Question

The equilibrium constant for the reaction $H _{2}( g )+ S ( s ) \rightleftharpoons H _{2} S ( g )$; is $18.5$ at $925\, K$ and $9.25$ at $1000\, K$ respectively. The enthalpy of the reaction will be :

• A. $-68000.05\, J\, mol ^{-1}$
• B. $-71080.57\, J\, mol ^{-1}$
• C. $- 80071.75\, J\, mol ^{-1}$
• D. $57080.75\, J\, mol ^{-1}$

Answer 1

Answer: (B)

$H _{2}( g )+ S ( s ) \rightleftharpoons H _{2} S ( g )$
$\log \frac{ K _{2}}{ K _{1}}=\frac{\Delta H }{2.303\, R }\left(\frac{1}{ T _{1}}-\frac{1}{ T _{2}}\right)$
$\log \left(\frac{9.25}{18.5}=\frac{1}{2}\right)=\frac{\Delta H }{2.303 \times 8.314}\left(\frac{1}{925}-\frac{1}{1000}\right)$
$-0.3010=\frac{\Delta H }{2.30 \times 8.314}\left(\frac{75}{925 \times 1000}\right)$
$\Delta H =-71080.57 \,J / mole$