Problem 35 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following reaction will be favoured at low pressure? (a) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) (b) \(\mathrm{H}_{2}+\mathrm{I}_{2}=2 \mathrm{HI}\) (c) \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\) (d) \(\mathrm{N}_{2}+\mathrm{O}_{2}=2 \mathrm{NO}\)

Step-by-Step Solution

Answer

Reaction (c) is favoured at low pressure as it results in an increase in moles.

Step 1: Identify the Change in Moles for Each Reaction

For each reaction, calculate the change in the number of moles from reactants to products to determine how the equilibrium position will shift with pressure changes.

Step 2: Calculate Change in Moles for Reaction (a)

In reaction (a), \( \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} \). The total initial moles on the left side are 4 (1 for \(\mathrm{N}_{2}\) and 3 for \(\mathrm{H}_{2}\)), and on the right side, it is 2 (2 for \(\mathrm{NH}_{3}\)). The change in moles, \( \Delta n = 2 - 4 = -2 \).

Step 3: Calculate Change in Moles for Reaction (b)

In reaction (b), \( \mathrm{H}_{2} + \mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\). The total initial moles are 2 on both sides (1 for each gas), so \( \Delta n = 2 - 2 = 0 \).

Step 4: Calculate Change in Moles for Reaction (c)

In reaction (c), \( \mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3} + \mathrm{Cl}_{2} \). The total initial moles are 1 on the left and 2 on the right (1 for \(\mathrm{PCl}_{3}\) and 1 for \(\mathrm{Cl}_{2}\)), Thus, \( \Delta n = 2 - 1 = +1 \).

Step 5: Calculate Change in Moles for Reaction (d)

In reaction (d), \( \mathrm{N}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO} \). The total initial moles are 2 on both sides (1 for each gas), so \( \Delta n = 2 - 2 = 0 \).

Step 6: Determine Which Reaction is Favoured at Low Pressure

Low pressure favours the reaction where the number of moles of products is greater than the number of moles of reactants (i.e., \( \Delta n > 0 \)). From Step 4, only reaction (c) has \( \Delta n = +1 \), so it is favoured at low pressure.

Key Concepts

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept to predict how the equilibrium position of a reaction will respond to changes in conditions. When a change such as pressure, temperature, or concentration occurs, the system adjusts to minimize the effect of that change. This principle ensures that the reaction in equilibrium tries to counterbalance the disturbance.

For instance, when the concentration of a reactant is increased, the equilibrium shifts towards the products to reduce the concentration of the reactant. Conversely, reducing a reactant's concentration shifts the balance towards the reactants.

• If pressure is increased, the equilibrium favors the side with fewer gas molecules, as this effectively reduces the system's pressure.
• Decreasing pressure has the opposite effect, favoring the side with more gas molecules.

This principle provides a predictive framework but requires calculations, such as those involving change in moles, to predict responses accurately.

Reaction Moles Calculation

Calculating the change in the number of moles helps predict the direction of equilibrium shifts when pressure changes. This involves counting the moles of gases on the reactants and products sides and determining the difference.

For a reaction aA + bB ⇌ cC + dDwe calculate the change in moles, \( \Delta n = (c + d) - (a + b) \).

• If \( \Delta n > 0 \), there are more moles of products than reactants, indicating the reaction favors a shift toward products at low pressure.
• If \( \Delta n < 0 \), there are fewer product moles, and high pressure favors the reaction toward this side.
• If \( \Delta n = 0 \), pressure changes do not affect the equilibrium.

Reaction mole calculations, like those shown in the step-by-step solution, help determine which conditions favor each side of the equilibrium in the given reactions.

Pressure Effect on Equilibrium

Pressure changes significantly impact gaseous reactions, influencing the equilibrium position based on the number of moles of gases involved. Under Le Chatelier's Principle, a change in pressure affects how equilibrium can shift based on the number of gas molecules or moles.

In a system involving gases, increasing pressure causes equilibrium to favor the side with fewer moles, as the system attempts to reduce pressure. For example, in a reaction where reactants have more moles than the products, increasing pressure will drive the equilibrium towards the products.
Meanwhile, when pressure decreases, the equilibrium shifts toward the side with more moles to counteract the pressure change. Understanding how pressure affects equilibrium allows one to manipulate reactions favorably, such as in industrial applications where yield is optimized by altering pressure.

• For instance, in the original exercise, reaction (c), which shows an increase in moles from reactants to products, is favored at lower pressures since \( \Delta n = +1 \).
• Other reactions with \( \Delta n = 0 \) are not directly influenced by pressure changes.

These insights assist in controlling reactions to generate desired outcomes efficiently.