Chemical Kinetics Question 38

Question: The activation energy for a reaction is 9.0,K,cal/mol. The increase in the rate constant when its temperature is increased from 298K to 308K is [JIPMER 2000]

Options:

A) 63%

B) 50%

C) 100%

D) 10%

Show Answer

Answer:

Correct Answer: A

Solution:

2.303log⁡K2K1=EaR[T2−T1T1T2]

log⁡K2K1=9.0×1032.303×2[308−298308×298]

K2K1=1.63,;K2=1.63,K1;,1.63K1−K1K1×100=63.0

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