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For the reaction , A to B , k1=10^8 e^(-...

For the reaction , `A to B , k_1=10^8 e^(-6000/"8.34 T")` and `P to Q , k_2=10^10 e^(-8000/"8.34T")`
The temperature at which `k_1=k_2` is

A

386 K

B

221 K

C

26 K

D

52 K

Text Solution

Verified by Experts

The correct Answer is:
D
`k_1=k_2`
`10^8 e^(-6000/"8.34 T")=10^10 e^(-8000/"8.34 T")`
`10^8/10^10 = e^((-2000)/(8.34 T))`
ln `10^(-2) = (-2000)/(8.34 T) rArr -ln 100=(-2000)/(8.34 T)`
`rArr` 2.303 log 100= `2000/(8.34T) rArr T= 2000/(2.303xx2xx8.34)`= 52 K
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