For a certain thermochemical reaction $\mathrm{M} \rightarrow \mathrm{N}$ at $$\mathrm{T}=400 \mathrm{~K}, \Delta \mathrm{H}^{\ominus}=77.2 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{S}=122 \mathrm{~JK}^{-1}, \log$$ equilibrium constant $(\log K)$ is __________ $\times 10^{-1}$.
Answer (integer)
37
Solution
<p>$$\begin{aligned}
& \Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ} \\
& =77.2 \times 10^3-400 \times 122=28400 \mathrm{~J} \\
& \Delta \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \log \mathrm{K} \\
& \Rightarrow 28400=-2.303 \times 8.314 \times 400 \log \mathrm{K} \\
& \Rightarrow \log \mathrm{K}=-3.708=-37.08 \times 10^{-1}
\end{aligned}$$</p>
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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