Chemical Kinetics Question 272
Question: A catalyst lowers the activation enegy of a reaction from $ 20kJmo{l^{-1}} $ to $ 10kJmo{l^{-1}} $ . The temperature at which the uncatalyzed reaction will have the same rate as that of the catalyzed at $ 27{}^\circ C $ is
Options:
A) $ -123{}^\circ C $
B) $ 327{}^\circ C $
C) $ 32.7{}^\circ C $
D) $ +23{}^\circ C $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \frac{E’a}{T_1}=\frac{E’a}{T_2}=\frac{10}{300}=\frac{20}{T_2} $
$ \therefore {T_2}=600K=327{}^\circ C $
