The rate of a reaction doubles when the tempeature changes from 300 K to 310 K. Activation energy for the reaction is: `(R = 8.314 JK^(-1) mol^(-1), log2 = 0.3010)`

Options

A. `60.5 kJ mol^(-1)`
B. `53.6 kJ mol^(-1)`
C. `48.6 kJ mol^(-1)`
D. `58.5 kJ mol^(-1)`

Answer

B

Explanation

b) According to the available data:
`k_(2)/k_(1)=2, T_(1)=27+273=300K`
`T_(2)=2, T_(1)=27+273=300K`
`T_(2)=37+273=310K, R=8.314 JK`
log `k_(2)/k_(1) = E_(a)/(2.303R)[1/T_(1)-1/T_(2)]`
`log2= E_(a)/(2.303 xx 8.314(JK^(-1)mol^(-1)))[1/(300K) -1/(310K)]`
`0.3010=E_(a)/(2.303 xx (8.314 Jmol^(-1))) xx (310-300)/(300xx310)`
`0.3010 = E_(a)/(2.303 xx (8.314J mol^(-1))) xx (10)/(300 xx 310)`
`E_(a)=(0.3010xx2.303xx8.314(Jmol^(-1))xx300 xx310)/(10)`
`=53598.5 Jmol^(-1)=53.6kJmol^(-1)`