JEE MAIN - Chemistry (2024 - 27th January Evening Shift - No. 30)
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
37
Explanation
$$\begin{aligned}
& \Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ} \\
& =77.2 \times 10^3-400 \times 122=28400 \mathrm{~J} \\
& \Delta \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \log \mathrm{K} \\
& \Rightarrow 28400=-2.303 \times 8.314 \times 400 \log \mathrm{K} \\
& \Rightarrow \log \mathrm{K}=-3.708=-37.08 \times 10^{-1}
\end{aligned}$$
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