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

d) The unit of rate constant (k) indicates reaction to be of first order.
`therefore` rate = `k[N_(2)O_(5)]`
`[N_(2)O_(5)] = ("Rate")/k = (2.4 xx 10^(-5)mol L^(-1)s^(-1))/(3.0 xx10^(-5)s)`=0.8