Question

A mixture of ${N_2}$ and ${H_2}$ are in the molar ratio $1:3$ is allowed to attain equilibrium when 50% of mixture has reacted. If ${\rm P}$ is the equilibrium pressure then partial pressure of $N{H_3}$ formed is:
A) $\dfrac{P}{6}$
B) $\dfrac{P}{2}$
C) $\dfrac{P}{3}$
D) $\dfrac{P}{4}$

Answer 1

Hint: First of all, write the reaction of formation of $N{H_3}$ using ${N_2}$ and ${H_2}$. Then, calculate the number of moles left in the reaction. Find the total moles at the equilibrium and then the mole fraction. Use the formula of partial pressure.

Formula used: The formula for calculating the partial pressure can be –
$p = X \times \operatorname{P} $
where, $p$ is the partial pressure
$X$ is the mole fraction
$P$ is the total pressure

Complete answer:
The equilibrium can be defined as the state in which reactants and products are present in concentrations and the system does not show any change with respect to time. The chemical equilibrium is achieved when the rate of forward reaction is equal to the rate of reverse reaction.
From the question, the overall reaction can be written as –

	${N_2}$	$3H_2$	$\to$	$2NH_3$
Initial	1mole	3mole		0
At equilibrium	$1 - x$	$3 - 3x$		$2x$

From the above reaction, we can see that, out of the 4 moles, 2 moles have reacted but 2 moles are still left.
Now, calculating the value for $x$ -
$
  \therefore 1 - x + 3 - 3x = 2 \\
   \Rightarrow 4x = 4 - 2 \\
   \Rightarrow 4x = 2 \\
   \Rightarrow x = 0.5 \\
 $
Now, we got the value of $x$ as 0.5. Therefore, calculating the value of total moles at the equilibrium –
$
  \therefore 1 - x + 3 - 3x + 2x \\
   \Rightarrow 4 - 2x \\
   \Rightarrow 4 - 1 \\
   \Rightarrow 3 \\
 $
Hence, the total number of moles at the equilibrium are 3.
Number of moles of $N{H_3}$ at the equilibrium $ = 2x = 2 \times 0.5 = 1$
Hence, there is only 1 mole of $N{H_3}$ at the equilibrium.
Now, calculating the mole fraction for $N{H_3}$ -
We know that, mole fraction is given by the formula –
${X_A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}}$
where, A and B are the two components and ${n_A}\& {n_B}$ are the number of moles of the components A and B respectively.
$\therefore {X_{N{H_3}}} = \dfrac{1}{3}$
Now, the partial pressure is calculated by the formula –
$p = X \times \operatorname{P} $
Putting the values in the above formula, we get –
$
   \Rightarrow p = \dfrac{1}{3} \times P \\
  \therefore p = \dfrac{P}{3} \\
 $

Hence, the correct answer is option (C).

Note: The pressure that is exerted by one among the mixture of gasses if it occupies the same volume on its own is known as Partial pressure. It is the measure of thermodynamic activity of gas molecules.