Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Complex Reactions · Exam: NET JUNE 2018 · Marks: 2 · Difficulty: Medium
Difference between-activation energies of the reverse and forward steps of a reversible reaction is 9.212 RT . If the pre-exponential factor of the forward reaction is double that of the reverse reaction at the same temperature, the equilibrium constant for the reaction at that temperature will be $(\ln 10=2.303)$
(a)$1 \times 10^{4}$
(b)$2 \times 10^{4}$
(c)$1 \times 10^{-4}$
(d)$2 \times 10^{-4}$
Answer
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