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 rate`=-1/2(Delta[N_(2)O_(5)])/(Deltat)=-1/2([2.08-2.33])/(184)`
`=6.79xx10^(-4) M min^(-1)`
`=6.79xx10^(-4)xx1/60 M s^(-1)`
`=1.13xx10^(-5) M s^(-1)`
`=6.79xx1-^(-4)xx60 M hr^(-1)`
`=4.076xx10^(-2) M hr^(-1)`
Also `=1/4(Delta[NO_(2)])/(Deltat)=-1/2(Delta[N_(2)O_(5)])/(Deltat)`
`(Delta[NO_(2)])/(Deltat)=-1/2(Delta[N_(2)O_(5)])/(Deltat)xx4`
`=6.79xx10^(-4)xx4`
`=2.716xx10^(-3) M min^(-1)`