The value of log _10 K for a reaction A leftharpoons B is (Given : Δ_r H_298 O^°=-54.07 ~kJ ~mol^-1 , Δ_r S_298 ~K^°=10 JK^-1 ~mol^-1 and R=8.314 JK^-1 mol^-1 ; 2.303 × 8.314 × 298=5705)

Correct answer: (B) 10. A leftharpoons B Δ G^°=Δ H^°-T Δ S^° Δ G^°=-2.303 RT log _10 ~K -2.303 RT log _10 ~K=Δ H^°-T Δ S^° 2.303 RT log _10 ~K=T Δ S^°-Δ H^° log _10 ~K=fracT Δ S^°-Δ H^°2.303 R T =(298 × 10+54.07 × 1000)/(2.303 × 8.314 × 298) =9.998 ≈ 10

Q23 / 75 · VITEEE 2021 · Chemistry

The value of $\log _{10} K$ for a reaction $\mathrm{A} \rightleftharpoons \mathrm{B}$ is (Given : $\Delta_{\mathrm{r}} \mathrm{H}_{298 \mathrm{O}}^{\circ}=-54.07 \mathrm{~kJ} \mathrm{~mol}^{-1}$, $\Delta_{\mathrm{r}} \mathrm{S}_{298 \mathrm{~K}}^{\circ}=10 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ and $\mathrm{R}=8.314 \mathrm{JK}^{-1}$ $\mathrm{mol}^{-1}$; $2.303 \times 8.314 \times 298=5705)$

A
5
Correct answer
B
10
C
95
D
100

Solution

$$ \begin{aligned} & \text { } \mathrm{A} \rightleftharpoons \mathrm{B} \\ & \Delta G^{\circ}=\Delta H^{\circ}-\mathrm{T} \Delta S^{\circ} \\ & \Delta G^{\circ}=-2.303 \mathrm{RT} \log _{10} \mathrm{~K} \\ & -2.303 \mathrm{RT} \log _{10} \mathrm{~K}=\Delta H^{\circ}-T \Delta S^{\circ} \\ & 2.303 \mathrm{RT} \log _{10} \mathrm{~K}=T \Delta S^{\circ}-\Delta H^{\circ} \\ & \log _{10} \mathrm{~K}=\frac{T \Delta S^{\circ}-\Delta H^{\circ}}{2.303 R T} \\ & =\frac{298 \times 10+54.07 \times 1000}{2.303 \times 8.314 \times 298} \\ & =9.998 \approx 10 \\ & \end{aligned} $$

Hence, the correct answer is option: 'B'.

Solution

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