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 \(K_{c}\) in \(m o l L^{-1}\) is (a) \(0.08\) (b) \(0.016\) (c) \(0.004\) (d) \(0.160\)

Step-by-Step Solution

Answer

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

Step 1: Determine the Known Values

Start by noting the given information for the reaction at equilibrium. We have:- 1 mole of \(\text{H}_2\text{S}\)- 0.2 moles of \(\text{H}_2\)- 0.8 moles of \(\text{S}_2\)- All substances are in a 2-liter flask.

Step 2: Calculate Concentrations

Calculate the concentration of each substance by using the formula \( \text{Concentration} = \frac{\text{moles}}{\text{volume}} \).\[[\text{H}_2\text{S}] = \frac{1}{2} = 0.5 \, \text{mol/L}\]\[[\text{H}_2] = \frac{0.2}{2} = 0.1 \, \text{mol/L}\]\[[\text{S}_2] = \frac{0.8}{2} = 0.4 \, \text{mol/L}\]

Step 3: Write the Expression for Kc

The equilibrium constant \( K_c \) expression for the reaction \(2\mathrm{H}_{2}\mathrm{S}(\mathrm{g}) \rightleftharpoons 2\mathrm{H}_{2}(\mathrm{g})+\mathrm{S}_{2}(\mathrm{g})\) is:\[ K_c = \frac{[\text{H}_2]^2[\text{S}_2]}{[\text{H}_2\text{S}]^2} \]

Step 4: Substitute Concentrations into Kc Expression

Substitute the concentrations calculated in the previous steps into the \( K_c \) expression.\[ K_c = \frac{(0.1)^2(0.4)}{(0.5)^2} \]

Step 5: Calculate Kc Value

Perform the calculation to find the value of \( K_c \):\[ K_c = \frac{0.01 \times 0.4}{0.25} = \frac{0.004}{0.25} = 0.016 \]

Key Concepts

Equilibrium Constant (Kc)

In chemical reactions that reach a state of equilibrium, the equilibrium constant, denoted as \( K_c \), is a significant value that helps us understand the proportions of reactants and products at equilibrium. It is derived from the balanced equation of the reaction and represents the ratio of the products' concentrations to the reactants' concentrations, each raised to the power of their respective coefficients in the balanced equation.
For example, in the reaction \( 2 ext{H}_2 ext{S} ightleftharpoons 2 ext{H}_2 + ext{S}_2 \), the \( K_c \) expression is:

• \( K_c = \frac{[ ext{H}_2]^2[ ext{S}_2]}{[ ext{H}_2 ext{S}]^2} \)

This equation tells us how the concentrations of the different chemical species relate to one another at equilibrium. A higher \( K_c \) value indicates a higher concentration of products compared to reactants at equilibrium, while a lower \( K_c \) suggests more reactants compared to products.

Reaction Quotient

Before a reaction reaches equilibrium, we can calculate the reaction quotient, \( Q_c \), using the same formula as \( K_c \). However, unlike \( K_c \), which uses equilibrium concentrations, \( Q_c \) is calculated with the concentrations at any point during the reaction. This provides a way to determine whether a reaction is at equilibrium or in which direction it will proceed to reach equilibrium.

• If \( Q_c = K_c \), the reaction is at equilibrium and no net change will occur in the concentrations of reactants and products.
• If \( Q_c < K_c \), the reaction will proceed in the forward direction, producing more products.
• If \( Q_c > K_c \), the reaction will move in the reverse direction, producing more reactants.

Thus, \( Q_c \) is an important tool for predicting and understanding the dynamics of chemical reactions as they approach equilibrium.

Mole Concept

The mole is a fundamental concept in chemistry that relates the mass of a substance to the number of entities, such as atoms, molecules, or ions. One mole equals Avogadro's number (approximately \( 6.022 imes 10^{23} \)) of these entities.
In equilibrium calculations, like the one described in the original exercise, the concept of moles is used to determine concentrations. By converting the given moles of each species to moles per liter, or molarity, we gain valuable insights into the equilibrium state of the system. For instance, in a 2-liter flask:

• 1 mole of \( ext{H}_2 ext{S}\) becomes 0.5 mol/L.
• 0.2 moles of \( ext{H}_2\) becomes 0.1 mol/L.
• 0.8 moles of \( ext{S}_2\) becomes 0.4 mol/L.

Understanding the mole concept allows us to correctly calculate and interpret these concentrations in relation to the equilibrium constant.

Concentration in Equilibrium Reactions

Concentration is a crucial parameter in determining the behavior of chemical reactions at equilibrium. In equilibrium reactions, the concentration of each reactant and product remains constant over time, though they are not necessarily equal.
To calculate the concentration of a substance in a reaction at equilibrium, use the formula:

• Concentration \( = \frac{\text{moles}}{\text{volume}} \)

Once the concentrations are established, they can be substituted into the \( K_c \) expression to better understand the equilibrium state. For example, the concentration of \( ext{H}_2\) in the exercise is calculated as \(0.1 ext{ mol/L}\) given 0.2 moles are present in a 2-liter flask.
Accurate concentration measurements are vital for calculating \( K_c \) and predicting the extent and direction of reactions. This helps chemists control and optimize reactions by manipulating variables such as concentration and temperature.