For the reaction, `2N_(2)O_(5)to4NO_(2)+O_(2)` rate and rate constant are `1.02xx10^(-4) M sec^(-1)` and `3.4xx10^(-5)sec^(-1)` respectively, the concentration of `N_(2)O_(5)`, at that time will be

Options

A. `3.4 xx 10^(-4)`
B. 3
C. 5.2
D. `3.2 xx 10^(-5)`

Answer

B

Explanation

`[N_(2) O_(5)] = ("Rate")/(k) = (1.02 xx 10^(-4) mol L^(-1) s^(-1))/(3.4 xx 10^(-5) s^(-1))`
`3 mol L^(-1)`