The rate constant of a reaction is `1.5xx10^7 s^(-1) " at" 50^@C and 4.5xx10^7 s^(-1) " at " 100^@C` . Calculate the value of activation energy, `E_a` for the reaction.

Explanation

The Arrehenius equation is `k = A exp(-E_(a)//RT)`
For the two temperatures, we get
`log.(k_(2))/(k_(1)) = (E_(a))/(2.303R)((1)/(T_(2)) - (1)/(T_(1)))`
Substituting the given data, we get
`log.(4.5 xx 10^(7))/(1.5 xx 10^(7)) = (E_(a))/((2.303)(8.314 JK^(-1) mol^(-1)))((1)/(373 K)-(1)/(323K))`
or `E_(a) = ((2.303)(8.314)(373)(323))/(50)log(3.0)`
`= 22012.7 J mol^(-1) = 22.013 kJ mol^(-1)`
The value of `A` would be
`A = kexp(E_(a)//RT)`
`= (1.5 xx 10^(7) s^(-1))exp[(220127 J mol^(-1))/((8.314 J K^(-1) mol^(-1))(323 K))]`
`= (1.5 xx 10^(7) s^(-1)) (3630.44) = 5.45 xx 10^(10) s^(-1)`