The reaction `2NO_(2) rarr2NO+O_(2)` is second order respect to `NO_(2)`. If the intial concentration of `NO_(2)(g)` is `6.54 xx10^(-4) mol L^(-1)` and the intial reaction rate is `4.42 xx10^(-7) mol L^(-1) s^(-1)` what is the half life of the reaction.

Options

A. `1.48 xx10^(3)` s
B. `2.63 xx10^(4)` s
C. `1.96 xx10^(2)` s
D. `3.06 xx10^(3)` s

Answer

A

Explanation

Rate =`k[NO_(2)]^(2)` to find the rate constant solve the rate law with the given data
The half life of the system is found using the half life for a second order reaction
`t_(1//2)=(1)/(kxx("initinal concentration of reactant" ))=(1)/((1.03 L mol^(-1) s^(-1)(6.54 xx10^(-4) mol LA^(-1))=1.48 xx10^(3) S`