For a reaction `2N_2O_5 rarr 4NO_2 + O_2`, rate and rate constant are `1.02 xx 10^(-4)" mol " L^(-1)s^(-1)` and `3.4 xx 10^(-5)s^(-1)` The concentration of `N_2O_5` at that time will be

Options

A. 1.732 mol `L^(-1)`
B. 3 mol `L^(-1)`
C. `1.02xx10^(-4) " "mol" "L^(-1)`
D. `3.2xx10^5" mol "L^(-1)`

Answer

B

Explanation

Rate = rate constant `xx` [concentration`]^n`
`1.02xx10^(-4)=3.4xx10^(-5)xx[N_2O_5]` (unit of rate constant shows that its a first order reaction)
`[N_2O_5]=(1.02xx10^(-4))/(3.4xx10^(-5))=3` M