The decomposition of N₂O₅ in CCl₄ at 318 K has been studied by monitoring the concentration of N₂O₅ in the solution. Initially the concentration of N₂O₅ is 2.33 M and after 184 minutes, it is reduced to 2.08 M. The reaction takes place according to the equation:

$2 N_{2} O_{5} (g) \to 4 {NO}_{2} (g) + O_{2} (g)$

Calculate the average rate of this reaction in terms of hours, minutes and seconds. What is the rate of production of NO₂ during this period?

Solution

\frac{1}{2} [- ∆ [N_{2} O_{5}] / ∆ (t)] = - \frac{1}{2} [2.08 - 2.33] M / 184 \min

= 6.79 \times 10^{- 4} M / \min = 6.79 \times 10^{- 4} M / \min \times 60 \min / 1 hr . = 4.07 \times 10^{- 2} M / hr = 6.79 \times 10^{- 4} M \times 1 \min / 60 s = 1.13 \times 10^{- 5} M / s

Again,

Rate = \frac{1}{4} d [{NO}_{2}] / dt

d [{NO}_{2}] / dt = 6.79 \times 10^{- 4} \times 4 {ML}^{- 1} / \min = 2.72 \times 10^{- 3} {ML}^{- 1} / \min

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