Question

The equilibrium constant for the reaction
$H_2(g)+S(g)\rightleftharpoons H_{2}S(g)$
is $18.5$ at $925$ and $9.25$ at $1000$ respectively. What is the enthalpy of the reaction?

• A. -142.16 kJ/mole
• B. -71.08 kJ/mole
• C. -35.54 kJ/mole
• D. none of these

Answer 1

Answer: (B)

$\log \frac{K_2}{K_1}=\frac{\Delta H}{2.303R}\cdot \frac{\left[T_2-T_1\right]}{T_1{T}_2}$
$\log \frac{9.25}{18.5}=\frac{\Delta H}{2.303\times 8.314}\times \frac{75}{925\times 1000}$
$-0.0310=\frac{\Delta H\times 75}{2.303\times 8.314\times 925\times 1000}$
$\Delta H=-71080.57J/mole$
$=-71.08$ kJ/mole