Thermodynamics and Thermochemistry - Result Question 28
####28. The value of $\log _{10} K$ for a reaction $A \rightleftharpoons B$ is
(Given : $\Delta _r H^{\circ}{ } _{298 K}=-54.07 kJ mol^{-1}$,
$\Delta _r S^{\circ}{ } _{298 K}=10 JK^{-1} mol^{-1}$ and $R=8.314 JK^{-1} mol^{-1}$;
$2.303 \times 8.314 \times 298=5705$ )
(2007, 3M)
(a) 5
(b) 10
(c) 95
(d) 100
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Solution:
- $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}=-54.07 \times 10^{3} J-298 \times 10 J$
$$ =-57.05 \times 10^{3} J $$
Also, $\quad \Delta G^{\circ}=-2.303 R T \log K$
$$ \begin{aligned} \Rightarrow \quad \log K & =\frac{-\Delta G^{\circ}}{2.303 R T} \ & =\frac{57.05 \times 10^{3}}{5705}=10 \end{aligned} $$
