Solid ${NH}_{4} I$ on rapid heating in a closed vessel at 357° C develops pressure of 300 mm Hg owing to partial decomposition ofNH4 I into NH3 and HI but .he pressure gradually increases further (when the excess solid residue remains in the vessel) owing to the dissociation of HI. Calculate the final pressure (in mm of Hg) developed at equilibrium. $K_{p}$ for HI dissociation is 2 at 357° C. $[HI ⇌ \frac{1}{2} H_{2} + \frac{1}{2} I_{2}]$ .

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answer is 900.

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Detailed Solution

In the reaction ammonia is obtained;

$\begin{array}{l} {NH}_{4} I (s) ⇌ {NH}_{3} (g) + HI (g) \\ 150 mm 150 mm \\ K_{P} = 150 \times 150 \\ HI ⇌ \frac{1}{2} H_{2} + \frac{1}{2} I_{2} \\ 150 - P \frac{P}{2} \frac{P}{2} \end{array}$

$\begin{array}{l} \Rightarrow \frac{\frac{P}{2}}{150 - P} = 2 = K_{P} \\ P = 600 - 4 P \\ 5 P = 600 \\ P = 120 \\ P_{HI} = 150 - 120 = 30 \\ 150 \times 150 = P_{{NH}_{3}} \times 30 \Rightarrow P_{{NH}_{3}} = 750 mm \\ Final total P = P_{{NH}_{3}} + P_{HI} + P_{H_{2}} + P_{I_{2}} \\ = 750 + 30 + 60 + 60 = 900 mm \end{array}$