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# A manometer attached to a flask containing ammonia gas has no difference in mercury level initially as shown in diagram. After sparking into the flask, ammonia is partially dissociated as $2N{H_3}(g) \to {N_2}(g) + 3{H_2}(g)$ now it have difference of 6 cm in mercury level in two columns, what is partial pressure of ${H_2}(g)$  at equilibrium?(A) 6 mm Hg(B) 18 mm Hg(C) 27 mm Hg (D) None of these

Last updated date: 01st Mar 2024
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Hint: According to ideal gas law, partial pressure is inversely proportional to volume. It is directly proportional to the number of moles and temperature. Then the pressure at initial and pressure at equilibrium will be found.

We have been given the question that a manometer attached to the flask contains ammonia gas.
Then we have been given that ammonia is partially dissociated to nitrogen gas and hydrogen gas as follows
$2N{H_3}(g) \to {N_2}(g) + 3{H_2}(g)$
It is given that ammonia is at $76$ cm Hg pressure. Therefore the initial pressure will be 76 cm Hg
At equilibrium, the pressure will be $76 - 2x,x,3x$ respectively. So the total pressure will be written as $76 - 2x + x + 3x = 76 + 2x$
So the total pressure at equilibrium is $76 + 2x$
We can observe from here that the increase in pressure is $76 - (76 + 2x) = 2x$
It is given that difference in mercury level is six cm
$\Rightarrow 2x = 6$
$\Rightarrow x = 3$ cm
So the partial pressure of hydrogen gas $= 3x$
$\Rightarrow 3 \times 3 = 9\;cm\;Hg$
So the partial pressure of hydrogen is $9\;cm\;Hg$

So, the correct answer is Option D.

The partial pressure of gas $=$ Total pressure $\times$ mole fraction of gas
$\Rightarrow P = {P_T}{X_A}$