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. 0.8
B. 0.7
C. 1.2
D. 1

Answer

B

Explanation

Rate `= K [N_(2) O_(5)]` (first order as unit of rate constant is `s^(-1)`)
`[N_(2)O_(5)] = ("rate")/(k) = (1.4 xx 10^(-5) mol L^(-1) s^(-1))/(2 xx 20^(-5) s^(-1)) = 0.7 mol L^(-1)`