The rate constant for the reaction :
`2N_(2) O_(5) rarr 4NO_(2) +O_(2)`
is `3.0xx10^(-5) "sec"^(-1)` . If the rate is `2.40xx10^(-5)` mol `"litre"^(-1) "sec"^(-1)` then the concentration of `N_(2)O_(5)` (in mol `"litre"^(-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.