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»

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

SOLUTION :`10^(16) e^((-2000)/(T)) = 10^(15) e^((-1000)/(T))`
`10 = e^((-1000)/(T))/(e^(-2000)/(T)) implies 10 e^((1)/(T) xx 1000) implies "ln" 10 = (1000)/(T)`
`implies 2.303` log 10 = `(1000)/(T) implies T = (1000)/(2.303) K` .


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