For the reaction at 298 K , 2 A + B rightarrow C . Δ H =400 kJ mol -1 and Δ S =0.2 kJ mol -1 K -1 . The reaction will become spontaneous above \ \ \ \ \ \ \ \ \ K

Chemistry · Thermodynamics & Thermochemistry

JEE Main 2024 — 6 April, Shift 2 — Question 82

For the reaction at $298 \mathrm{~K}, 2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C} . \Delta \mathrm{H}$ $=400 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta \mathrm{S}=0.2 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ . The reaction will become spontaneous above $\_\_\_\_\_\_\_\_\_$ K

Answer: 2000

Numerical answer — enter this value.

Step-by-step solution

$\Delta \mathrm{G}=0$

$\mathrm{T}=\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}=\frac{400}{0.2}=2000 \mathrm{~K}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Gibbs Free Energy - Relation with Equilibrium, Metallurgy and Electrochemistry