Question

The equilibrium for $2H_{2}S\left(g\right)\rightleftharpoons 2H_2\left(g\right)+S_2\left(g\right)$ is 0.0118 at 1300 K while the heat of dissociation is 597.4 kJ. The standard equilibrium constant of the reaction at 1200 K is :

• A. $1.180\times {10}^{-4}$
• B. 11.8
• C. 118
• D. cannot be calculated from given data

Answer 1

The correct option is A $1.180\times {10}^{-4}$

Using Arrhenius equation and Equilibrium constant,

$log\frac{K_2}{K_1}=\frac{-\Delta H}{2.303R}\times [\frac{1}{T_1}-\frac{1}{T_2}]$

where, $K_{2}and{K}_1$ are equilibrium constant.

$\Delta H=heatofdissociation=597.4KJ$

$T_1=1300Kand{T}_2=1200K$

$log\frac{K_2}{0.0118}=\frac{-597400}{-2.303\times 8.314}-[\frac{1}{1300}-\frac{1}{1200}]$

$K_2=1.180\times {10}^{-4}$

Hence, option (A) is correct.