Question

The value of ${\log }_{10}K$ for a reaction $A⟶B$ is :

$(Given:{\Delta }_r{H}_{298K}^o=-54.07kJmo{1}^{-1},{\Delta }_r{S}_{298K}^o=10J{K}^{-1}mo{1}^{-1}and$

$R=8.314J{K}^{-1}mo{1}^{-1};2.303\times 8.314\times 298=5705)$

• A. 5
• B. 10
• C. 95
• D. 100

Answer 1

Answer: (B)

10

Given that for a reaction $A\to B$,

${\Delta }_r{H}_{298K}^o=-54.07$ kJ/mol,

${\Delta }_r{S}_{298K}^o=10$ kJ/molK and

$R=8.314$ J/mol K and the value of $2.303\times 8.314\times 298=5705$

Now,

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

$=-57050=-5705{\log }_{10}K=-2.303{\text{RTlog}}_{10}\text{K}$

$\Rightarrow {\log }_{10}K=10$

Hence, option ($B$) is correct.