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Rate constant k = 1.2 xx 10^(3) mol^(-1)...

Rate constant `k = 1.2 xx 10^(3) mol^(-1) L s^(-1)` and `E_(a) = 2.0 xx 10^(2) kJ mol^(-1)`. When `T rarr oo`:

A

`A=2.0xx10^(2)kJ"mol"^(-1)`

B

`A=1.2xx10^(3)" mol "L^(-1) S^(-1)`

C

`A=1.2xx10^(3)" mol"^(-1)LS^(-1)`

D

`A=2.4xx10^(3)kJ" mol"^(-1)S^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`k=A_e^(-E_a//RT)`
when, `Trarroo,k=Ae^(o)=A`
`A=k=1.2xx10^(3)"mol "Ls^-1`
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MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-CHEMICAL KINETICS -Exercise 2
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