For two reactions (i) `P rarr` Product and (ii) `Q rarr` Product, the order of reaction (i) is `1` while that of reaction (ii) is `2`. At `347 K`, the energy of activation of reaction (i) is `55 kJ mol^(-1)` but whenever this reaction is carried out in the presence of catalyst at the same temperature, the energy of activation is `53 kJ mol^(-1)`. Moreover, for reaction (ii), when the temperature is increased form `298 K` to `308 K`, the rate of reaction increases as many times as for reaction (i) in the presnece of catalyst. Calculate the rate constant of reaction (ii) at `318 K`, if the pre-exponential factor for reaction (ii) is `3.56 xx 10^(9) mol^(-1) L s^(-1)`.
Note: Arrhenius equation does not depend on the order of reaction.

Explanation

(a) For reaction (i) `P rarr` Product,
`k_(1) = Ae^(-E_(1)//RT)` [In absence of catalyst]
`k_(2) = Ae^(-E_(2)//RT)` [In absence of catalyst]
`:. ln(k_(2))/(k_(1)) = ((E_(2) - E_(1)))/(RT)`
`= ((55-53)kJ mol^(-1))/(8.314 xx 10^(-3) kJ mol^(-1) K^(-1) xx 347 K)`
`= 0.694`
`:. 2.303 log.(k_(2))/(k_(1)) = 0.694`
`log.(k_(2))/(k_(1)) = (0.694)/(2.303) ~~ 0.3`
`:. (k_(2))/(k_(1)) = 2`
(b) For reaction (ii) `Q rarr` Product
`log((k_(308))/(k_(298))) = (E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`:. (k_(308))/(k_(298)) = (k_(2))/(k_(1)) = 2` (Also temperature coeffiecient is `2`)
Substitute the value of `k_(308)//k_(298)` in Eq. (ii),
`:. log 2 = (E_(a))/(2.303 xx 8.314 xx 10^(-3)kJ mol^(-1)K^(-1)) xx (10)/(298 xx 308)`
Solve for `E_(a)`:
`:. E_(a) = 68.127 kJ mol^(-1)`
(c ) Given: `A = 3.56 xx 10^(9)mol^(-1) L s^(-1), T = 318 K`
`:. log k = log A - (E_(a))/(2.303 RT)`
`:. log k = log(3.56 xx 10^(9)) - (68.127 kJ mol^(-1))/(2.303 xx 8.314 xx 318)`
Solve for `k`:
`:. k = 7.33 mol^(-1) L s^(-1)`