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Rate constant K varies with temperature ...

Rate constant K varies with temperature by equation log `K("min"^(-1))=5-((2000))/T`.We can conclude that `(R=8.314J"mol"^(-1)K^(-1)` (or) cal `"mol"^(-1)K^(-1)`)

A

Pre exponential factor A is 5

B

Ea is 4 k cal/mol

C

Pre exponential factor, A is `10^(5)`

D

Ea is 9.212 Kcal/mol

Text Solution

Verified by Experts

The correct Answer is:
C, D
`K=A.e^(-Ea//RT),logK=logA-(epsilon_(a))/(2.303R)xx1/T`
`logA=5,A=10^(5),(E_(a))/(2.303RT)=2000/T,E_(a)=(2000xx2.303xx2)/1,E_(a)=9.212KCal`
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