Question

\(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) For the reaction initially the mole ratio was \(1: 3\) of \(\mathrm{N}_{2}: \mathrm{H}_{2} .\) At equilibrium \(50 \%\) of each has reacted. If the equilibrium pressure is \(p\), the partial pressure of \(\mathrm{NH}_{3}\) at equilibrium is : (a) \(\frac{P}{3}\) (b) \(\frac{P}{4}\) (c) \(\frac{P}{6}\) (d) \(\frac{P}{8}\) \(\therefore\)

Short answer

The partial pressure of \( \mathrm{NH}_{3} \) at equilibrium is \( \frac{P}{3} \).

Step-by-step solution

- Understanding the Chemical Reaction

Note that the given reaction is: \( \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \). Initially, the mole ratio of \( \mathrm{N}_{2} \) to \( \mathrm{H}_{2} \) is 1:3. According to the stoichiometry of the reaction, 1 mole of \( \mathrm{N}_{2} \) reacts with 3 moles of \( \mathrm{H}_{2} \) to produce 2 moles of \( \mathrm{NH}_{3} \).

- Determining the Amount of Reactants Reacted

At equilibrium, 50% of each reactant has reacted. If we start with 1 mole of \( \mathrm{N}_{2} \) and 3 moles of \( \mathrm{H}_{2} \), then at equilibrium, 0.5 moles of \( \mathrm{N}_{2} \) and 1.5 moles of \( \mathrm{H}_{2} \) will have reacted to form \( \mathrm{NH}_{3} \).

- Calculating the Moles of Ammonia Produced

Since 0.5 moles of \( \mathrm{N}_{2} \) react to form \( \mathrm{NH}_{3} \), and the reaction produces 2 moles of \( \mathrm{NH}_{3} \) for every mole of \( \mathrm{N}_{2} \) used, it will produce 1 mole of \( \mathrm{HN}_{3} \) (because 0.5 x 2 = 1).

- Relating Moles and Partial Pressure

The total moles of gas at equilibrium will be the sum of unreacted and reacted species. This is 0.5 moles \( \mathrm{N}_{2} \) + 1.5 moles \( \mathrm{H}_{2} \) + 1 mole \( \mathrm{NH}_{3} \) = 3 moles. Since the equilibrium pressure is denoted as \(p\), and we have a total of 3 moles of gas where 1 mole is \( \mathrm{NH}_{3} \), the mole fraction of \( \mathrm{NH}_{3} \) is \(\frac{1}{3} \) of the total number of moles.

- Calculating the Partial Pressure of Ammonia

The partial pressure of \( \mathrm{NH}_{3} \) at equilibrium is equal to the mole fraction of \( \mathrm{NH}_{3} \) multiplied by the total pressure \( p \). Therefore, the partial pressure of \( \mathrm{NH}_{3} \) is \( \frac{1}{3}p \).

Key concepts

Le Chatelier's Principle

Le Chatelier's principle is a central concept in chemical equilibrium that describes how a system at equilibrium responds to changes in concentration, temperature, or pressure. Essentially, it states that if a stress is applied to a system in equilibrium, the system will adjust in a way to counteract the change and restore a new equilibrium.

For example, in the synthesis of ammonia (NH_3) from nitrogen (N_2) and hydrogen (H_2), if more N_2 is added to the reaction mixture, Le Chatelier's principle predicts that the system will produce more NH_3 to reduce the concentration of N_2 to its new equilibrium level.

Partial Pressure

Partial pressure is a measure of the pressure that a single gas in a mixture of gases contributes to the total pressure. Each gas in a mixture exerts a pressure as if it were alone in the volume, and the total pressure is the sum of all these individual pressures.

When dealing with chemical reactions involving gases, like the synthesis of ammonia, understanding partial pressure is crucial. Given that the equilibrium pressure is p in our exercise, the partial pressure of NH_3 would be the fraction of p that NH_3 contributes. Calculating this requires knowledge of the mole fraction of NH_3 in the equilibrium mixture.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It is essential for predicting the amounts of materials consumed and produced in a reaction.

In our example, stoichiometry tells us that 1 mole of N_2 reacts with 3 moles of H_2 to form 2 moles of NH_3. If we start with 1 mole of N_2 and 3 moles of H_2, after 50% of each has reacted, the stoichiometry allows us to calculate the moles of ammonia produced and consequently its partial pressure.

Equilibrium Calculations

Equilibrium calculations are performed to determine the concentrations or partial pressures of reactants and products in a reaction at equilibrium. They are based on the balanced chemical equation, the initial concentrations or pressures, and the equilibrium constant (K).

These calculations can be straightforward if the values needed to complete them are known. In the case of the ammonia synthesis exercise, we apply stoichiometry and understanding of partial pressures to determine that at equilibrium, the partial pressure of NH_3 is one-third of the total pressure p, as it makes up one-third of the total moles of gas present.