In general it is observed that the rate of a chemical reaction becomes double for every `10^@` rise in temperature. If this generalisation holds for a reaction in the temperature range 2908 K to 398 K, what would be the value of activation energy for the reaction.
(R=8.314 J `K^(-1)mol^(-1)`)

Explanation

`k_2//k_1=2,T_1=300K`, (say), `T_2=310K`
R=8.314 J `K^(-1)mol^(-1),E_a`=?
log`(k_2)/(k_1)=(E_a)/(2.303R)[(T_2-T_1)/(T_1T_2)]`
log 2=`(E_a)/(2.303xx8.314 JK^(-1)mol^(-1))[(310-300)/(310xx300K)]`
`0.3010=(E_a)/(19.15 JK^(-1)mol^(-1))xx(10)/(310xx300K)`
`E_a=0.3010xx19.15xx310xx30" J mol"^(-1)`
`E_a=53606" J mol"^(-1)=53.61J mol^(-1)`