दिया है - `(K_(t+10))/(K_t)=1.75`
`therefore" "T_1=273+25=298K, " "T_2=273+35=308K`
`therefore" "2.303log_(10)""K_2/K_1=E_a/R([T_2-T_1])/(T_1T_2)`
`therefore" "2.303log_(10)""1.75=E_a/(1.987)xx[(308-298)/(308xx298)]`
`E_a=10206.2xx10^3` कैलोरी/मोल
=10206.2 किलो-कैलोरी-मोल