The rate constant for the reaction, `2N_(2)O_(5) to 4NO_(2) + O_(2)` is `2 xx 10^(-5) s^(-1)`. If rate of reaction is `1.4 xx 10^(-5) mol L^(-1), s^(-1)` What will be the concentration of `N_(2)O_(5)` in 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),`