Question

An amount of 3 moles of $N_2$ and some $H_2$ is introduced into an evacuated vessel. The reaction starts at t = 0 and equilibrium is attained at t = $t_1$. The amount of ammonia at t = $2t_1$ is found to be 34 g. It is observed that $\frac{w(N_2)}{w(H_2)} = \frac{14}{3}$ at t = $\frac{t_1}{3}$ and t=$\frac{t_1}{2}$. The only correct statement is

• A. w($N_2$) + w($H_2$) + w($NH_3$) = 118 g at t = $t_1$
• B. w($N_2$) + w($H_2$) + w($NH_3$) = 102 g at t = $2t_1$
• C. w($N_2$) + w($H_2$) + w($NH_3$) = 50 g at t = $t_1/3$
• D. w($N_2$) + w($H_2$) + w($NH_3$) cannot be predicted

Answer 1

Answer: (B)

w($N_2$) + w($H_2$) + w($NH_3$) = 102 g at t = $2t_1$

Answer 2

The reaction is $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$. Molar masses: $M(N_2) = 28$ g/mol, $M(H_2) = 2$ g/mol, $M(NH_3) = 17$ g/mol.

The mass ratio $\frac{w(N_2)}{w(H_2)} = \frac{14}{3}$ implies $\frac{n(N_2) \times 28}{n(H_2) \times 2} = \frac{14}{3}$, which simplifies to $\frac{n(N_2)}{n(H_2)} = \frac{1}{3}$. This mole ratio is the same as the stoichiometric ratio, meaning the initial mole ratio of $N_2$ to $H_2$ must have been $1:3$. Given initial moles of $N_2$ = 3 mol, the initial moles of $H_2$ must be $3 \times 3 = 9$ mol.

Initial mass of $N_2 = 3 \text{ mol} \times 28 \text{ g/mol} = 84 \text{ g}$. Initial mass of $H_2 = 9 \text{ mol} \times 2 \text{ g/mol} = 18 \text{ g}$. Total initial mass = $84 \text{ g} + 18 \text{ g} = 102 \text{ g}$.

By the law of conservation of mass, the total mass of reactants and products in a closed system is constant. Thus, $w(N_2) + w(H_2) + w(NH_3) = 102 \text{ g}$ at all times.