At $300 \mathrm{~K}, \Delta_{\mathrm{r}} \mathrm{G}^{\circ}$ for the reaction $\mathrm{A}_2(\mathrm{~g})…

At $300 \mathrm{~K}, \Delta_{\mathrm{r}} \mathrm{G}^{\circ}$ for the reaction $\mathrm{A}_2(\mathrm{~g}) \rightleftharpoons \mathrm{B}_2(\mathrm{~g})$ is -11.5 kJ $\mathrm{mol}^{-1}$. The equilibrium constant at 300 K is approximately $\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$
  1. 10
  2. 100
  3. 1000
  4. 25

Solution

We know that, $\Delta_{\mathrm{r}} \mathrm{G}^0=-\mathrm{RT} \ln \mathrm{~K}_{\mathrm{eq}}$ given, $\begin{aligned} & \Delta_{\mathrm{r}} \mathrm{G}^0=-11.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \quad \mathrm{~T}=300 \mathrm{~K} \\ & \quad \mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\ & \therefore \quad-11.5 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1} \\ & =-8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K} \times \ln \mathrm{K}_{\mathrm{eq}} \end{aligned}$ $\text { or, } 4.6106=2.303 \log _{10} \mathrm{~K}_{\mathrm{eq}}$ $\begin{aligned} \therefore & \log _{10} \mathrm{~K}_{\mathrm{eq}}=2 \\ \Rightarrow & \mathrm{~K}_{\mathrm{eq}}=(10)^2 \\ & \mathrm{~K}_{\mathrm{eq}}=100 \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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