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_(5)` 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

Rate of reaction
`= -(Delta[N_(2)O_(5)])/(Delta t)`
`= -((2.33 - 2.08)mol L^(-1))/(184 min)`
Note: `-ve` isgn is not included in calculation.
`= 1.36 xx 10^(-3)mol L^(-1) min^(-1)`
If time `= 184 min = 3.067 hr`
Rate `= -((2.33 - 2.08) mol L^(-1))/(3.067 hr)`
`= 8.16 xx 10^(-2) mol L^(-1) hr^(-1)`
If time `= 184 xx 60 = 11040 s`
Rate `= -((2.33 - 2.08) mol L^(-1))/(11040 s)`
`= 2.26 xx 10^(-5) mol L^(-1) s^(-1)`
Rate `= -(Delta[N_(2)O_(5)])/(2Delta t) = (Delta[NO_(2)])/(4Delta t)`
`:. (Delta(NO_(2)))/(Delta t) = -(4)/(2)([N_(2)O_(5)])/(Delta t)`
`= 2 xx (-[N_(2)O_(5)])/(Delta t) = 2 xx 1.36 xx 10^(-3) mol L^(-1) min^(-1)`
`= 2.72 xx 10^(-3) mol. L^(-1) min^(-1)`