Question

For the \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\), the initial mole ratio of \(\mathrm{N}_{2}: \mathrm{H}_{2}\) is \(1: 3 .\) If at equilibrium only \(50 \%\) has reacted and equilibrium pressure is \(\mathrm{P}\). Find the value of \(\mathrm{P}_{\mathrm{NH}_{3}}\) at equilibrium. (a) \(\frac{\mathrm{P}}{3}\) (b) \(\frac{\mathrm{P}}{5}\) (c) \(\frac{\mathrm{P}}{9}\) (d) \(\frac{\mathrm{P}}{6}\)

Short answer

The value is (a) \(\frac{\mathrm{P}}{3}\).

Step-by-step solution

Understand the Problem

The chemical reaction given is \ \(\mathrm{N}_2 + 3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3\). We know the initial mole ratio of \(\mathrm{N}_2:\mathrm{H}_2\) is \(1:3\) and 50% of the reactants have reacted at equilibrium. We need to find the partial pressure of \(\mathrm{NH}_3\) at equilibrium given that the total equilibrium pressure is \(\mathrm{P}\).

Set Initial Moles for Reaction

Assuming 1 mole of \(\mathrm{N}_2\), there will be 3 moles of \(\mathrm{H}_2\) initially as per the ratio. Initial moles of \(\mathrm{NH}_3\) will be 0 because no product would have formed at the start.

Calculate Change in Moles Due to Reaction

Since 50% of \(\mathrm{N}_2\) reacts, \(0.5\) moles of \(\mathrm{N}_2\) are consumed. For every mole of \(\mathrm{N}_2\) consumed, \(3\) moles of \(\mathrm{H}_2\) are consumed and \(2\) moles of \(\mathrm{NH}_3\) are formed. So, \(1.5\) moles of \(\mathrm{H}_2\) are consumed and \(1\) mole of \(\mathrm{NH}_3\) is formed at equilibrium.

Determine Moles At Equilibrium

At equilibrium, moles of \(\mathrm{N}_2 = 1 - 0.5 = 0.5\), moles of \(\mathrm{H}_2 = 3 - 1.5 = 1.5\), and moles of \(\mathrm{NH}_3 = 1\). Total moles at equilibrium are \(0.5 + 1.5 + 1 = 3\).

Apply Dalton's Law of Partial Pressures

The partial pressure of each gas is proportional to its mole fraction times the total pressure. The mole fraction of \(\mathrm{NH}_3\) is \(\frac{1}{3}\), so \(\mathrm{P}_{\mathrm{NH}_3} = \frac{1}{3} \times \mathrm{P} = \frac{\mathrm{P}}{3}\).

Match with Given Choices

From the calculation, we find that the value of \(\mathrm{P}_{\mathrm{NH}_3}\) is \(\frac{\mathrm{P}}{3}\), which corresponds to option (a).

Key concepts

Stoichiometry

Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. It allows us to understand how much of each substance is involved in a reaction.
In the reaction \(_2 + 3 \h_2 \rightleftharpoons 2 h_3\), stoichiometry tells us that one mole of nitrogen (2) reacts with three moles of hydrogen ( ext{H}_2 ext{H}_2) to produce two moles of ammonia ( ext{NH}_3 ext{NH}_3).
This relationship is essential for predicting the amounts of products formed or reactants consumed in the reaction.

• Analyzing the reaction through stoichiometry helps us understand the balance of the equation.
• By knowing the initial amounts and the percentage reacted, we can calculate the amounts at equilibrium.

In our exercise, understanding stoichiometry allows us to determine that 0.5 moles of \text{N}_2 ext{N}_2) react to produce 1 mole of \text{NH}_3 ext{NH}_3 ext{NH}_3) at equilibrium.

Partial Pressure

The concept of partial pressure is crucial in understanding gas behavior in a mixture. Each gas in a mixture exerts its own pressure, known as the partial pressure.
The partial pressure is proportional to the number of moles of the gas present and its behavior is based on Dalton's Law.

• Partial pressure is an important property because it helps analyze how each gas contributes to the total pressure in a mixture.
• In reactions like our \(\text{N}_2 + 3 \text{H}_2 \rightleftharpoons 2 \text{NH}_3\), calculating the partial pressure helps in understanding how the reaction equilibrium is established under given conditions.

In our solution, the total pressure at equilibrium \(P\) includes contributions from the partial pressures of \(\text{N}_2\), \(\text{H}_2\), and \(\text{NH}_3\). Calculating the partial pressure of \(\text{NH}_3\) helps evaluate the reaction outcome at equilibrium.

Mole Fraction

Mole fraction is a way to express the concentration of a component in a mixture. It is the ratio of the number of moles of one component to the total number of moles in the mixture.
This is especially useful in assessing mixtures of gases, as it allows us to determine each gas's contribution to the total pressure.

• The mole fraction provides insight into the prevalence of a particular gas in a mixture.
• Mathematically, it is expressed as \( \text{Mole fraction of X} = \frac{\text{Moles of X}}{\text{Total moles}}\).

In the step-by-step solution from the problem, the mole fraction of \(\text{NH}_3\) is calculated as \(\frac{1}{3}\) at equilibrium. This fraction enables us to find its partial pressure, thus contributing to understanding the gas's behavior in the mixture.

Dalton's Law

Dalton's Law 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.
This principle allows chemists to dissect the contributions of different gases in a mixture to the total pressure, crucial for tasks like determining equilibrium states in reactions.

• Dalton's Law assumes no interactions between the gases, meaning each gas behaves independently, contributing its own partial pressure to the total.
• In our exercise, Dalton's Law aids in calculating the partial pressure of \(\text{NH}_3\) in terms of total equilibrium pressure \(P\).

Understanding this law helps to predict outcomes and make accurate adjustments in reactions, ensuring the correct calculation of partial pressures like \(\frac{P}{3}\) found in our problem.