The specific rate constant for a reaction increases by a factor if the temperature is changed from `27^(@)`C to `47^(@)`C. Find the activation energy for the reaction. (R = `198` cal )

Explanation

We know, that the activation energy at two different temperature, can be calculated by taking the help of the following equation
`log"" (k_2)/(k_1) = (E_a)/( 2.303 R) [(T_2 - T_1)/( T_2 T_1 ) ]`
Hence, log `(4k_1)/(k_1) = (E_a)/(2.303 xx 1.98 ) [(320 - 300)/(320 xx 300)]`
or, `0.6021 = (E_a)/(2.303 xx 1.98) xx (20)/(320 xx 4800)`
`therefore E_a = 0.6021 xx 2.303 xx 1.98 xx 4800`
` =13180" cal" = 13.18 " kcal"`.