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. `1.732`
B. `3`
C. `1.02 xx 10^(-4)`
D. `3.4 xx 10^(-5)`

Answer

B

Explanation

`2N_(2) O_(5) to 4NO_(2) + O_(2)`
`- (-d [N_2 O_5])/( dt) = k , [N_2 O_5]`
`1.02 xx 10^(-4) = 3.4 xx 10^(-5)" s"^(-1) xx [N_2 O_5]`
`[N_2 O_5] = (1.02 xx 10^(-4) )/( 3.4 xx 10^(-5) ) = 3`