Question

In general, it is observed that the rate of a chemical reaction doubles with every `10^(@)` rise in temperature. If the generalisation holds for a reaction in the temperature range 295 K to 305 K, what would be the value of activation energy for the reaction?

Answer 1

According to Arrheneius equation:
`logk_(2)/k_(1) = E_(a)/(2.303R)[1/T_(1)-1/T_(2)]`
`k_(2)//k_(1)=2, T_(1)=295 K, T_(2)= 305K, R=8.314J K^(-1)mol^(-1)`
`log2= E_(a)/(2.303 xx (8.314 JK^(-1)mol^(-1)))[1/295K -1/305k]`
or `0.3010 = E_(a)/(2.303 xx (8.314Jmol^(-1))) xx (10)/(295 xx 305)`
`E_(a) = (0.310 xx 2.303 xx 8.314 xx 295 xx 305)/(10) (J mol^(-1))= 51855 J mol^(-1)`
`=51.855 kJ mol^(-1)`