The decomposition of $N_{2}O_{5}$ in $CCl_{4}$ 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 $2.33 \ mol \ L^{-1}$ and after $184 \ minutes$,it is reduced to $2.08 \ mol \ L^{-1}$. The reaction takes place according to the equation:
$2N_{2}O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$
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?

(N/A) Average Rate $= \frac{1}{2} \left\{ -\frac{\Delta[N_{2}O_{5}]}{\Delta t} \right\} = -\frac{1}{2} \left[ \frac{(2.08-2.33) \ mol \ L^{-1}}{184 \ min} \right] = 6.79 \times 10^{-4} \ mol \ L^{-1} \ min^{-1}$
In terms of hours:
Rate $= (6.79 \times 10^{-4} \ mol \ L^{-1} \ min^{-1}) \times (60 \ min / 1 \ h) = 4.07 \times 10^{-2} \ mol \ L^{-1} \ h^{-1}$
In terms of seconds:
Rate $= (6.79 \times 10^{-4} \ mol \ L^{-1} \ min^{-1}) / (60 \ s / 1 \ min) = 1.13 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$
Rate of production of $NO_{2}$:
Rate $= \frac{1}{4} \frac{\Delta[NO_{2}]}{\Delta t} \implies \frac{\Delta[NO_{2}]}{\Delta t} = 4 \times \text{Average Rate}$
$= 4 \times 6.79 \times 10^{-4} \ mol \ L^{-1} \ min^{-1} = 2.72 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$