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JEE Main 201912 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2019 · 12 April · Evening Shift · Chemistry · Thermodynamics (C)

The incorrect match in the following is:

The incorrect match in the following is:

Options

  1. AΔG0<0,K>1\Delta{G}^{0} < 0, K > 1
  2. BΔG0>0,K<1\Delta{G}^{0} > 0, K < 1
  3. CΔG0=0,K=1\Delta{G}^{0} = 0, K = 1
  4. DΔG0<0,K<1\Delta{G}^{0} < 0, K < 1

Correct answer

D. ΔG0<0,K<1\Delta{G}^{0} < 0, K < 1

Step-by-step solution

We know that \DeltaG°=−RTlnK\DeltaG ° = - RTlnK

\DeltaG°=−2.303RTlogK\DeltaG ° = - 2.303RTlog K

so for k=1k = 1 , logK=0log K = 0

hence \DeltaG°=0;\DeltaG ° = 0;

for k>1k > 1 , logK>0log K > 0

\DeltaG°=−ve\DeltaG ° = - ve

for k<1k < 1 , logK<0log K < 0

\DeltaG°>0\DeltaG ° > 0

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