What would be the activation energy of a reaction when the temperature is increased from `27^(@)C` to `37^(@)C` ?

Options

A. `534"kJ mol"^(-1)`
B. `53.4 "kJ mol"^(-1)`
C. `5.34 "kJ mol"^(-1)`
D. None of these

Answer

B

Explanation

Rate of reaction is doubled
`ln.(k_(2))/(k_(1))=(E_(a))/(R )[(T_(2)-T_(1))/(T_(1)-T_(2))]`
`ln.(k_(2))/(k_(1))=(E_(a))/(R )[(10)/(300xx310)]`
`E_(a)=9300 "R ln"_(2)=53.4 "kJ mol"^(-1)`