For a certain thermochemical reaction M rightarrow N at T =400 K , Δ H ominus =77.2 kJ mol -1 , Δ S =122 JK -1 , log equilibrium constant (log K) is - × 10 -1

Chemistry · Thermodynamics & Thermochemistry

JEE Main 2024 — 27 January, Shift 2 — Question 79

For a certain thermochemical reaction $M \rightarrow N$ 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 - $\qquad$ $\times 10^{-1}$

Answer: 37

Numerical answer — enter this value.

Step-by-step solution

$\quad \Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$

$=77.2 \times 10^{3}-400 \times 122=28400 \mathrm{~J}$

$\Delta \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \log \mathrm{K}$

$\Rightarrow 28400=-2.303 \times 8.314 \times 400 \log \mathrm{~K}$

$\Rightarrow \log \mathrm{K}=-3.708=-37.08 \times 10^{-1}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 2
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Gibbs Free Energy - Relation with Equilibrium, Metallurgy and Electrochemistry