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 $ .



NCERT Chapter Video Solution

Dual Pane