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On introducing a catalyst at 500 K, the ...

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:

A

`+1`

B

`-1`

C

`+1000`

D

`-100`

Text Solution

Verified by Experts

`("Rate in presence of catalyst")/("Rate in absence of catalyst") = "Antilog"[(+Delta E)/(2.303 RT)]`
`1.718 "Antilog " (E_(a) - E_(p))/(2.303 xx 8.314 xx 500) , E_(a) - E_(p) = 2.25 kJ`
`E_(a) = E_(p) + 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 xx 1000)/(8.3) = -1000`
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