For reaction A `rarr` B , the rate constant `K_(1)=A_(1)(e^(-E_(a_(1))//RT))` and the reaction `X rarr Y` , the rate constant `K_(2)=A_(2)(e^(-E_(a_(2))//RT))` . If `A_(1)=10^(9)`, `A_(2)=10^(10)` and `E_(a_(1))`=1200 cal/mol and `E_(a_(2))` =1800 cal/mol , then the temperature at which `K_(1)=K_(2)` is : (Given , R=2 cal/K-mol)

Options

A. 300K
B. `300 xx 2.303 K`
C. `(303)/(2.303)K`
D. None of these

Answer

C

Explanation

`k=Ae^(-E_(a) // RT)`
If `k_(1)=k_(2)`
`A_(1)e^(-E_(a1) // RT)=A_(2)e^(-E_(a2) // RT)`
`(A_(2))/(A_(1))=e^((E_(a2)-E_(a1)) // RT)`
`10=Exp ((600)/(RT)), R=2 cal // K-mol`
`l n 10=(600)/(2T), T=(300)/(2.303)K`