Question

A nitrogen-hydrogen mixture initially in the molar ratio of \(1: 3\) reached equilibrium to form ammonia when \(25 \%\) of the \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) had reacted. If the total pressure of the system was 21 atm, the partial pressure of ammonia at the equilibrium was: (a) \(4.5\) atm (b) \(3.0 \mathrm{~atm}\) (c) \(2.0 \mathrm{~atm}\) (d) \(1.5 \mathrm{~atm}\)

Short answer

3.0 atm

Step-by-step solution

Understand the Chemical Reaction

For the synthesis of ammonia, the balanced chemical equation is \[\mathrm{N_{2}(g)} + 3\mathrm{H_{2}(g)} \rightleftharpoons 2\mathrm{NH_{3}(g)}.\] This indicates that 1 mole of nitrogen reacts with 3 moles of hydrogen to form 2 moles of ammonia.

Calculate Initial Moles

Assume an arbitrary amount of moles for \(\mathrm{N_{2}}\) and \(\mathrm{H_{2}}\) based on their molar ratio of 1:3, for instance, 1 mole of \(\mathrm{N_{2}}\) and 3 moles of \(\mathrm{H_{2}}\). At equilibrium, 25% of both \(\mathrm{N_{2}}\) and \(\mathrm{H_{2}}\) reacted, thus leaving 0.75 moles of \(\mathrm{N_{2}}\) and 3 * 0.75 = 2.25 moles of \(\mathrm{H_{2}}\).

Calculate Moles of Ammonia Produced

Since 25% of \(\mathrm{N_{2}}\) and \(\mathrm{H_{2}}\) have reacted, the number of moles of \(\mathrm{NH_{3}}\) produced at equilibrium will be \(0.25 * 2 = 0.5\) moles, following the stoichiometry of the balanced chemical equation.

Calculate the Total Moles at Equilibrium

Add the moles of nitrogen, hydrogen, and ammonia that are present at equilibrium: \(0.75 \text{ moles of } \mathrm{N_{2}} + 2.25 \text{ moles of } \mathrm{H_{2}} + 0.5 \text{ moles of } \mathrm{NH_{3}} = 3.5 \text{ moles}\).

Calculate Partial Pressure of Ammonia

Utilize the formula for partial pressure, \(P_{i} = (n_{i} / n_{total}) * P_{total}\), where \(n_{i}\) is the moles of ammonia, \(n_{total}\) is the total moles at equilibrium, and \(P_{total}\) is the total pressure. Substituting \(n_{i} = 0.5\), \(n_{total} = 3.5\), and \(P_{total} = 21 \text{ atm}\): \(P_{\mathrm{NH_{3}}} = (0.5 / 3.5) * 21 \text{ atm} = 3 \text{ atm}\).

Key concepts

Chemical Equilibrium

In the realms of chemistry, chemical equilibrium represents a state in which the rate of the forward reaction equals the rate of the reverse reaction, resulting in no overall change in the concentrations of the reactant and product over time. It's critical to understand that being at equilibrium does not mean the reactants and products are present in equal amounts, but rather that their ratios are fixed and stable.

In the context of ammonia synthesis from nitrogen and hydrogen, \[{\mathrm{N_{2}(g)} + 3\mathrm{H_{2}(g)} \rightleftharpoons 2\mathrm{NH_{3}(g)}}\], reaching equilibrium implies that the system has found a balance between the formation of ammonia and the decomposition back into nitrogen and hydrogen. The position of equilibrium, which tells us the relative amounts of reactants and products at equilibrium, is governed by the equilibrium constant, which is not covered in the exercise but is a pivotal concept in understanding chemical equilibria. Equilibrium is dynamic, meaning that even though we see no changes macroscopically, there are ongoing microscopic processes where reactants form products and vice versa.

Stoichiometry

The study of the quantitative relationships or ratios between the amounts of reactants used and products formed by a chemical reaction is known as stoichiometry. It is based on the conservation of mass and the law of definite proportions. For our ammonia formation, the balanced chemical equation \[{\mathrm{N_{2}(g)} + 3\mathrm{H_{2}(g)} \rightleftharpoons 2\mathrm{NH_{3}(g)}}\] suggests that 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia gas. This is a stoichiometric relationship of 1:3:2.

Understanding this relationship allows us to calculate how much reactant is needed to form a certain amount of product or vice versa. When the problem states that 25% of the \(\mathrm{N}_{2}\)\ and \(\mathrm{H}_{2}\)\ reacted, stoichiometry enables us to deduce that the moles of ammonia produced will be half the moles of nitrogen that reacted, maintaining the stoichiometric ratios.

Partial Pressure Calculations

The concept of partial pressure is crucial when dealing with gases. It is defined as the pressure that a gas would exert if it alone occupied the entire volume of the mixture at the same temperature. Dalton's Law of Partial Pressures states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of individual gases.

For our ammonia equilibrium exercise, calculating the partial pressure involves understanding that the total pressure is the sum of pressures contributed by each gas. As shown in the step-by-step solution, by applying the formula \[{P_{i} = (n_{i} / n_{total}) \cdot P_{total}}\] we can find the partial pressure of ammonia. Here, \(n_{i}\) is the moles of ammonia (0.5 moles), \(n_{total}\) is the total number of moles present at equilibrium (3.5 moles), and \(P_{total}\) is the total pressure of the system (21 atm). This calculation helps us understand how the properties of a gas mixture relate to its components. Answer (b) 3.0 atm is obtained, illustrating the practical use of stoichiometry in conjunction with the laws governing gas behavior.