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Standard X

Chemistry

Industrial Preparation of Ammonia

Solid NH4I ...

Question

Solid $N H_{4} I$ on rapid heating in a closed vessel at $357^{\circ}$ develops a constant pressure of 275 mm Hg owing to partial decomposition of $N H_{4} I$ into $N H_{3}$ and HI but the pressure gradually increases further (when the excess solid residue remains in the vessel) owing to the dissociation of HI. Calculate the final pressure developed at equilibrium. $N H_{4} I (s) ⇌ N H_{3} (g) + H I (g)$ $2 H I (g) ⇌ H_{2} (g) + I_{2} (g), K_{c} = 0.065$ at $357^{\circ} C$ .

369 mm Hg

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375 mm Hg

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405 mm Hg

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412 mm Hg

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Solution

The correct option is D 369 mm Hg
The partial pressures of ammonia and HI are equal. They are one half of total pressure or 137.5 mm Hg.
$K_{p} = (137.5)^{2}$ .
For the dissociation of HI, the equilibrium concentration will be $(137.5 - 2 x), x a n d x$ respectively.
The expression for the equilibrium constant will be $\frac{x^{2}}{(137.5 - x)^{2}} = 0.065.$
$x = 27.95.$
Thus the pressure of $H I$ remaining is $137.5 - 2 (27.95) = 137.5 - 55.9 = 81.6$ .
The pressure of ammonia will be $(137.5)^{2} = 81.6 \times P_{N H_{3}}$
$P_{N H_{3}} = 231.7.$
The total pressure will be $231.7 + 81.6 + 55.9 = 369 m m H g .$

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