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 M
B. 3M
C. `1.02xx10^(-4)` M
D. `3.5xx10^(5)M`

Answer

B

Explanation

`r=k[N_(2)O_(5)]^(1)`
`1.02xx10^(4)=3.4xx10^(-5)[N_(2)O_(5)]`
`rArr[N_(2)O_(5)]=(1.02xx10^(-4))/(3.4xx10^(-5))=3M`