Gaseous `N_(2)O_(5)` decomposes according to the following equation:
`N_(2)O_(5)(g) rarr 2NO_(2)(g) + (1)/(2)O_(2)(g)`
The experimental rate law is `-Delta [N_(2)O_(5)]//Delta t = k[N_(2)O_(5)]`. At a certain temerature the rate constant is `k = 5.0 xx 10^(-4) sec^(-1)`. In how seconds will the concentration of `N_(2)O_(5)` decrease to one-tenth of its initial value ?

Options

A. `2.0 xx 10^(3)s`
B. `4.6 xx 10^(3)s`
C. `2.1 xx 10^(2)s`
D. `1.4 xx 10^(3) s`

Answer

B

Explanation

`t = (2.303)/(5 xx 10^(-4)) "log"(A_(0))/(A_(0)//10)`
`= 4.6 xx 10^(3) sec`.