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the rate constants k(1) and k(2) for...

the rate constants `k_(1) and k_(2)` for two different reactions are `10^(16).e^(-2000//T)` and `10^(15).e^(-1000//T)`, respectively the temperature at which `k_(1) = k_(2) ` is

A

2000 K

B

`1000//2.303K`

C

1000 K

D

`2000//2.303K`

Text Solution

Verified by Experts

The correct Answer is:
B
`k_(1)=10^(16)e^(-2000//T)`
`k_(2)=10^(15)e^(-1000//T)`
The temperature at which `k_(1)=k_(2)` will be
`10^(16)e^(-2000//T)=10^(15)e^(-1000//T)`
`(e^(-2000//T))/(e^(-1000//T))=(10^(15))/(10^(16))`
`e^(-1000//T)=10^(-1)`
`log_(e)e^(-1000//T)=log_(e)10^(-1)`
`2.303xxlog_(10)E^(-1000//T)=2.303xxlog_(10)10^(-1)`
`(-1000)/(T)xxlog_(10)e=-1`
On solving, we get
`T=1000//2,303K`
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