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JEE Main 202124 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
24 February
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2021 · 24 February · Morning Shift · Chemistry · Thermodynamics (C)

For the reaction A _ (g) → B _ (g) , the value of the equilibrium...

For the reaction A(g)→B(g){A}_{\left(g\right)} \to {B}_{\left(g\right)}, the value of the equilibrium constant at 300Kand1atm300K and 1atm is equal to 100.0.100 . 0 . The value of ΔGo{\Delta G}^{o} for the reaction at 300Kand1atminJmol−1300K and 1atm in {Jmol}^{- 1} is −xR,- xR, where xx is (Rounded off to the nearest integer) (R=8.31Jmol−1K−1andln10=2.3)\left(R = 8 . 31J{mol}^{- 1}{K}^{- 1}andln10 = 2 . 3\right)

Correct answer

0

Step-by-step solution

ΔG°=−RTlnKp\Delta G ° = - RTlnKp

=−R(300)(2)ln⁡(10)= - R\left(300\right)\left(2\right)\ln \left(10\right)

=−R(300×2×2.3)= - R\left(300 \times 2 \times 2 .3\right)

ΔG°=−1380R\Delta G ° = - 1380R

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