1.

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

Answer»

`1000 K`
`(2000)/(2.303)K`
`2000 K`
`(1000)/(2.303)K`

Solution :`k_1=10^16eF^(-2000//T), k_2=10^(15)e^(-1000//T`
When `k_1=k_2,10^16e^(-2000//T)=10^15e^(-1000//T)`
or`10E^(-2000//T)=e^(=-1000//T)`
Taking natural logarithm of both SIDES, we get
In `10-2000/T=(-1000)/T" or "2*303-2000/T=(-1000)/T`
or `1000/T=2*303` or `T=1000/(2*303)K`


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