The decomposition of `N_(2)O_(5)` in `C Cl_(4)` solution at `318 K` has been studied by monitoring the concentration of `N_(2)O_(5)` in the solution. Initially, the concentration of `N_(2)O` is `2.33 M` and after `184 min`, it is reduced to `2.08 M`. The reaction takes place according to the equation:
`2N_(2)O_(5) rarr 4NO_(2) + O_(2)`
Calculate the average rate of this reaction in terms of hours, minutes, and seconds. What is the rate of Production of `NO_(2)` during this period?

Explanation

Average Rate `=1/2 {-(Delta[N_(2)O_(5)])/(Deltat)}=-1/2[((2.08-2.33)"mol"^(-1))/184"min"]`
`= 6.79xx 10^(-4)"mol"L^(-1)//min =(6.79xx10^(-4)"mol"L^(-1)min L^(-1))xx(60 min//h)`
`= 4.07xx 10^(-2)"mol"L^(-1)//h`
`= 6.79xx 10^(-4)"mol""L"^(-1)xx1 min//60s`
`=1.13xx10^(-5)"mol"L^(-1)s^(-1)`
It may be remembered that
Rate`=1/4{(Delta[NO_(2)])/(Deltat)}`
`(Delta[NO_(2)])/(Deltat)=6.79xx10(_4)xx"mol""L"^(-1)"min"^(-1)=2.72xx10^(-3)"molL"^(-1)"min"^(-1)`