Problem 15 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following equilibrium, in gaseous phase, would be unaffected by an increase in pressure? (a) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) (b) \(\mathrm{N}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons=2 \mathrm{NO}_{2}\) (d) \(\mathrm{CO}_{2}^{2}+1 / 2 \mathrm{O}_{2}=\mathrm{CO}_{2}\)

Step-by-Step Solution

Answer

Reaction (b) is unaffected by an increase in pressure.

Step 1: Understanding the Effect of Pressure on Equilibrium

In any gaseous equilibrium, the change in pressure affects the position of equilibrium if the number of moles of gaseous reactants differs from the number of moles of gaseous products. According to Le Chatelier's Principle, an increase in pressure will shift the equilibrium position to the side with fewer gaseous moles.

Step 2: Analyzing Reaction (a)

Reaction (a) is \(\mathrm{N}_{2} + 3\mathrm{H}_{2} \rightleftharpoons 2\mathrm{NH}_{3}\). The moles of gases are: \(1+3=4\) moles on the reactant side and \(2\) moles on the product side. An increase in pressure would shift the equilibrium to the right, towards fewer moles of gas.

Step 3: Analyzing Reaction (b)

Reaction (b) is \(\mathrm{N}_{2} + \mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}\). The moles of gases are \(1+1=2\) moles on the reactant side, and \(2\) moles on the product side. Since the moles of gas are equal on both sides, the equilibrium would be unaffected by a change in pressure.

Step 4: Analyzing Reaction (c)

Reaction (c) is \(\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}\). The moles of gases are \(1\) mole on the reactant side and \(2\) moles on the product side. Increasing pressure will favor the side with fewer gas moles (left side reaction).

Step 5: Analyzing Reaction (d)

Reaction (d) is \(\mathrm{CO}_{2}^{2} + 1/2 \mathrm{O}_{2} = \mathrm{CO}_{2}\). The moles of gases are \(2+0.5=2.5\) moles on the reactant side and \(1\) mole on the product side. An increase in pressure would shift the equilibrium to the right.

Key Concepts

Equilibrium Shift

Equilibrium is a state where the forward and reverse reactions occur at the same rate. This means the concentrations of reactants and products remain constant over time. However, when a system at equilibrium is subjected to a change, like pressure or concentration, the system shifts to counteract the change, according to Le Chatelier's Principle.
This shift in equilibrium can be imagined as a balance. For example, in a seesaw, if the weight on one side increases (like a change in condition), the seesaw tilts towards the heavier side until it reaches a new balance. In chemical reactions, if pressure is increased, the system adjusts by shifting the equilibrium towards the side with fewer gas molecules to minimize the effect of increased pressure. Hence, equilibrium shift is essentially the system's way of restoring balance in reaction dynamics after a change is introduced.
This concept is crucial in predicting and controlling the outcomes and yields of chemical reactions in industrial processes.

Pressure Effects on Equilibrium

Pressure can significantly affect the position of equilibrium, especially in reactions involving gases. In accordance with Le Chatelier's Principle, when pressure in a system is increased, the equilibrium shifts towards the side with fewer moles of gas. This shift aims to reduce the pressure increase, by decreasing the number of gaseous molecules.
Let's consider a scenario: in the reaction \( f ext{N}_{2} + 3 ext{H}_{2} ightarrow 2 ext{NH}_{3}\). Here, increasing pressure causes a shift towards the product side, which has fewer moles of gas compared to the reactant side. On the other hand, if there's an equal amount of moles on both the reactant and product sides, as in \( ext{N}_{2} + ext{O}_{2} ightarrow 2 ext{NO}\), the equilibrium remains _unaffected_ by changes in pressure, because shifting in either direction does not reduce the overall gas pressure.
Understanding this topic is key for controlling reactions in chemical engineering and ensuring that desired conditions for product formation are achieved efficiently.

Mole Calculation in Reactions

The concept of 'moles' is essential in understanding how reactions proceed and how they are affected by changes, such as pressure. A 'mole' is a unit of measurement that represents a specific number of atoms or molecules, similar to how a dozen stands for twelve items.
In the context of chemical reactions, balancing moles on both sides of an equation unveils the stoichiometry of the reaction. The stoichiometric coefficients tell us how much of each reactant is required and how much of each product is formed. For example, in the reaction \(\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3\), the balanced equational representation shows that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia.
Accurate mole calculations are fundamental when determining the equilibrium position in reactions. They help chemists know which direction the reaction might shift under different conditions, like pressure changes. Thus, mastering mole calculations allows one to better predict and control chemical reaction outcomes in various processes, from laboratory experiments to large-scale industrial applications.