Problem 111 - A Complete Resource Book in Chemistry for JEE Main

Question

An equilibrium mixture for the reaction, \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{S}_{2}(\mathrm{~g})\) had 1 mole of \(\mathrm{H}_{2} \mathrm{~S}, 0.2\) mole of \(\mathrm{H}_{2}\) and \(0.8\) mole of \(\mathrm{S}_{2}\) in a 2 litre flask. The value of \(\mathrm{K}_{\varepsilon}\) in \(\mathrm{mol} \mathrm{L}^{-1}\) is (a) \(0.08\) (b) \(0.016\) (c) \(0.004\) (d) \(0.160\)

Step-by-Step Solution

Answer

The value of \(K_c\) is (b) 0.016.

Step 1: Write the Equilibrium Expression

For the balanced chemical reaction \(2 \mathrm{H}_{2} \mathrm{S}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{~g}) + \mathrm{S}_{2}(\mathrm{~g})\), the equilibrium constant expression \(K_c\) is given by:\[K_c = \frac{[\mathrm{H}_2]^2[\mathrm{S}_2]}{[\mathrm{H}_2\mathrm{S}]^2}\] where square brackets \([]\) denote the molarity (concentration in \(\mathrm{mol/L}\)).

Step 2: Calculate Molarity of Each Species

Calculate the molarity of each species at equilibrium, given a 2-liter flask. The molarity is the number of moles divided by the volume in liters.- \(\mathrm{H}_2\mathrm{S} = \frac{1 \, \text{mol}}{2 \, \text{L}} = 0.5 \, \mathrm{mol/L}\)- \(\mathrm{H}_2 = \frac{0.2 \, \text{mol}}{2 \, \text{L}} = 0.1 \, \mathrm{mol/L}\)- \(\mathrm{S}_2 = \frac{0.8 \, \text{mol}}{2 \, \text{L}} = 0.4 \, \mathrm{mol/L}\)

Step 3: Substitute the Molarities into the Equilibrium Expression

Using the calculated molarity values, substitute them into the expression for \(K_c\):\[K_c = \frac{(0.1)^2 \times 0.4}{(0.5)^2}\]

Step 4: Solve for Equilibrium Constant

Calculate \(K_c\) using the values from the substitution:\[K_c = \frac{0.01 \times 0.4}{0.25} = \frac{0.004}{0.25} = 0.016\]

Step 5: Select the Correct Answer from Options

The calculated value of \(K_c\) is \(0.016\). Therefore, the correct answer is (b) \(0.016\).

Key Concepts

Equilibrium Constant

In a chemical reaction, not all reactants turn completely into products. Some reactions stop when the forward and reverse reactions balance each other out, creating a situation called equilibrium. The equilibrium constant, represented as \( K_c \), is a value that indicates the ratio of product concentrations to reactant concentrations at equilibrium. The concentrations are raised to the power of their coefficients from the balanced equation.

For the reaction \(2 \text{H}_2\text{S}(\text{g}) \rightleftharpoons 2 \text{H}_2(\text{g}) + \text{S}_2(\text{g})\), \( K_c \) is calculated using the expression:
\[K_c = \frac{[\text{H}_2]^2[\text{S}_2]}{[\text{H}_2\text{S}]^2}\]
This equation compares the concentrations of hydrogen gas and sulfur gas to hydrogen sulfide. A higher \( K_c \) value means more products at equilibrium, while a lower \( K_c \) signifies more reactants. This concept is essential for predicting reaction behavior under various conditions.

Molarity Calculations

Molarity is a measure of concentration that expresses the number of moles of a solute per liter of solution. Understanding molarity is key in calculations related to chemical equilibrium. In the given equilibrium reaction:

• \(\text{H}_2\text{S} = \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ mol/L}\)
• \(\text{H}_2 = \frac{0.2 \text{ mole}}{2 \text{ L}} = 0.1 \text{ mol/L}\)
• \(\text{S}_2 = \frac{0.8 \text{ mole}}{2 \text{ L}} = 0.4 \text{ mol/L}\)

It's crucial to be consistent with units when calculating molarity, ensuring the volume is always in liters and the number of moles correctly accounts for the amount in the reaction mixture.

Converting moles to molarity allows one to use these concentrations directly in the equilibrium expression, making it possible to solve for the equilibrium constant \( K_c \).

Equilibrium Expression

The equilibrium expression is a way to quantify the concentrations of reactants and products in a balanced chemical reaction at equilibrium. It's constructed by placing the product concentration terms in the numerator, and the reactant terms in the denominator, with each raised to the power of their stoichiometric coefficients. For the given reaction, the equilibrium expression is:
\[K_c = \frac{[\text{H}_2]^2[\text{S}_2]}{[\text{H}_2\text{S}]^2}\]
When establishing an equilibrium expression, ensure that all substances are gases or aqueous, as only these states are included in \( K_c \).

Using this expression helps predict how changes in concentration, pressure, or temperature could shift the equilibrium position. It's a foundational tool in understanding how a system can be manipulated in practice.

Reaction Quotient

The reaction quotient \( Q \) serves as a "snapshot" of a reaction's progress at any given moment, unlike the equilibrium constant \( K_c \), which is only valid at equilibrium. \( Q \) uses the same expression as \( K_c \):
\[Q = \frac{[\text{H}_2]^2[\text{S}_2]}{[\text{H}_2\text{S}]^2}\]
The value of \( Q \) compared to \( K_c \) determines the direction the reaction should go to reach equilibrium:

• If \( Q < K_c \), the forward reaction is favored, producing more products.
• If \( Q > K_c \), the reverse reaction is favored, forming more reactants.
• If \( Q = K_c \), the system is at equilibrium.

Understanding \( Q \) helps in monitoring reaction progress and adjusting conditions to achieve desired yields in industrial and laboratory settings.