The rate constant `k`, for the reaction `N_(2)O_(5)(g) rarr 2NO_(2) (g) + (1)/(2) O_(2)(g)` is `2.3 xx 10^(-2) s^(-1)`. Which equation given below describes the change of `[N_(2)O_(5)]` with time ? `[N_(2)O_(5)]_(0)` and `[N_(2)O_(5)]_(t)` correspond to concentration of `N_(2)O_(5)` initially and at time, `t` ?

Options

A. `[N_(2)O_(5)]_(t) = [N_(2)O_(5)]_(0) + kt`
B. `[N_(2)O_(5)]_(0) = [N_(2)O_(5)]_(t)e^(kt)`
C. `log_(10) [N_(2)O_(5)]_(t) = log_(10) [N_(2)O_(5)]_(0) - kt`
D. In `([N_(2)O_(5)]_(0))/([N_(2)O_(5)]_(t)) = kt`

Answer

D

Explanation

Rate constant `= 2.3 xx 10^(-2) sec^(-1)`
It means it is a first order reaction (because unit of rate constant is `sec^(-1)`)
For first order reaction`K = (1)/(t)ln(a)/(a-x)`
`Kt = ln(a)/(a-x) = ln[N_(2)O_(5)]_(0)/([N_(2)O_(5)]_(x)`