Verified Solution • 2025–26 Session moderate

QJEEM24Chb71b: JEE Main 2024 Chemistry - Class 11 JEEM24Chb71b Solution: For a certain thermochemical reaction at equilibrium constant…

For a certain thermochemical reaction M→N\mathrm{M} \rightarrow \mathrm{N} at T=400 K,ΔH⊖=77.2 kJ mol−1,ΔS=122 JK−1,log⁡\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)(\log K) is __________ ×10−1\times 10^{-1} .

Step-by-Step Verified Solution

amp;ΔG∘=ΔH∘−TΔS∘amp;=77.2×103−400×122=28400 Jamp;ΔG∘=−2.303RTlog⁡Kamp;⇒28400=−2.303×8.314×400log⁡Kamp;⇒log⁡K=−3.708=−37.08×10−1\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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