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

To solve the problem, we will follow these steps:

### Step 1: Calculate the change in concentration of \( N_2O_5 \)
The initial concentration of \( N_2O_5 \) is given as \( 2.33 \, M \) and the final concentration after \( 184 \, \text{min} \) is \( 2.08 \, M \).

\[ \Delta [N_2O_5] = [N_2O_5]_{\text{final}} - [N_2O_5]_{\text{initial}} = 2.08 \, M - 2.33 \, M = -0.25 \, M \]
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