Question

For a. reaction $A+B⟶C+D,K_{eq}$ represents equilibrium constant, The graph of $\ln K_{\text{eq }}$ versus $\frac{1}{T}$ is a straight line with a slope equal to $300$ . Also at $T=300K$, the reaction is at equilibrium with
$[A]=[B]=[C]=1m$ and $[D]10M.$ Calculate the value of $\left(1000\times \Delta S^{\circ }\right)$ in cal/Kelvin.
$[R=2cal/molK]$

Answer 1

Answer: 2606

$\ln K_{eq}=\frac{-\Delta H^{\circ }}{RT}+\frac{\Delta S^{\circ }}{R}$
$\therefore$ Slope of $\ln K_{\text{eq }}$ vs $\frac{1}{T}=\frac{-\Delta H^{\circ }}{R}$
$\Rightarrow \Delta H^{\circ }=-300R\Rightarrow -600cal$
at $300K{K}_{eq}=10$
$\therefore \Delta G^{\circ }=-RT\ln 10$
$\Rightarrow -2.303\times 300\times 2$
$\therefore \Delta S^{\circ }=\frac{\Delta H^{\circ }-\Delta G^{\circ }}{T}=\frac{-600+600\times 2.3C3}{300}$
$\Rightarrow 2.606cal/K$
$\therefore 1000\Delta S^{\circ }=2606cal/K$