Question

The value of $log_{10}$ K for a reaction $A\rightleftharpoons B$ is

(Given: $\Delta _{r}H^{o}_{298 K}=-54.07$ $kJmol^{- 1},\Delta _{r}S^{o}_{298 K}=10J$ $K^{- 1}mol^{- 1}$

and $R=8.314$ $JK^{- 1}mol^{- 1};2.303\times 8.314\times 298=5705$ )

• A. 90
• B. 100
• C. 5
• D. 10

Answer 1

Answer: (D)

$\Delta G^{o}=\Delta H^{o}-T\Delta S^{o}=-54.07\times 1000-298\times 10=-57050$ $Jmol^{- 1}$

$\Delta G^\circ =-2.303RTlog_{10}K$

$-57050=-5705log_{10}K$

$log_{10}K=10$