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:

$ 2.303\log \frac{K_2}{K_1}=\frac{E_{a}}{R}[ \frac{T_2-T_1}{T_1T_2} ] $

$ \log \frac{K_2}{K_1}=\frac{9.0\times 10^{3}}{2.303\times 2}[ \frac{308-298}{308\times 298} ] $

$ \frac{K_2}{K_1}=1.63,;K_2=1.63,K_1;,\frac{1.63K_1-K_1}{K_1}\times 100=63.0% $



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