Consider the following reaction at $$298 \mathrm{~K} \cdot \frac{3}{2} \mathrm{O}_{2(g)} \rightleftharpoons \mathrm{O}_{3(g)} \cdot \mathrm{K}_{\mathrm{P}}=2.47 \times 10^{-29}$$. $\Delta_r G^{\ominus}$ for the reaction is _________ $\mathrm{kJ}$. (Given $\mathrm{R}=8.314 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}$)
Answer (integer)
163
Solution
<p>$$\begin{aligned}
& \frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{O}_{3(\mathrm{~g})} \cdot \mathrm{K}_{\mathrm{P}}=2.47 \times 10^{-29} . \\
& \Delta_{\mathrm{r}} \mathrm{G}^{\Theta}=-\mathrm{RT} \ln \mathrm{K}_{\mathrm{P}} \\
& =-8.314 \times 10^{-3} \times 298 \times \ln \left(2.47 \times 10^{-29}\right) \\
& =-8.314 \times 10^{-3} \times 298 \times(-65.87) \\
& =163.19 \mathrm{~kJ}
\end{aligned}$$</p>
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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