The decomposition of $2 N_{2} O_{5} \to 2 N_{2} O_{4} + O_{2}$ is at $200^{o} C$ . If the initial pressure is 114 mm and after 25 min. of the reaction the total pressure of gaseous mixture is 133 mm. Calculate the average rate of the reaction in
a) $a t m m i n^{- 1}$ b) $m o l L^{- 1} s^{- 1}$ respectively.

0.002, 8.58 \times 10^{- 7}

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0.001, 8.58 \times 10^{- 7}

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0.002, 8.58 \times 10^{- 4}

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0.001, 8.58 \times 10^{- 3}

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Solution

The correct option is A $0.002, 8.58 \times 10^{- 7}$
The decomposition reaction is $2 N_{2} O_{5} \to 2 N_{2} O_{4} + O_{2} .$
The inital number of moles of $N_{2} O_{5}, N_{2} O_{4}$ and $O_{2}$ are $a - x, x$ and $\frac{x}{2}$ respectively.
After 25 min, the number of moles are $a + \frac{x}{2}$ . The pressure is 133 mm.
Thus the pressure due to $\frac{x}{2}$ moles corresponds to $133 - 114 = 19$ mm Hg.
The pressure due to x moles corresponds to $19 \times 2 = 38$ mm Hg.
Pressure due to (a-x) moles $= 114 - 76 = 38$ mm Hg.
The average rate of the reaction is $= \frac{38}{25 \times 760} = 0.002 a t m m i n^{- 1} .$
The ideal gas equation is $P V = n R T .$
$\frac{n}{V} = \frac{0.002}{0.0821 \times 473 \times 60} = 8.58 \times 10^{- 7} m o l l i t^{- 1} s^{- 1} .$

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