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

`"Rate in the presence of catalyst"/("Rate in the absence of catalyst")="Antilog"[(+DeltaE)/(2.303RT)]`
`1.78="Antilog"=(E_(a)-E_(P))/(2.303xx8.314xx500)`
Where `E_(a)` and `E_(p)` are energy of activation in absenced and presence of catalyst.
`E_(a)-E_(p)= 8.3xx500xx10^(-3)`
`E_(a)=E_(p)+(8.3xx500xx10^(-3))`
`E_(a)=E_(p)+(8.3xx500xx10^(3))`
`=4.15+(8.3xx500xx10^(-3))`
`=8.3KJ mol^(-1)`
In `k=In A=(E_(a))/(R )xx(1)/(T)`
Slope `=(-E_(a))/(R )= (-8.3xx1000)/(8.3)= -1000`