Q. The value of $log_{10}$ K for a reaction $A\rightleftharpoonsB$ is
(Given: $\Delta _{r}H^{o}_{298 K}=-54.07$ $kJmol^{- 1},\Delta _{r}S^{o}_{298 K}=10J$ $K^{- 1}mol^{- 1}$
and $R=8.314$ $JK^{- 1}mol^{- 1};2.303\times 8.314\times 298=5705$ )

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A

90

B

100

C

5

D

10

Solution:

$\Delta G^{o}=\Delta H^{o}-T\Delta S^{o}=-54.07\times 1000-298\times 10=-57050$ $Jmol^{- 1}$
$\Delta G^\circ =-2.303RTlog_{10}K$
$-57050=-5705log_{10}K$
$log_{10}K=10$

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