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)molL^(-1))/(1*84min)=6*79xx10^(-4)molL^(-1)min^(-1)`
`=(6*79xx10^(-4)molL^(-1))/(min)xx(1min)/(60s)=1.13xx10^(-5)molL^(-1)s^(-1)`
`=(6*79xx10^(-4)molL^(-1))/(min)xx(60min)/(1hr)=4*07xx10^(-2)molL^(-1)hr^(-1)`
Rate `=(1)/(4)(Delta[NO_(2)])/(Deltat)=-(1)/(2)(Delta[N_(2)O_(5)])/(Deltat)=6*79xx10^(-4)molL^(-1)min^(-1)" "("calculated above")`
`:." Rate of production of "NO_(2),(Delta[NO_(2)])/(Deltat)=4xx6*79xx10^(-4)molL^(-1)min^(-1)=2*72xx10^(-3)molL^(-1)min^(-1)`