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JEE Main 202118 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
18 March
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2021 · 18 March · Evening Shift · Chemistry · Thermodynamics (C)

The gas phase reaction 2A(g) ⇌ A _ 2 (g) at 400K has Δ G ^ o = + 25...

The gas phase reaction

2A(g)⇌A2(g)2A\left(g\right) ⇌ {A}_{2}\left(g\right)

at 400K400K has ΔGo=+25.2kJmol−1\Delta{G}^{o} = + 25 . 2kJ{mol}^{- 1}.

The equilibrium constant KC{K}_{C} for this reaction is ___ ×10−2\times {10}^{- 2}. (Round off to the Nearest integer)

Use : R=8.3Jmol−1K−1,ln⁡10=2.3log⁡102=0.30,1atm=1barR = 8 . 3J{mol}^{- 1}{K}^{- 1}, \ln 10 = 2 . 3\log_{10}2 = 0 . 30, 1atm = 1bar

antilog (−0.3)=0.501( - 0 . 3 ) = 0 . 501

Correct answer

0

Step-by-step solution

Using formula

ΔrG0=−RTlnKp{\Delta}_{r}{G}^{0} = - {RTlnK}_{p}

25200=−2.3×8.3×400log⁡(Kp)25200 = - 2 . 3 \times 8 . 3 \times 400\log \left({K}_{p}\right)

Kp=10−3.3=10−3×0.501{K}_{p} = {10}^{- 3 . 3} = {10}^{- 3} \times 0 . 501

=5.01×10−4Bar−1= 5 . 01 \times {10}^{- 4}{Bar}^{- 1}

=5.01×10−9Pa−1= 5 . 01 \times {10}^{- 9}{Pa}^{- 1}

=KC8.3×400= \dfrac{{K}_{C}}{8 . 3 \times 400}

KC=1.66×10−5m3/mole{K}_{C} = 1 . 66 \times {10}^{- 5}{m}^{3} / mole

=1.66×10−2L/mol= 1 . 66 \times {10}^{- 2}L / mol

=2= 2

Gibbs free energy can be visualized as the amount of useful energy present in a thermodynamic system that can be utilized to perform some useful work.

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