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A mixture of ${NH}_{3} (g)$ and $N_{2} H_{4} (g)$ is placed in a sealed container at 300 K. The total pressure of the gas is 0.5 atm. The container is heated to 1200 K, where the following decomposition reactions take place.
$2 {NH}_{3} (g) ⟶ N_{2} (g) + 3 H_{2} (g)$ and $N_{2} H_{4} (g) ⟶ N_{2} (g) + 2 H_{2} (g)$
The pressure of the vessel at this stage becomes 4.5 atm. The mole per cent of $N_{2} H_{4} (g)$ in the original mixture was

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Detailed Solution

Let x be the mole fraction of N₂H₄ in the original mixture.

Amount of gas to start with $= 1 m o l$

Amount of gas at the end $= [3 x + 2 (1 - x)] mol = (2 + x) mol$

Now $p V = n R T$ gives

$\frac{p 1}{p_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}}$ ; or $(\frac{0.5 atm}{4.5 atm}) = (\frac{1}{2 + x}) (\frac{300 K}{1200 K})$

Hence, $x = \frac{4.5}{0.50 \times 4} - 2 = 0.25$

Mole per cent of $N_{2} O_{4} = 25 %$

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