Question

Ammonia gas at $76cm\,Hg$ pressure was connected to a manometer. After sparking in the flask, ammonia is partially dissociated as follows:
$2N{H_3}\left( g \right) \rightleftharpoons {N_2}\left( g \right) + 3{H_2}\left( g \right)$
The level in the mercury column of the manometer was found to show a difference of $18cm$. What is the partial pressure of ${H_2}\left( g \right)$at equilibrium?
(A) $18cm\,Hg$
(B) $9cm\,Hg$
(C) $27cm\,Hg$
(D) $24cm\,Hg$

Answer 1

According to the ideal gas law, partial pressure is inversely proportional to volume. It is also directly proportional to moles and temperature.

Complete step by step answer:
According to the question, ammonia is partially dissociated to nitrogen gas and hydrogen gas as follows:
$2N{H_3}\left( g \right) \rightleftharpoons {N_2}\left( g \right) + 3{H_2}\left( g \right)$
It is given that ammonia $N{H_3}$ is at $76cm\,\,Hg$ pressure
Therefore, we can consider $76cm\,\,Hg$ as the initial pressure.
Initial, $\mathop {N{H_3}\left( g \right)}\limits_{76cm\,\,Hg} \rightleftharpoons \mathop {{N_2}\left( g \right)}\limits_0 + \mathop {3{H_2}\left( g \right)}\limits_0$
Let at time $= t$equilibrium is obtained and the dissociation (let) be $x$
At time = t$$$$\begin{array}{*{20}{c}} {76 - 2x}& &{3x} \end{array}
(Equilibrium)
Thus, we have written taking in view the stoichiometric coefficients.
The total pressure after dissociation $= 76 - 2x + x + 3x = 76 + 2x$
So, the increase in pressure $= 76 - 2x - 76 = 2x$
Increase in pressure = $18cm\,Hg$
(as given in the question)
$2x = 18$
$x = 9cm\,Hg$
We know that, the partial pressure of ${H_2}\left( g \right)$is $3x$ so the answer is $3 \times 9 = 27cm$

So, the correct answer is “Option C”.

Additional Information:
The total pressure of a mixture of gases can be defined as the sum of pressures of each individual gas
$P\,{\text{Total = }}{{\text{P}}_1} + {{\text{P}}_2} + {{\text{P}}_3} + {{\text{P}}_4}......$

Note:
The partial pressure of an individual gas is equal to the total pressure multiplied by mole fraction of that gas.
${\text{Partial pressure of Gas A = Total pressure }} \times {\text{mole fraction of Gas A}}$
${{\text{P}}_{\text{A}}}{\text{ = }}{{\text{P}}_{\text{T}}}{{\text{X}}_{\text{A}}}$