For dissociation of ${NH}_{3} giving N_{2} and H_{2}$ gases, the partial pressures at equilibrium are 100, 80, 80 torr respectively. If some $N_{2}$ gas is removed and at new equilibrium partial pressure of $H_{2}$ becomes 128 torr then the partial pressure of $N_{2}$ remaining will be approx:

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71 torr

72 torr

8 torr

9 torr

answer is A.

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Detailed Solution

The reaction takes place as;

$2 {NH}_{3} ⇌ N_{2} + 3 H_{2} K_{P} = \frac{80 \times 80^{3}}{100^{2}}$

$\begin{array}{ccc} 100 & 80 & 80 \\ 100 - P & ? & 80 + \frac{3 P}{2} = 128 \\ = 68 & \Rightarrow P = 32 \end{array}$

$\begin{array}{l} \Rightarrow \frac{80 \times 80^{3}}{100^{2}} = \frac{P_{N_{2}} \times (128)^{s}}{68^{2}} \\ \Rightarrow P_{N_{2}} = 9 atm \end{array}$

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