The rate constant for the reaction `2N_(2)O_(5)rarr 4NO_(3)+O_(2)` is `3.0 xx10^(-4)s^(-1)`. If start made with `1.0 mol L^(-1)` of `N_(2)O_(5)`. Calculate the rate of formation of `NO_(2)` at the moment of the reaction when concentration of `O_(2)` is `0.1 mol L^(-1)`

Options

A. `2.7 xx 10^(-4)mol L^(-1)s^(-1)`
B. `2.4xx10^(-4)mol L^(-1)s^(-1)`
C. `4.8 xx 10^(-4) mol L^(-1)s^(-1)`
D. `9.6xx10^(-1)mol L^(-1)s^(-1)`

Answer

4

Explanation

`Mol L^(-1)` of `N_(2)O_(5)` reacted `=2xx0.1=0.2,[N_(2)O_(5)]`left`=1.0-0.2=0.8mol L^(-1)`
Rate of reaction `=kxx[N_(2)O_(5)]=3.0xx10^(-4)xx0.8=2.4xx10^(-4)mol L^(-1)s^(-1),`
Rate of formation of `NO_(2)=4xx2.4xx10^(-4)=9.6xx10^(-4)mol L^(-1)s^(-1),`