Problem 71 - A Complete Resource Book in Chemistry for JEE Main

Question

In the reaction \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{S}_{2}(\mathrm{~g})\) the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is \(0.5 \mathrm{~mol} \mathrm{~L}^{-1}\) and concentration of \(\mathrm{H}_{2}\) is \(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\) while concentration of \(\mathrm{S}_{2}\) is \(0.4\) \(\operatorname{mol} \mathrm{L}^{-1}\) in one litre vessel. The value of equilibrium constant of the reaction is (a) \(0.016\) (b) \(0.013\) (c) \(0.020\) (d) \(0.030\)

Step-by-Step Solution

Answer

The equilibrium constant \(K_c\) is 0.016, option (a).

Step 1: Write the Equilibrium Constant Expression

For the given 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 \([\text{ ] }\) denotes concentration of each species at equilibrium.

Step 2: Substitute the Concentrations into the Expression

Substitute the given concentrations into the equilibrium constant expression:\([\mathrm{H}_{2} \mathrm{~S}] = 0.5 \mathrm{~mol L}^{-1}\), \([\mathrm{H}_{2}] = 0.1 \mathrm{~mol L}^{-1}\), and \([\mathrm{S}_{2}] = 0.4 \mathrm{~mol L}^{-1}\).Replace these values into the equation:\[ K_c = \frac{(0.1)^2 \times 0.4}{(0.5)^2} \]

Step 3: Calculate the Values of the Numerator and Denominator

Evaluate the numerator and denominator separately:- **Numerator:** \((0.1)^2 \times 0.4 = 0.01 \times 0.4 = 0.004\)- **Denominator:** \((0.5)^2 = 0.25\)

Step 4: Compute the Equilibrium Constant \(K_c\)

Calculate \(K_c\) by dividing the numerator by the denominator:\[ K_c = \frac{0.004}{0.25} = 0.016 \]

Step 5: Compare Calculated \(K_c\) with Given Options

Match the calculated \(K_c\) value of 0.016 with the options provided. It matches option (a), which is 0.016.

Key Concepts

Equilibrium Expressions

Equilibrium expressions play an integral role in understanding how chemical reactions achieve balance between the reactants and products. In our example, the reaction is:\[2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{S}_{2}(\mathrm{~g})\]To express this reaction mathematically, we use an equilibrium constant expression, denoted as \(K_c\). This expression is a formula that relates the concentrations of products and reactants when a reaction is at equilibrium. For the given reaction, the expression for \(K_c\) is:\[ K_c = \frac{[\mathrm{H}_{2}]^2 [\mathrm{S}_{2}]}{[\mathrm{H}_{2} \mathrm{~S}]^2} \]Here’s a breakdown of the terms:

• Numerator: It includes the concentrations of the products (\([\mathrm{H}_{2}]^2\) and \([\mathrm{S}_{2}]\)).
• Denominator: It involves the concentrations of the reactants (\([\mathrm{H}_{2} \mathrm{~S}]^2\)).

Concentration values are raised to the power of their coefficients in the balanced equation. Understanding this structure is key to calculating equilibrium constants in various chemical reactions.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction reaches a state where the concentrations of reactants and products no longer change with time. This does not mean that the reactants and products are equal in concentration, but rather that their rates of formation are equal, resulting in a steady state.In the provided reaction:2 \(\text{H}_{2} \text{S}(\text{g}) \rightleftharpoons 2 \text{H}_{2}(\text{g})+\text{S}_{2}(\text{g})\)The system will eventually stabilize such that there is no net change in the concentrations of \(\text{H}_{2} \text{S}\), \(\text{H}_{2}\), and \(\text{S}_{2}\).Key points to remember about chemical equilibrium:

• Dynamic Process: At equilibrium, reactions continue to occur, but there is a balance between the forward and reverse reactions.
• Dependent on Temperature and Pressure: The position of equilibrium can shift if external conditions such as temperature or pressure are changed.
• Equilibrium Constant \(K_c\): This value tells us how far the reaction will proceed, based on the ratio of product concentrations to reactant concentrations at equilibrium.

Mastering the concept of chemical equilibrium is crucial for predicting the behavior of reactions under different conditions.

Mole Concentration

The term 'mole concentration,' also known as molarity, is crucial in understanding and performing equilibrium calculations. Molarity is the measure of the moles of a solute present in a liter of solution, typically expressed as \(\text{mol L}^{-1}\).In our example, you are given:

• \([\text{H}_{2} \text{S}] = 0.5 \text{ mol L}^{-1}\): This means there are 0.5 moles of \(\text{H}_{2} \text{S}\) per litre of solution.
• \([\text{H}_{2}] = 0.1 \text{ mol L}^{-1}\): This indicates 0.1 moles of \(\text{H}_{2}\) per litre.
• \([\text{S}_{2}] = 0.4 \text{ mol L}^{-1}\): This reflects 0.4 moles of \(\text{S}_{2}\) per litre.

Understanding mole concentration is essential for setting up reliable equilibrium calculations. It allows you to apply these concentrations into equilibrium expressions and compare different scenarios of a reaction under equilibrium conditions. Calculating these accurately ensures precise determination of the equilibrium constant \(K_c\).