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))/(1.84 min)`
`=6.79 xx 10^(-4) mol L^(-1) min^(-1)`
`=(6.79 xx 10^(-4) mol L^(-1))/(min) xx (1min)/(60s) = 1.13 xx 10^(-5) mol L^(-1)s^(-1)`
`=(6.79 xx 10^(-4) mol L^(-1))/(min) xx (60min)/(1hr) = 4.07 xx 10^(-2) mol L^(-1)hr^(-1)`
Rate =`1/4 (Delta[NO_(2)])/(Deltat)= -1/2(Delta[N_(2)O_(5)])/(Deltat) = 6.79 xx 10^(-4) mol L^(-1)min^(-1)`
`therefore` Rate of production of `NO_(2)`, `(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)`