Question

In what mole ratio, a mixture of $N_2$ and $NH_3$ gases should be taken such that the mole-fraction of $N_2$ at any instant remains constant, even after dissociation of $NH_3$ into $N_2$ and $H_2$? Mole fraction is the ratio of mole of that gas and the total mole of gases.

• A. 1 : 1
• B. 1 : 2
• C. 1 : 3
• D. 3 : 2

Answer 1

Answer: (A)

1 : 1

Answer 2

Let the initial moles of $N_2$ and $NH_3$ be $N$ and $A$ respectively. The initial mole fraction of $N_2$ is $\frac{N}{N+A}$.

When $NH_3$ dissociates according to $2NH_3 \rightleftharpoons N_2 + 3H_2$, let the degree of dissociation be $\delta$.

The moles of $NH_3$ that dissociate are $\delta A$.

The moles of $N_2$ produced are $\frac{1}{2} \delta A$. The moles of $H_2$ produced are $\frac{3}{2} \delta A$.

At any instant, the moles of gases are: $n_{N_2} = N + \frac{1}{2} \delta A$, $n_{NH_3} = A(1-\delta)$, $n_{H_2} = \frac{3}{2} \delta A$.

The total moles are $n_{total} = (N + \frac{1}{2} \delta A) + A(1-\delta) + \frac{3}{2} \delta A = N + A + \delta A$.

The mole fraction of $N_2$ at this instant is $x_{N_2} = \frac{N + \frac{1}{2} \delta A}{N + A + \delta A}$.

For the mole fraction of $N_2$ to remain constant, $x_{N_2}$ must be equal to the initial mole fraction $x_{N_2, 0}$ for any $\delta \in [0, 1]$.

$\frac{N + \frac{1}{2} \delta A}{N + A + \delta A} = \frac{N}{N + A}$

Cross-multiplying gives $(N + \frac{1}{2} \delta A)(N + A) = N(N + A + \delta A)$.

Expanding gives $N(N+A) + \frac{1}{2}\delta A(N+A) = N(N+A) + N\delta A$.

Subtracting $N(N+A)$ from both sides gives $\frac{1}{2}\delta A(N+A) = N\delta A$.

$\frac{1}{2}\delta AN + \frac{1}{2}\delta A^2 = N\delta A$.

For this equation to hold for any $\delta \in (0, 1]$ (assuming $A > 0$), we can divide by $\delta A$:

$\frac{1}{2}N + \frac{1}{2}A = N$.

Solving for $A$: $\frac{1}{2}A = N - \frac{1}{2}N = \frac{1}{2}N$, which gives $A = N$.

Thus, the initial number of moles of $N_2$ must be equal to the initial number of moles of $NH_3$. The mole ratio of $N_2$ to $NH_3$ is $N : A = N : N = 1 : 1$.