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a) The rate of a particular reaction dou...

a) The rate of a particular reaction doubles when the temperature changes from 300 K to 310 K. Calculate the energy of activation of the reaction. [Given : `R="8.314 JK"^(-1)" mol"^(-1)`].

A

`53.6 "kJ mol"^(-1)`

B

`48.6 "kJ mol"^(-1)`

C

`58.5 "kJ mol^(-1)`

D

`60.5"kJ mol"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
( A) log`(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`
log `2 = (E_(a))/(2.303xx8.314)((310-300)/(310xx300))`
`E_(a)= (0.3010xx2.303xx8.314xx310xx300)/(10)`
= 53.598 kJ/mol
or 53.6 kJ / mol .
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