On introducing a catalyst at 500 K, the rate of a first order reaction increases by 1.718 times. The activation energy in the presence of catalyst is 6.05 kJ `mol^(-1)` . The slope of the plot of In k`(sec^(-1))` against 1/T in the absence of catalyst is:

Options

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

Answer

D

Explanation

`("Rate in presence of catalyst")/("Rate in absence of catalyst")=Anitilog [+triangle]/[2.303RT]`
`E_(a)=E_(q)+2.25 =6.05 +2.25 =8.30 kj mol^(-1)`
ln k=ln`-A(E_(a))/(R )xx(1)/(T )`
slope =`(-E_(a))/(R )=(-8.3 xx1000)/(8.3)=-1000`