Problem 12 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

In which of the following reactions, an increase in the volume of the container will favour the formation of products? (a) \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}\) (l) (b) \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) (c) \(3 \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{O}_{3}(\mathrm{~g})\) (d) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

Step-by-Step Solution

Answer

Reaction (b) will favor product formation with increased volume.

Step 1: Analyze the Reaction Volume Changes

When the volume of the container increases, the pressure decreases favoring the reaction where the number of gas moles increases. To determine this, calculate the change in moles for each reaction.

Step 2: Evaluate Reaction (a)

The reaction is: \(4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightleftharpoons 4 \text{NO}(g) + 6 \text{H}_2\text{O}(l)\). The number of moles of gases on the reactant side is 9 (4+5) and on the product side is 4 (liquid water doesn't count as gas). The net change in moles of gas is \(-5\).

Step 3: Evaluate Reaction (b)

The reaction is: \(2 \text{NO}_2(g) \rightleftharpoons 2 \text{NO}(g) + \text{O}_2(g)\). The number of moles of gases on the reactant side is 2 and on the product side is 3. The net change in moles of gas is \(+1\).

Step 4: Evaluate Reaction (c)

The reaction is: \(3 \text{O}_2 \rightleftharpoons 2 \text{O}_3(g)\). The number of moles of gases on the reactant side is 3 and on the product side is 2. The net change in moles of gas is \(-1\).

Step 5: Evaluate Reaction (d)

The reaction is: \(\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2 \text{HI}(g)\). The number of moles of gases on both sides is 2 (1+1=2 for reactants, and 2 for products). The net change in moles of gas is \(0\).

Step 6: Determine the Favored Reaction

An increase in volume favors reactions with an increase in the number of moles of gases. Only reaction (b) has a net increase of moles of gas \(+1\). Therefore, increasing the volume will favor reaction (b).

Key Concepts

Reaction Volume Changes

When the volume of a container increases, there is more space available for the gas particles to occupy. This change in volume can affect the chemical equilibrium of a reaction, as per Le Chatelier's Principle. According to this principle, if a system at equilibrium is disturbed, the system will adjust itself to counteract the disturbance and restore a new equilibrium. In the context of volume change:

• Increasing the volume decreases pressure.
• The system compensates by favoring the reaction that produces more gas moles.
• This is because more gas moles mean a higher pressure, which counterbalances the pressure decrease caused by the increase in volume.

Therefore, an increase in volume favors the side of the reaction with more moles of gas.

Gas Moles Calculation

A key step in applying Le Chatelier's Principle to chemical reactions is calculating the change in moles of gas. This involves:1. Counting the total moles of gaseous reactants and products.2. Determining the difference between them.Here's how it's done:

• Write down the balanced equation of the reaction.
• Identify and sum up the moles of gaseous reactants. For example, in reaction (b), there are 2 moles of NO\(_2\) gas.
• Identify and sum up the moles of gaseous products. For example, in reaction (b), there are 2 moles of NO and 1 mole of O\(_2\), making a total of 3 moles of products.
• Subtract the total reactant moles from the total product moles: 3 - 2 = +1.

The net positive change (+1 in this case) indicates that increasing the volume of the container will favor the formation of more gas molecules, thus shifting the equilibrium to the right.

Chemical Equilibrium

Chemical equilibrium occurs in reversible reactions where the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of reactants and products remain constant over time. However, this balance can be disrupted by changes in conditions such as pressure, temperature, or volume. Le Chatelier's Principle helps predict how equilibrium will shift in response to these disturbances. For volume changes:

• If the volume is increased, the equilibrium shifts towards the side with more gaseous moles.
• If the volume is decreased, the system shifts to the side with fewer gaseous moles.

This principle allows predictions of how the reaction will adapt, ensuring its dynamic balance within the new conditions. Understanding this concept can help in controlling industrial chemical processes and optimizing reactions for desired outcomes.