At \( 298 \mathrm{~K}, 3 \mathrm{M} \) of hydrogen gas and 1 M of nitrogen gas were added to the reaction vessel. At equilibrium, 1 M of hydrogen gas was left. Determine the concentration of nitrogen gas and ammonia at equilibrium.

To solve this problem, we start with the following balanced reaction for the synthesis of ammonia: \[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \] Initially, we have: - \([H_2] = 3 \, \text{M}\) - \([N_2] = 1 \, \text{M}\) - \([NH_3] = 0 \, \text{M}\) At equilibrium, it is given that the concentration of hydrogen gas left is 1 M. This means that 2 M of hydrogen was consumed (3 M initial - 1 M equilibrium). According to the stoichiometry of the reaction: - For every 3 M of \(H_2\) that reacts, 1 M of \(N_2\) is consumed and 2 M of \(NH_3\) is produced. - Therefore, if 2 M of \(H_2\) reacts, it will consume \(\frac{2 M}{3} = \frac{2}{3} M\) of \(N_2\) and produce \(2 \cdot \frac{2}{3} M = \frac{4}{3} M\) of \(NH_3\). Now, let’s calculate the equilibrium concentrations: - \([N_2] = 1 \, \text{M} - \frac{2}{3} \, \text{M} = \frac{1}{3} \, \text{M}\) - \([NH_3] = 0 + \frac{4}{3} \, \text{M} = \frac{4}{3} \, \text{M}\) So, at equilibrium, the concentration of nitrogen gas is \(\frac{1}{3} \, \text{M}\) and the concentration of ammonia is \(\frac{4}{3} \, \text{M}\).