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_(3)])/(Deltat)} =-1/2 [(2.08 - 2.33 mol L^(-1))/(184 "min")]`
`=6.79 xx 10^(-4) "mol" L^(-1)//"min` It may be understand that
Rate `=1/4 (Delta[NO_(2)])/(Deltat) = 6.79 xx 10^(-4) "mol"L^(-1)//min`
`(Delta[NO_(2)])/(Deltat) = 4 xx 6.79 xx 10^(-4) mol L^(-1) min^(-1)`
`=2.72 xx 10^(-3) mol L^(-1) min^(-1)`