Chapter 14, Problem 14.125QA

Interpretation Introduction

To find:

$K_P$ for reaction $4 {CH}_{4\left(g\right)}+ S_{8\left(S\right)}\leftrightarrow {4 CS}_{2\left(g\right)}+ {8H}_{2\left(g\right)}$ at ${25}^{0}C$ and ${500}^{0}C$

Answer to Problem 14.125QA

Solution:

a) $K_P$ for reaction $4 {CH}_{4\left(g\right)}+ S_{8\left(S\right)}\leftrightarrow {4 CS}_{2\left(g\right)}+ {8H}_{2\left(g\right)}$ at ${25}^{0}C$ is $5.44 \times {10}^{-82}.$

b) $K_P$ for reaction $4 {CH}_{4\left(g\right)}+ S_{8\left(S\right)}\leftrightarrow {4 CS}_{2\left(g\right)}+ {8H}_{2\left(g\right)}$ at ${500}^{0}C$ is $4.70 \times {10}^{-32}.$

Explanation of Solution

1) Concept:

Equilibrium constant expression for $K_P$ is the ratio of the equilibrium partial pressures of the product divided by the equilibrium partial pressure of the reactants each raised to respective stoichiometric coefficient in the balanced equation.

If the reaction temperature is higher, it increases the equilibrium constant of an endothermic reaction and decreases the equilibrium constant for an exothermic reaction.

2) Given:

$∆G_f^0\left(C{H}_4\right)=\mathit{ }-50.8\frac{kJ}{mol} ,∆G_f^0\left(C{S}_2\right)=65.1\frac{kJ}{mol} , ∆G_f^0\left(H_2\right)=0 ,∆G_f^0\left(S_8\right)=0$

3) Formula:

Relation between $∆G_{ren}^0$ and $∆G_f^0\left(C{H}_4\right) ,∆G_f^0\left(C{S}_2\right) ,∆G_f^0\left(H_2\right) ,∆G_f^0\left(S_8\right)$ is given below;

$∆G_{rxn}^0=\{4\times ∆G_f^0\left(C{S}_2\right)+8\times ∆G_f^0\left(H_2\right)\}-\left\{\left[4\times ∆G_f^0\left(C{H}_4\right)\right]+\left[1 \times ∆G_f^0\left(S_8\right)\right]\right\}$

The relation between $K_p$ and $∆G_{rxn}^0$ is given below;

$K_p= e^{-∆G^0/RT}$

Where $K_p$ is equilibrium constant,$∆G^0$ is Gibbs free energy, $R$ is gas constant,$T$ is absolute temperature.

4) Calculations:

a)

We are given the temperature in degree Celsius. So we first convert it to Kelvin as shown;

$T\left(K\right)= T^{0}C+273$

$T\left(K\right)= 25+273$

$T\left(K\right)= 298 K$

$∆G_{rxn}^0$ is calculated as follows;

$∆G_{rxn}^0=\{4\times ∆G_f^0\left(C{S}_2\right)+8\times ,∆G_f^0\left(H_2\right)\}-\left\{\left[4\times ∆G_f^0\left(C{H}_4\right)\right]+\left[1 \times ∆G_f^0\left(S_8\right)\right]\right\}$

$∆G_{rxn}^0=\left\{\right(4 mol\times 65.1\frac{kJ}{mol})+(8 mol\times 0\left)\right\}-\left\{\left[4 mol\times \left(-50.8\frac{kJ}{mol}\right)\right]+\left[1 mol\times 0 \right]\right\}$

$∆G_{rxn}^0=\left( 260.4 kJ\right)-\left(-203.2 kJ\right)$

$∆G_{rxn}^0=463.6 kJ$

From the relation between $K_p$ and $∆G^0$ we can calculate $K_p$ at ${25}^{0}C$ as

$K_p= e^{-\frac{∆G^0}{RT}}$

$K_P= e^{-∆G^0/RT}$

$K_P= e^{\frac{-463600}{8.314\times 298}}$

$K_P= e^{-187.119}$

$K_P=5.44 \times {10}^{-82}$

b)

We are given temperature in degree Celsius; so we first convert it to Kelvin as:

$T\left(K\right)= T^{0}C+273$

$T\left(K\right)= 500+273$

$T\left(K\right)= 773 K$

So value of $K_p$ at 773 K is calculated as follows:

$K_p= e^{-\frac{∆G^0}{RT}}$

$K_P= e^{-∆G^0/RT}$

$K_P= e^{\frac{-463600}{8.314\times 773}}$

$K_P= e^{-72.136}$

$K_P=4.70 \times {10}^{-32}$

Conclusion:

As the reaction is endothermic, the equilibrium constant value increases as the temperature raises from ${25}^{0}C$ to ${500}^{0}C$.