Chapter 12, Problem 16QAP

At 800°C, $K=2.2\times {10}^{-4}$ for the following reaction

$2{\text{H}}_2\text{S}\left(g\right)\rightleftharpoons 2{\text{H}}_2\left(g\right)+{\text{S}}_2\left(g\right)$

Calculate K at 8000C for

(a) the synthesis of one mole of H₂S from H₂ and S₂ gases.

(b) the decomposition of one mole of H₂S gas.

Interpretation Introduction

(a)

Interpretation:

The equilibrium constant for the formation of 1 mol of ${\text{H}}_{\text{2}}\text{S}$ gas need to be determined.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general reaction as follows:

$\text{A}\left(g\right)+\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The expression for the equilibrium constant is represented as follows:

$K=\frac{\left(P_C\right)\left(P_D\right)}{\left(P_A\right)\left(P_B\right)}$

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

Answer to Problem 16QAP

The equilibrium constant for the synthesis of 1 mol of ${\text{H}}_{\text{2}}\text{S}$ is 67.42

Explanation of Solution

The given equilibrium reaction is as follows:

$2H_{2}S\left(g\right)\rightleftharpoons 2{\text{H}}_{\text{2}}\left(g\right)+{\text{S}}_{\text{2}}\left(g\right)$

The value of equilibrium constant at $800{}^{\circ }\text{C}$ is $2.2\times {10}^{-4}$.

The expression for the equilibrium constant will be:

$K=\frac{P_{{\text{S}}_{\text{2}}}{\left(P_{{\text{H}}_{\text{2}}}\right)}^2}{{\left(P_{{\text{H}}_{\text{2}}\text{S}}\right)}^2}$

Thus,

$K=\frac{P_{{\text{S}}_{\text{2}}}{\left(P_{{\text{H}}_{\text{2}}}\right)}^2}{{\left(P_{{\text{H}}_{\text{2}}\text{S}}\right)}^2}=2.2\times {10}^{-4}$.. ...... (1)

Now, the reaction for the synthesis of 1 mol of ${\text{H}}_{\text{2}}\text{S}$ will be as follows:

${\text{H}}_{\text{2}}\left(g\right)+\frac{1}{2}{\text{S}}_{\text{2}}\left(g\right)\rightleftharpoons H_{2}S\left(g\right)$

The expression for the equilibrium constant of the above reaction will be:

$K^{\text{'}}=\frac{\left(P_{{\text{H}}_{\text{2}}\text{S}}\right)}{{\left(P_{{\text{S}}_{\text{2}}}\right)}^{1/2}\left(P_{{\text{H}}_{\text{2}}}\right)}$

Since,

$\frac{P_{{\text{S}}_{\text{2}}}{\left(P_{{\text{H}}_{\text{2}}}\right)}^2}{{\left(P_{{\text{H}}_{\text{2}}\text{S}}\right)}^2}=2.2\times {10}^{-4}$

Taking reciprocal of the above equation:

$\frac{{\left(P_{{\text{H}}_{\text{2}}\text{S}}\right)}^2}{P_{{\text{S}}_{\text{2}}}{\left(P_{{\text{H}}_{\text{2}}}\right)}^2}=\frac{1}{2.2\times {10}^{-4}}$

Now, taking square root:

$\frac{\left(P_{{\text{H}}_{\text{2}}\text{S}}\right)}{{\left(P_{{\text{S}}_{\text{2}}}\right)}^{1/2}\left(P_{{\text{H}}_{\text{2}}}\right)}=\sqrt{\frac{1}{2.2\times {10}^{-4}}}=67.42$

Therefore, equilibrium constant for the synthesis of 1 mol of ${\text{H}}_{\text{2}}\text{S}$ is 67.42.

Interpretation Introduction

(b)

Interpretation:

The equilibrium constant for the decomposition of 1 mol of ${\text{H}}_{\text{2}}\text{S}$ gas need to be determined.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general reaction as follows:

$\text{A}\left(g\right)+\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The expression for the equilibrium constant is represented as follows:

$K=\frac{\left(P_C\right)\left(P_D\right)}{\left(P_A\right)\left(P_B\right)}$

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

Answer to Problem 16QAP

The equilibrium constant for the decomposition of 1 mol of ${\text{H}}_{\text{2}}\text{S}$ gas is 0.015.

Explanation of Solution

The given equilibrium reaction is as follows:

$2H_{2}S\left(g\right)\rightleftharpoons 2{\text{H}}_{\text{2}}\left(g\right)+{\text{S}}_{\text{2}}\left(g\right)$

The value of equilibrium constant at $800{}^{\circ }\text{C}$ is $2.2\times {10}^{-4}$.

The expression for the equilibrium constant will be:

$K=\frac{P_{{\text{S}}_{\text{2}}}{\left(P_{{\text{H}}_{\text{2}}}\right)}^2}{{\left(P_{{\text{H}}_{\text{2}}\text{S}}\right)}^2}$

Thus,

$K=\frac{P_{{\text{S}}_{\text{2}}}{\left(P_{{\text{H}}_{\text{2}}}\right)}^2}{{\left(P_{{\text{H}}_{\text{2}}\text{S}}\right)}^2}=2.2\times {10}^{-4}$.. ...... (1)

The reaction for the decomposition of 1 mol of ${\text{H}}_{\text{2}}\text{S}$ gas will be:

$H_{2}S\left(g\right)\rightleftharpoons {\text{H}}_{\text{2}}\left(g\right)+\frac{1}{2}{\text{S}}_{\text{2}}\left(g\right)$

The expression for the equilibrium constant will be:

$K\text{'}\text{'}=\frac{{\left(P_{{\text{S}}_{\text{2}}}\right)}^{1/2}\left(P_{{\text{H}}_{\text{2}}}\right)}{\left(P_{{\text{H}}_{\text{2}}\text{S}}\right)}$

Since,

$\frac{P_{{\text{S}}_{\text{2}}}{\left(P_{{\text{H}}_{\text{2}}}\right)}^2}{{\left(P_{{\text{H}}_{\text{2}}\text{S}}\right)}^2}=2.2\times {10}^{-4}$

Taking square root:

$\frac{{\left(P_{{\text{S}}_{\text{2}}}\right)}^{1/2}\left(P_{{\text{H}}_{\text{2}}}\right)}{\left(P_{{\text{H}}_{\text{2}}\text{S}}\right)}=0.015$

Therefore, the equilibrium constant for the decomposition of 1 mol of ${\text{H}}_{\text{2}}\text{S}$ gas is 0.015.