What is the activation energy for a reaction if its rate doubles when the temperature is raised from `20^(@)C` to `35^(@)C`? `(R = 8.314 J "mol K"^(-))`

Options

A. `269 kJ mol^(-1)`
B. `34.7 kJ mol^(-1)`
C. `15.1 kJ mol^(-1)`
D. `342 kJ mol^(-1)`

Answer

B

Explanation

Using Arrhenius equation, `k=Ae^(-E_(a)//RT)` , we derive a relationship between the activation energy `E_(a)` and the specific rate constant that are measured at any two temperatures. If the temperatures are `T_(1)` and `T_(2)` and the specitic rate constants are `k_(1)` and `k_(2)` the relationship is
`log(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
or, solving for `E_(a):E_(a)=2.303R[(T_(1)T_(2))/(T_(2)-T_(1))][logk_(2)//k_(1)]`
Since rate doubles, we have `k_(2)=2k_(1)` . Thus
`E_(a)=(2.303)(8.314Jmol^(-1)K^(-1))`
`xx[((293K)(308K)/((308K-293K))][log(2k_(1))/(k_(1))]`
`=(2.303)(8.314Jmol^(-1)k^(-1))(6.016xx10^(3)k)(log2)`
`=34.7KJmol^(-1)(log2=0.301)`