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

Hint: 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.
\[{\text{Partial pressure of Gas A = Total pressure }} \times {\text{mole fraction of Gas A}}\]
\[{{\text{P}}_{\text{A}}}{\text{ = }}{{\text{P}}_{\text{T}}}{{\text{X}}_{\text{A}}}\]