The decompoistion of `N_(2)O_(5)` in `C CI_(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 reaction rate `= -(1)/(2) ((Delta[N_(2)O_(5)])/(Delta t))`
`= -(1)/(2) (([2.08-2.33]mol L^(-1))/(184 min))`
`= -(1)/(2) ((-0.25 mol L^(-1))/(184 min))`
`= 6.79xx10^(-4)mol L^(-1) min^(-1)`
Average reaction rate in seconds
`= 6.79xx10^(-4) (mol L^(-1))/(min)xx(1min)/(60s)`
`= 0.113xx10^(-4) mol L^(-1)s^(-1)`
`= 1.13xx10^(-5)mol L^(-1)s^(-1)`
Average reaction rate in hours
`= 6.79xx10^(-4)(mol L^(-1))/(min)xx(60min)/(1h)`
`= 407.4xx10^(-4)mol L^(-1)h^(-1)`
`= 4.07xx10^(-2) mol L(-1)h^(-1)`