Question

For the reaction: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}), \Delta \mathrm{H}=-93.6 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\), the concentration of \(\mathrm{H}_{2}\) at equilibrium can be increased by: (1) Lowering the temperature (2) Increasing the volume of the system (3) Adding \(\mathrm{N}_{2}\) at constant volume (4) Adding \(\mathrm{H}_{2}\) at constant volume (a) (ii) and (iv) are correct (b) Only (ii) is correct (c) (i), (ii) and (iii) are correct (d) (iii) and (iv) are correct

Short answer

Options (ii) and (iv) are correct, so answer (a) is correct.

Step-by-step solution

Analyzing the Reaction

The given reaction is \(\mathrm{N}_{2} (\mathrm{~g}) + 3\mathrm{H}_{2} (\mathrm{~g}) \rightleftharpoons 2\mathrm{NH}_{3} (\mathrm{~g})\), with \(\Delta H = -93.6 \mathrm{~kJ}\ \mathrm{mol}^{-1}\). It is an exothermic reaction as the enthalpy change is negative, meaning heat is released when the reaction proceeds from left to right.

Effect of Lowering Temperature (Exothermic Reaction)

According to Le Chatelier's principle, for an exothermic reaction, reducing the temperature shifts the equilibrium to the right to produce more heat (favoring product formation). This will decrease the concentration of \(\mathrm{H}_{2}\), so option (i) does not increase \(\mathrm{H}_{2}\) at equilibrium.

Effect of Increasing Volume

Increasing the volume of the system decreases the pressure. The reaction will shift toward the side with more moles of gas to increase the pressure. Since there are 4 moles of gas on the left and 2 on the right, increasing the volume favors the reactants, increasing \(\mathrm{H}_{2}\), so option (ii) is correct.

Adding \(\mathrm{N}_{2}\)

Adding \(\mathrm{N}_{2}\) at constant volume increases the concentration of a reactant, shifting the equilibrium to the right to consume \(\mathrm{N}_{2}\) and produce more \(\mathrm{NH}_{3}\). This decreases \(\mathrm{H}_{2}\) concentration, so option (iii) does not increase \(\mathrm{H}_{2}\) at equilibrium.

Adding \(\mathrm{H}_{2}\)

Adding \(\mathrm{H}_{2}\) directly increases its concentration since we are adding more of it. At constant volume, the equilibrium will adjust; however, initially, \(\mathrm{H}_{2}\) concentration is increased, so option (iv) is correct.

Conclusion on Correct Options

Based on the above analysis, options (ii) and (iv) result in an increase in the equilibrium concentration of \(\mathrm{H}_{2}\). Hence, answer (a) is correct.

Key concepts

Equilibrium Shift

When a chemical system is at equilibrium, it means the rate of the forward reaction is equal to the rate of the reverse reaction. If there are any changes in conditions such as concentration, pressure, or temperature, Le Chatelier's Principle tells us that the system will adjust to counteract those changes and re-establish equilibrium.

For the reaction \[ \mathrm{N}_{2} (\mathrm{~g}) + 3\mathrm{H}_{2} (\mathrm{~g}) \rightleftharpoons 2\mathrm{NH}_{3} (\mathrm{~g}),\]an equilibrium shift can occur. This reaction has more moles of gas on the reactant side (4 moles) than on the product side (2 moles). So, depending on the condition changes like volume or pressure, the equilibrium can shift either to the right, favoring products, or to the left, favoring reactants.

When analyzing the shift, consider:

• Adding more reactants or removing products usually shifts the equilibrium toward the products.
• Decreasing volume shifts equilibrium toward the side with fewer gas moles.
• Any shift aims to reach a new state that restores balance under the changed conditions.

Understanding this principle is crucial for predicting the outcome of a reaction when its conditions are altered.

Exothermic Reactions

For many reactions, the direction of the equilibrium shift is influenced by whether the reaction is exothermic (releases heat) or endothermic (absorbs heat).

In our reaction \[\Delta H = -93.6 \mathrm{~kJ} \ \mathrm{mol}^{-1}, \]which indicates it's exothermic. This means, as the reaction proceeds from reactants to products, heat is released into the surroundings.

In the case of exothermic reactions, lowering the temperature will shift the equilibrium to the right, as the system will attempt to generate heat and counterbalance the lower temperature. Conversely, if the temperature is increased, the equilibrium will shift to the left to absorb some of that additional heat.

This principle helps us control how much of product or reactant is formed by adjusting the temperature:

• Lowering temperature favors product formation as the system tries to produce more heat.
• Raising temperature can lead to more reactants as the system tries to absorb the excess heat.

This is a vital concept when maximizing yields in industrial chemical processes.

Effect of Pressure Changes

The pressure aspect can also impact equilibrium in reactions involving gases. Changes in system pressure mainly occur through changes in volume. According to Le Chatelier's Principle, changing the pressure of a gaseous system causes the equilibrium to shift toward the side with fewer or more gas moles, depending on whether pressure is increased or decreased.

For the equilibrium reaction we consider:\[\mathrm{N}_{2} (\mathrm{~g}) + 3\mathrm{H}_{2} (\mathrm{~g}) \rightleftharpoons 2\mathrm{NH}_{3} (\mathrm{~g}),\]increasing the volume of the container lowers the pressure. The system responds by favoring the formation of more moles. Here, the left side (4 moles) is favored, so more \(\mathrm{H}_{2}\) is produced.

Key points about pressure changes include:

• Increasing pressure shifts equilibrium toward the side with fewer gas moles.
• Decreasing pressure by increasing volume shifts equilibrium toward the side with more gas moles.

This makes control of pressure an effective strategy to direct the outcome of a reaction involving gases.

Reaction Enthalpy

Enthalpy, represented as \( \Delta H \), is a measure of the total energy change in a chemical reaction. It tells us if a reaction is exothermic or endothermic by indicating the heat absorbed or released during the reaction.

For our reaction \[\mathrm{N}_{2} (\mathrm{~g}) + 3 \mathrm{H}_{2} (\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3} (\mathrm{~g}), \Delta H = -93.6 \mathrm{~kJ} \ \mathrm{mol}^{-1},\]the negative sign indicates an exothermic process, meaning it releases heat when ammonia (NH\(_3\)) is being formed.

Understanding reaction enthalpy is important because it helps predict how temperature changes could affect equilibrium:

• Exothermic reactions release heat, so they are typically more favored at lower temperatures.
• Endothermic reactions absorb heat, thus they become more favored at higher temperatures.

By manipulating conditions such as temperature, chemists can drive reactions in the desired direction to optimize yields. This knowledge is particularly crucial in industrial applications where specific products are targeted.