Problem 86 - A Complete Resource Book in Chemistry for JEE Main

Question

One mole of \(\mathrm{A}(\mathrm{g})\) is heated to \(300^{\circ} \mathrm{C}\) in a closed one litre vessel till the following equilibrium is reached. \(\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})\) The equilibrium constant of this reaction at \(300^{\circ} \mathrm{C}\) is 4 . What is the concentration of \(\mathrm{B}\) (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) at equilibrium? (a) \(0.2\) (b) \(0.6\) (c) \(0.75\) (d) \(0.1\)

Step-by-Step Solution

Answer

The concentration of \(\mathrm{B}\) is 0.8 mol L⁻¹, but correct options do not match.

Step 1: Write the Expression for Equilibrium Constant

The equilibrium constant expression for the reaction \( \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g}) \) is given by: \[ K_c = \frac{[B]}{[A]} \] where \([B]\) and \([A]\) are the equilibrium concentrations of \(\mathrm{B}\) and \(\mathrm{A}\), respectively. It's known that \( K_c = 4 \) at \(300^{\circ} \mathrm{C}\).

Step 2: Set Up Initial and Change Variables

Initially, we have 1 mole of \(\mathrm{A}\) in a 1 L vessel, so \([A]_0 = 1 \ \mathrm{mol} \ \mathrm{L}^{-1}\). No \(\mathrm{B}\) is present initially, so \([B]_0 = 0\). Suppose \(x\) moles of \(\mathrm{A}\) react to form \(x\) moles of \(\mathrm{B}\).

Step 3: Write the Equilibrium Concentrations

At equilibrium, the concentration of \(\mathrm{B}\) is \([B] = x \) and the concentration of \(\mathrm{A}\) is \([A] = 1 - x \).

Step 4: Substitute in the Equilibrium Expression

Substitute the equilibrium concentrations into the equilibrium expression: \[ K_c = \frac{x}{1 - x} = 4 \].

Step 5: Solve for x

To find \(x\), solve the equation: \[ \frac{x}{1 - x} = 4 \] which simplifies to \(x = 4(1 - x)\). Expand to get \(x = 4 - 4x\). Add \(4x\) on both sides to get \(5x = 4\), so \(x = \frac{4}{5} = 0.8\).

Step 6: Validate and Compare with Options

The concentration of \(\mathrm{B}\) at equilibrium, \([B] = x = 0.8 \, \mathrm{mol} \, \mathrm{L}^{-1}\). None of the options match, suggesting something went wrong. Review and repeat steps taking into account any overlooked factors.

Key Concepts

Equilibrium Constant

In chemical reactions, the **equilibrium constant** (\( K_c \)) is a fundamental concept that helps predict the extent to which a reaction will proceed. For a given reversible reaction, it describes the relative concentrations of products and reactants at equilibrium. For the reaction \( \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g}) \), the equilibrium constant is calculated using the formula:
\[ K_c = \frac{[B]}{[A]} \]
Here, \([B]\) and \([A]\) are the molar concentrations of species \( \mathrm{B} \) and \( \mathrm{A} \) at equilibrium.

• A large \( K_c \) value indicates that products are favored at equilibrium, meaning the reaction tends to proceed towards product formation.
• A small \( K_c \) value means reactants are more prominent at equilibrium, signifying that the reaction prefers to stay as reactants.

Knowing the equilibrium constant is critical for predicting reaction behavior under different conditions. It remains unchanged unless you alter the temperature of the reaction.

Equilibrium Concentration

The **equilibrium concentration** of a substance in a chemical reaction is the concentration of that substance when the reaction has reached a state of balance and its forward and reverse reactions occur at equal rates. Understanding how to determine these concentrations is essential for solving equilibrium problems. Initially, we have 1 mole of \( A \) in a 1-liter container, so the starting concentration is \( [A]_0 = 1 \mathrm{mol} \mathrm{L}^{-1} \).

• No \( B \) is present initially, so \([B]_0 = 0\).
• At equilibrium, \( x \) moles of \( A \) are converted into \( x \) moles of \( B \).
• The changes help set up an initial state, allowing the determination of final concentrations: \([B] = x \) and \([A] = 1 - x \). This explains the concentrations of each at equilibrium.

Knowing how concentrations change allows students to solve for unknowns using the equilibrium expression, as was needed in the original exercise.

Reaction Stoichiometry

**Reaction stoichiometry** is another key topic in understanding equilibrium. It relates to the quantitative relationships between reactants and products in a chemical reaction. In our reaction \( \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g}) \), the stoichiometric relationship is 1:1. This means:

• One mole of \( A \) produces one mole of \( B \) upon reaction, defining a direct and simple stoichiometry.
• Using this stoichiometric factor, we determine the changes in concentration as the reaction progresses toward equilibrium.

When solving problems, stoichiometry helps in setting up the change ("x") variables accurately. For example, if \( x \) moles of \( A \) turns into \( x \) moles of \( B \), stoichiometry ensures this transformation respects conservation of mass. It plays a crucial role in formulating the equilibrium expression and solving subsequent equations to find equilibrium concentrations.