For a reaction, consider the plot of In K versus `1//T` given in the figure. If the rate constant of this reaction at 400 K is `10^(-5)s^(-1)`, then the rate constant at `500 K` is :

Options

A. `10^(-4) s^(-1)`
B. `10^(-6) s^(-1)`
C. `4xx10^(-4) s^(-1)`
D. `2xx10^(-4) s^(-1)`

Answer

A

Explanation

In k = In `A- (E_(a))/(RT)`
`y = m x +c`
`"|Slope| "=m =(E_(a))/(R )=4606`
In `((k_(2))/(10^(-5)))=+4606xx((500-400))/(500xx400)rArr" In "((k_(2))/(10^(-5)))=2.303 rArr" In "((k_(2))/(10^(-5)))=" In 10"`
`k=10^(-4)`