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 (M)`
B. `1.2 (M)`
C. `0.04 (M)`
D. `0.8 (M)`

Answer

D

Explanation

Order of rection is 1st
`:. [N_(2)O_(5)] = ("rate")/("rate constant") = (2.4 xx 10^(-5))/(3 xx 10^(-5))`
`= 0.8 (M)`