On introfucing a catalyst at `500K`, the rate of a first order reaction increases by `1.718` times. The activation energy in the presence of a catalyst is `4.15 KJ mol^(-1)`. The slope of the polt of `k(s^(-1))` against `1//T` in the absence of catalyst is

Options

A. `+1`
B. `-1`
C. `+1000`
D. `-1000`

Answer

D

Explanation

`k = A e^(-E_(a)//RT)`
`k' = A e^(-E'_(a)//RT)`
(where `k =` rate constant for non catalysed reaction and `k' =` rate constant for catalysed reaction. `E_(a)=` activation energy for non-catalysed reaction and `E_(a')=` activation energy for catalysed reaction)
`(k')/(k) = e^(E_(a)-E_(a))`
Also given `k' = k + 1.718 k = 2.718 k`
`:. 2.718 = e^((E_(a) - E'_(a))/(RT))`
`loge 2.718 = (E_(a) - E'_(a))/(8.314 xx 10^(-3) xx 500)`
`E_(a) - E_(a') = 4.11`
`E_(a') = 4.15`
`:. E_(a) = 8.3 kJ// "mol"^(-1)`
`k = A e^(-E_(a)//RT)`
`:. log_(e)k = loge A - (E_(a))/(RT)`
This is an equation for straight line with slope
`= -(E_(a))/(R ) = -(8.3)/(8.3 xx 10^(-3))`
`= -1000`