Chemical Kinetics Question 209
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% $
BETA
