The rate constant of a certain reaction is given by:
`log k= 5.4-(212)/(T)+2.17 log T`
Calculate `E_(a) at 127^(@)C`.

Explanation

`log k= 5.4-(212)/(T)+2.17 log T`…..(i)
Multiply Eq. by `2,3`,
`2.3 logk= 5.4xx2.3-(212xx2.3)/(T)+2.17xx2.3 logT`
In `k=5.4xx2.3-(212xx2.3)/(T)+2.17In T`
`(d)/(dt)(In k)=0+(212xx2.3)/(T^(2))+(2.17)/(T)=(4.88.236+2.17T)/(T^(2))`.....(ii)
`(d)/(dt)(IN k)=E_(a)//RT^(2)`.....(iii)
Compare Eqs. (ii) and (iii).
`(E_(a))/(RT^(2))= (488.236+2.17)/(T^(2))`
`(E_(a))/(RT^(2))=488.236 +2.17T(R = 1.98~2 cal)`
`:. E_(a)= 2 cal[488.236+2.17xx400]= 2712 cal`
`= 2.712 kcal`