Problem 146 - A Complete Resource Book in Chemistry for JEE Main

Question

In a system \(\mathrm{A}(\mathrm{s}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})+3 \mathrm{C}(\mathrm{g})\) If the concentration of \(\mathrm{C}\) at equilibrium is increased by a factor of 2, it will cause the equilibrium concentration of \(\mathrm{B}\) to change to (a) One half of its original value (b) Two times the original value (c) \(\frac{1}{2 \sqrt{2}}\) times the original value (d) \(2 \sqrt{2}\) times its original value

Step-by-Step Solution

Answer

The concentration of \(\mathrm{B}\) changes to \(\frac{1}{2\sqrt{2}}\) times its original value.

Step 1: Write the equilibrium expression for the reaction

For the given equilibrium \(\mathrm{A}(\mathrm{s}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})+3 \mathrm{C}(\mathrm{g}) \), the equilibrium constant \(K\) can be written in terms of the concentrations of \(\mathrm{B}\) and \(\mathrm{C}\) as follows:\[K = [B]^2 [C]^3\]Note that \(\mathrm{A}\) is a solid and does not appear in the expression.

Step 2: Analyze the change in concentration of C

The concentration of \(\mathrm{C}\) is increased by a factor of 2. Let the original concentration of \(\mathrm{C}\) be \([C]_0\). The new concentration becomes \(2[C]_0\). We must now consider how this change affects the equilibrium condition.

Step 3: Substitute the change in concentration into the equilibrium expression

With the change in \([C]\) to \(2[C]_0\), substitute into the equilibrium expression:\[K = [B]^2 (2[C]_0)^3\]This becomes:\[K = [B]^2 \cdot 8[C]_0^3\]Since \(K\) remains constant, we relate this to the original condition \( [B]^2 [C]_0^3 \).

Step 4: Solve for the new equilibrium concentration of B

Because the equilibrium constant \(K\) does not change:\[[B]^2 \cdot 8[C]_0^3 = [B_0]^2 [C]_0^3\]Simplify:\[8[B]^2 = [B_0]^2\]\[[B]^2 = \frac{[B_0]^2}{8}\]Thus, solving for \([B]\):\[[B] = \frac{[B_0]}{\sqrt{8}} = \frac{[B_0]}{2\sqrt{2}}\]

Step 5: Conclusion and comparison to the given options

\([B]\) is reduced to \(\frac{1}{2\sqrt{2}}\) times its original concentration \([B_0]\). Compare this to the options provided:- (a) One half of its original value- (b) Two times the original value- (c) \(\frac{1}{2\sqrt{2}}\) times the original value- (d) \(2\sqrt{2}\) times its original valueTherefore, the correct answer is (c).

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is an important concept in chemistry that provides a quantitative measure of the extent of a chemical reaction at equilibrium. It is a ratio of the concentrations of products to reactants, each raised to the power of their respective coefficients in the balanced equation. For the reaction \( \mathrm{A}(\mathrm{s}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})+3 \mathrm{C}(\mathrm{g}) \), the equilibrium constant expression is given by

• \[ K = [B]^2 [C]^3 \]

The concentration of solids, such as \( \mathrm{A}(\mathrm{s}) \), is omitted from the expression because they do not change significantly during the reaction.
The value of \( K \) is constant for a given reaction at a specific temperature, which means it does not change even if the concentrations of \( \mathrm{B} \) and \( \mathrm{C} \) change. Instead, the system shifts to maintain the \( K \) value, following the reaction dynamics. The equilibrium constant tells us whether products or reactants are favored at equilibrium, with a large \( K \) indicating that products are favored.

Concentration Changes

Concentration changes can have significant impacts on a chemical equilibrium system. When we alter the concentration of a substance in an equilibrium reaction, the system will shift according to Le Chatelier's Principle to counteract the change. In our exercise, the concentration of \( \mathrm{C} \) was increased by a factor of 2.
This initial change, denoted as from \([C]_0\) to \(2[C]_0\), affected the balance of the system. To accommodate this alteration while maintaining the equilibrium constant, the concentration of \( \mathrm{B} \) had to adjust as seen in the equilibrium expression.

• The system responded by reducing \( [B] \) to \( \frac{1}{2\sqrt{2}} \) times its original amount, demonstrating the sensitivity of equilibrium systems to concentration shifts.

This shift is the system's way of restoring equilibrium by reducing the stress caused by the increase in \( \mathrm{C} \). Thus, concentration changes are essential in dictating how equilibrium positions adjust.

Chemical Equilibrium

Chemical equilibrium occurs in a reversible chemical reaction when the rates of the forward and reverse reactions are equal, and the concentrations of reactants and products no longer change with time.
In this state, although reactions continue to occur, they do so at a balance, achieving what is known as a dynamic equilibrium. For the reaction \( \mathrm{A}(\mathrm{s}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})+3 \mathrm{C}(\mathrm{g}) \), the system reaches equilibrium when the products \( \mathrm{B} \) and \( \mathrm{C} \), and the reactant \( \mathrm{A} \), are produced and consumed at the same rate.

• Even small changes, like altering the concentration of one product, can disturb this equilibrium leading to a shift to reestablish balance.
• Understanding chemical equilibrium helps predict how a system will respond to various stresses, like changes in pressure, temperature, or concentration.

The concept underscores the natural tendency of reactions to minimize disruptions and preserve the status quo, a crucial insight for the application of chemistry in real-world scenarios.