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In a 3L vessel, the following equilibrium partial pressures are measured;
$N_{2} = 100 mm of Hg,$ $H_{2} = 400 mm of Hg$ and ${NH}_{3} = 1000 mm of Hg$ .Nitrogen is removed from the vessel at constant temperature until the pressure of hydrogen at equilibrium is equal to $700 mm of Hg$ . The new equilibrium partial pressure (in mm of Hg) of $N_{2}$ is

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Detailed Solution

Given,

$P_{N_{2}} = 100 mm$

$P_{H_{2}} = 400 mm$

$P_{N_{3}} = 1000 mm$

At equilibrium, $P_{H_{2}} = 700 mm$

New equilibrium $P_{N_{2}} = ?$

$\underset{100}{N_{2}} + \underset{400}{3 H_{2}} ⇌ \underset{1000}{2 {NH}_{3}} \Rightarrow K_{P} = \frac{{(1000)}^{2}}{{(400)}^{3} \times {(100)}^{2}}$

After removal of N₂, eqm shifts backward and N₂, H₂ are formed

$P_{H_{2 (n e w)}} = 700 \Rightarrow P_{H_{2 (f o r m a l)}} = 300 m m$

$\Rightarrow P_{{NH}_{3}} decreases by 200 mm$

$∴ P_{H_{2}} = 700, P_{{NH}_{3}} = 800, If P_{N_{2}} = P$

Then, $\frac{{(1000)}^{2}}{{(400)}^{3} {(100)}^{2}} = \frac{{(800)}^{2}}{{(700)}^{3} . P} \Rightarrow P = \frac{8^{2} \times 4^{3}}{7^{3}}$

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