right]}^2\left[{\text{O}}_{\text{2}}\right]$

The expression for the rate of backward reaction is,

$r_{\text{b}}=k\text{'}{\left[{\text{SO}}_3\right]}^2$

The rate of forward reaction is equal to the rate of backward reaction at equilibrium.

$\begin{array}{c}{r}_{\text{f}}=r_{\text{b}}\\ k{\left[{\text{SO}}_2\right]}^2\left[{\text{O}}_{\text{2}}\right]=k\text{'}{\left[{\text{SO}}_3\right]}^2\\ K_{\text{c}}=\frac{k}{k\text{'}}\\ K_{\text{c}}=\frac{{\left[{\text{SO}}_3\right]}^2}{{\left[{\text{SO}}_2\right]}^2\left[{\text{O}}_{\text{2}}\right]}\end{array}$

Where,

• $K_{\text{c}}$ is the equilibrium constant.

Hence, the equilibrium constant $\left(K_{\text{c}}\right)$ for the given reversible reaction is,

$K_{\text{c}}=\frac{{\left[{\text{SO}}_3\right]}^2}{{\left[{\text{SO}}_2\right]}^2\left[{\text{O}}_{\text{2}}\right]}$

The equilibrium constant in gas phase that is $K_{\text{p}}$ is represented as,

$K_{\text{p}}=\frac{{\left[p_{{\text{SO}}_3}\right]}^2}{{\left[p_{{\text{SO}}_2}\right]}^2\left[p_{{\text{O}}_2}\right]}$

Where,

• $p$ is the partial pressures of the corresponding components involved in the above reversible reaction occurring in gas phase.

Conclusion

1. a) The equilibrium constant $\left(K_{\text{c}}\right)$ for the given reversible reaction is,

   $K_{\text{c}}=\frac{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\right]}{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right]\left[{\text{H}}_{\text{2}}\right]}$

The equilibrium constant for the given reaction in gas phase that is $K_{\text{p}}$ is represented as,

$K_{\text{p}}=\frac{\left[p_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\right]}{\left[p_{{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}}\right]\left[p_{{\text{H}}_2}\right]}$

1. b) The equilibrium constant $\left(K_{\text{c}}\right)$ for the given reversible reaction is,
2. c) $K_{\text{c}}=\frac{{\left[{\text{SO}}_3\right]}^2}{{\left[{\text{SO}}_2\right]}^2\left[{\text{O}}_{\text{2}}\right]}$

   The equilibrium constant for the given reaction in gas phase that is $K_{\text{p}}$ is represented as,

   $K_{\text{p}}=\frac{{\left[p_{{\text{SO}}_3}\right]}^2}{{\left[p_{{\text{SO}}_2}\right]}^2\left[p_{{\text{O}}_2}\right]}$