Chemical Kinetics Question 22
Question: The rate constant is doubled when temperature increases from 27°C to 37°C. Activation energy in kJ is [JEE Orissa 2004]
Options:
A) 34
B) 54
C) 100
D) 50
Show Answer
Answer:
Correct Answer: B
Solution:
$ \log \frac{K_2}{K_1}=\frac{E_{a}}{2.303R}[ \frac{1}{T_1}-\frac{1}{T_2} ] $ If $ \frac{K_2}{K_1}=2 $
$ \log 2=\frac{E_{a}}{2.303\times 8.314}[ \frac{1}{300}-\frac{1}{310} ] $
$ E_{a}=.3010\times 2.303\times 8.314( \frac{300\times 310}{10} ) $
$ =53598.59\ Jmo{l^{-1}} $
$ =54\ kJ $ .
