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A rarr B, k(A)=10^(15) e^(-2000//T) C ...

`A rarr B, k_(A)=10^(15) e^(-2000//T)`
`C rarr D, k_(c)=10^(14) e^(-1000//T)`
Temperature at which `k_(A)=k_(c)` is

A

1000 K

B

2000 K

C

`(2000//2.303)K`

D

`(1000//2.303)K`

Text Solution

Verified by Experts

The correct Answer is:
D
`k_(A)=k_(C )`
`10^(15)e^(-2000//T)=10^(14)e^(-1000//T)`
`10=e^(2000//T-1000//T)`
`10=e^(1000//T)`
`log_(e )10=(1000)/(T)log_(e )e`
`2.303 log_(10)10=(1000)/(T)`
`T=(1000)/(2.303)K`
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