Published by:
CGP EDU Academic Team
Published on: August 14, 2026
For a reaction, A ⎯→ B ; if
log 10 K (sec –1 ) = 14 – 
K, the frequency factor and energy of activation for the reaction are -
Text Solution
Verified by ExpertsThe correct answer is:
A
og 10 K(sec –1 ) = 14 – 
log 10 K = log 10 A – 
log 10 A = 14
A = 10 14 sec –1 and 
= 1.25 × 10 4 ;
E a = 1.25 × 2.303 × 10 4 × 8.314 J
≈ 239 × 10 3 J
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