For a certain reaction at 300 K , K =10 , then Δ G ° for the same reaction is - × 10 -1 kJ mol -1 . (Given R =8.314 JK -1 mol -1 )

Chemistry · Thermodynamics & Thermochemistry

JEE Main 2024 — 1 February, Shift 2 — Question 81

For a certain reaction at $300 \mathrm{~K}, \mathrm{~K}=10$ , then $\Delta \mathrm{G}^{\circ}$ for the same reaction is - $\qquad$ $\times 10^{-1} \mathrm{~kJ} \mathrm{~mol}^{-1}$ .

(Given $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )

Answer: 57

Numerical answer — enter this value.

Step-by-step solution

$\quad \Delta \mathrm{G}^{\circ}=-\mathrm{RT}$ ใn K

$=-8.314 \times 300 \ln (10)$

$=5744.14 \mathrm{~J} / \mathrm{mole}$

$=57.44 \times 10^{-1} \mathrm{~kJ} / \mathrm{mole}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 2
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Gibbs Free Energy - Relation with Equilibrium, Metallurgy and Electrochemistry