Chemical Kinetics Question 56
Question: A hydrogenation reaction is carried out at 500 K. $ CH_2=CH_2+H_2\xrightarrow[no,catalyst]{500k}CH_3-CH_3 $ Activation energy $ -,E_{a},KJ,mo{l^{-1}} $
$ CH_2=CH_2+H_2\xrightarrow{pd,400k}CH_3-CH_3 $ Activation energy = ( $ E_{a}-20 $ ) KJ $ mo{l^{-1}} $ If rate remains constant, then $ E_{a} $ is
Options:
A) $ 120kJmo{l^{-1}} $
B) $ 100kJmo{l^{-1}} $
C) $ 20kJmo{l^{-1}} $
D) $ 80kJmo{l^{-1}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ k=A{e^{-E_{a}/RT}} $
$ k=A{e^{-E_{a}/R\times 500}} $
$ k’=A{e^{-(E_{a}-20)/R\times 400}} $ Rates are equal, hence
$ \therefore A{e^{-E_{a}/R\times 500}}=A{e^{-(E_{a}-20)/R\times 400}} $
$ \therefore \frac{E_{a}}{500}=\frac{E_{a}-20}{400} $
$ 4E_{a}=5E_{a}-100 $
$ \therefore E_{a}=100KJmo{l^{-1}} $
