Thermodynamics and Thermochemistry - Result Question 103

####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:

  1. $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}=-54.07 \times 10^{3} J-298 \times 10 J$

$$ =-57.05 \times 10^{3} J $$

Also, $\quad \Delta G^{\circ}=-2.303 R T \log K$

$$ \begin{aligned} \Rightarrow \quad \log K & =\frac{-\Delta G^{\circ}}{2.303 R T} \ & =\frac{57.05 \times 10^{3}}{5705}=10 \end{aligned} $$



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