Question

The value of $\log _{10} K$ for a reaction $A \rightleftharpoons B$ is
(Given: $\Delta_{r} H_{298\, K}=-54.07 \,kJ\,mol ^{-1}$ and $\Delta_{r} S=10\,J\, K^{-1} mol ^{-1}$
$\left.R =8.314 \,J\,K ^{-1} mol ^{-1} ; 2.303 \times 8.314 \times 298=5705\right)$

• A. 5
• B. 10
• C. 95
• D. 100

Answer 1

Answer: (B)

For the reaction $A \rightleftharpoons B$ we have
$\Delta G ^{\circ}=\Delta H ^{\circ}- T \Delta S ^{\circ}=-54.07-298 \times 10$
$=-54070-2980\, J\,mol ^{-1}$
$=-57050 \,J\,mol ^{-1}$
$\Delta G ^{\circ}=- RT\, In\, K =-2.303 \,RT \log K$
$-5705=2.303 \times 8.314 \times 298 \log _{10} K$,
$-57050=-5705 \log _{10} K$
$\log _{10} K=10$