Chemical Kinetics Question 145
Question: A first order reaction is 50% completed in 20 minutes at $ 27{}^\circ C $ and in 5 minutes at $ 47{}^\circ C $ . The energy of activation of the reaction is:
Options:
A) 43.85 kJ/mol
B) 55.14 kJ/mol
C) 11.97 kJ/mol
D) 6.65 kJ/mol
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ {k_{1(300)}}=\frac{0.693}{20}; $
$ {k_{2(320)}}=\frac{0.693}{5} $ In $ \frac{{k_{2(320)}}}{{k_{1(300)}}}=\frac{E_{a}}{R}[ \frac{1}{T_1}-\frac{1}{T_2} ] $
$ E_{a}=\frac{2.303RT_1T_2}{(T_2-T_1)}\log \frac{k_2}{k_1} $
$ =\frac{2.303\times 8.314}{20\times 1000}\times 300\times 320\log 4 $
$ =55.14kJ/mol $
BETA
