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 kı = ky is

Answer»

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

Solution : `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^(6)e^(-2000//T)=10^(15)e^(-1000//T)`
`implies (e^(-2000//T))/(e^(-1000//T))=(10^(15))/(10^(16))`
`implies e^((-1000)/(T)) = 10^(-1) implies log_e e^((-1000)/(T)) = log_e10^(-1)`
`implies 2.303 log_(10) e^((-1000)/(T)) = 2.303xxlog_(10)10^(-1)`
`implies (-1000)/(T)xxlog_(10) e=-1 therefore T=(1000)/(2.303)K`


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