Problem 182 - A Complete Resource Book in Chemistry for JEE Main

Question

In the heterogeneous equilibrium \(\mathrm{A}(\mathrm{s}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})+\) \(3 \mathrm{C}(\mathrm{g})\). If conc. of \(\mathrm{C}\) at equilibrium is doubled, then after the equilibrium is re- established, the conc. of \(\mathrm{B}\) will be \(\left(\frac{1}{\mathrm{r}}\right)\) times its original value. The value of \(\mathrm{r}^{2}\) is

Step-by-Step Solution

Answer

The value of \( r^2 \) is 8.

Step 1: Write the Expression for Kc

For the given reaction \( \text{A}(\text{s}) \rightleftharpoons 2 \text{B}(\text{g}) + 3 \text{C}(\text{g}) \), the equilibrium constant expression \( K_c \) is based on the concentration of gaseous species because pure solids and liquids are not included in the expression. Thus, the expression is:\[K_c = [\text{B}]^2 [\text{C}]^3\]

Step 2: Analyze the Effect of Doubling [C]

If the concentration of \( [\text{C}] \) is doubled, the new concentration becomes \( 2[\text{C}] \). Substitute this into the equilibrium expression:\[K_c = [\text{B}]'^2 (2[\text{C}])^3\]This simplifies to:\[K_c = [\text{B}]'^2 \cdot 8[\text{C}]^3\]

Step 3: Relate New Equilibrium to Original

Since \( K_c \) remains constant, we equate the updated expression to the original expression:\[[\text{B}]_0^2 [\text{C}]^3 = [\text{B}]'^2 \cdot 8[\text{C}]^3\]By simplifying, we have:\[[\text{B}]_0^2 = 8[\text{B}]'^2\]

Step 4: Solve for [B] in terms of Original [B]

Solving the equation \([\text{B}]_0^2 = 8[\text{B}]'^2\) for \([\text{B}]'\):\[[\text{B}]'^2 = \frac{1}{8}[\text{B}]_0^2\]Taking the square root, we find:\[[\text{B}]' = \frac{1}{\sqrt{8}}[\text{B}]_0 = \frac{1}{2\sqrt{2}}[\text{B}]_0\]This implies:\[\frac{1}{r} = \frac{1}{2\sqrt{2}}\]So, \(r = 2\sqrt{2}\).

Step 5: Calculate r^2

Now compute \( r^2 \):\[r^2 = (2\sqrt{2})^2 = 4 \times 2 = 8\]

Key Concepts

Equilibrium Constant (Kc)

In chemical reactions that reach a state of balance or equilibrium, the equilibrium constant, noted by the symbol \( K_c \), plays a crucial role. It helps in understanding the ratio of the concentrations of products to reactants at equilibrium. However, it's important to remember that \( K_c \) only includes gases and aqueous solutions but not solids or pure liquids.

• For the given equilibrium \( \text{A} (\text{s}) \rightleftharpoons 2 \text{B} (\text{g}) + 3 \text{C} (\text{g}) \), the \( K_c \) expression is focused on \( \text{B} \) and \( \text{C} \) since \( \text{A} \) is a solid.
• The expression is written as: \[ K_c = [\text{B}]^2[\text{C}]^3 \]

This expression helps us predict how changes in conditions might affect the system, balancing reactions at equilibrium, and is a guiding factor in solving equilibrium problems.

Heterogeneous Equilibrium

Heterogeneous equilibrium involves reactions that include substances in different phases, such as solids, liquids, and gases. In our example, \( \text{A} \) is a solid, while \( \text{B} \) and \( \text{C} \) are gases. Heterogeneous equilibria focus on the gaseous and aqueous participants because the concentrations of solids and pure liquids are considered to be constant.This constant nature in uniform phases simplifies our calculations because:

• Solids and pure liquids do not appear in the \( K_c \) expression, which means the complexity is reduced.
• By focusing on gaseous products and reactants, it allows us to effectively measure and calculate outcomes.

These simplified considerations are critical when solving for \( K_c \), especially in mixed-phase reactions like the one given in the exercise.

Effect of Concentration Changes

When a change in concentration occurs within a reaction at equilibrium, it disrupts the equilibrium state. The system then adjusts itself to restore equilibrium, which is highlighted by Le Chatelier's Principle. In the context of our reaction, double the concentration of \( \text{C} \) initially disrupts the system, leading to interesting consequences:

• The increased concentration of \( \text{C} \) shifts the system to consume more \( \text{C} \) and produce more \( \text{B} \).
• This adjustment maintains the \( K_c \), keeping it constant because it is a property of the system at a given temperature and pressure.

Using this principle in calculations, it shows us that the new equilibrium concentration of \( \text{B} \) is less than its original because of the manner it needs to re-establish equilibrium with the doubled \( \text{C} \). Solving the exercise shows how these adjustments are mathematically represented, adjusting \( [\text{B}] \), and showcasing the dependency on initial and changed conditions.