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 }{2.303 R } \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} $
$\Delta H =-71080.57$ J / mole
$= -71.08 $ kJ/mole