The rate constant for the reaction, `2N_(2)O_(5) rarr 4NO_(2) + O_(2)` is `3.0 xx 10^(-5) s^(-1)`. If the rate is `2.40 xx 10^(-5) mol L^(-1) s^(-1)`, then the initial concentration of `N_(2)O_(5)` (in `mol L^(-1)`) is

Options

A. `1.4`
B. `1.2`
C. `0.04`
D. `0.8`

Answer

D

Explanation

Rate `= k[N_(2)O_(5)]`
`2.4 xx 10^(-5) mol L^(-1) s^(-1) = (3.0 xx 10^(-5) s^(-1)) [N_(2)O_(5)]`
`[N_(2)O_(5)] = (2.4 xx 10^(-5) mol L^(-1) s^(-1))/(3.0 xx 10^(-5) s^(-1)) = 0.8 mol L^(-1)`
Since the unit of `k` is `s^(-1)`, hence the decompoistion of `N_(2)O_(5)` is first order reaction.