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Question
A catalyst lowers the activation energy of a certain reaction from 83.314 to 75 kJ mol−1 at 500 K. What will be the rate of reaction as compared to uncatalysed reaction? Assume other things are equal.
Options
Double
28 times
7.38 times
7.38 × 103 times
MCQ
Solution
7.38 times
Explanation:
`"k"_2/"k"_1 = ("Ae"^(-"E"_(a_2)//"RT"))/("Ae"^(-"E"_(a_1)//"RT")) = "e"^(("E"_(a_1) - "E"_(a_2))//"RT")`
`2.303 log "k"_2/"k"_1 = ("E"_("a"_1) - "E"_"a")/"RT"`
= `((83.314 - 75) xx 10^3)/(8.314 xx 500)`
= 2
log k2 = `2/2.303` = 0.868
Taking Antilog
k2 = 7.38
shaalaa.com
Temperature Dependence of Reaction Rates
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