Problem 112 - A Complete Resource Book in Chemistry for JEE Main

Question

\(1.25\) moles of NOCl were placed in a \(2.50 \mathrm{~L}\) reaction chamberat \(427^{\circ} \mathrm{C}\). After equilibrium was reached, \(1.10\) moles of NOCl remained. Calculate the equilibrium constant \(K_{c}\) for the reaction, \(2 \mathrm{NOC} 1(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (a) \(1.6 \times 10^{-3}\) (b) \(5.6 \times 10^{-4}\) (c) \(2.6 \times 10^{-3}\) (d) \(4.6 \times 10^{-4}\)

Step-by-Step Solution

Answer

The equilibrium constant \( K_c \) is approximately \(5.6 \times 10^{-4}\).

Step 1: Determine Initial Concentration of NOCl

First, calculate the initial concentration of NOCl using the formula \( \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} \). The initial moles of NOCl is 1.25 moles and the volume of the container is 2.50 L, thus the initial concentration is \( \frac{1.25}{2.50} = 0.50 \text{ M} \).

Step 2: Determine Change and Equilibrium Concentration of NOCl

Since the equilibrium concentration of NOCl is 1.10 moles in 2.50 L, calculate its equilibrium concentration as \( \frac{1.10}{2.50} = 0.44 \text{ M} \). The change in concentration is then \( 0.50 - 0.44 = 0.06 \text{ M} \).

Step 3: Use Stoichiometry to Find Changes in NO and Cl2 Concentrations

According to the reaction, \(2\) moles of NOCl produce \(2\) moles of NO and \(1\) mole of Cl2. Thus, a change of \(0.06 \text{ M}\) in NOCl means a change of \(0.06 \text{ M} \) in NO and \(0.03 \text{ M}\) in Cl2.

Step 4: Calculate Concentrations at Equilibrium for NO and Cl2

The equilibrium concentration of NO is \(0 + 0.06 = 0.06 \text{ M} \) and for Cl2 is \(0 + 0.03 = 0.03 \text{ M} \).

Step 5: Calculate the Equilibrium Constant Kc

The expression for the equilibrium constant \( K_c \) is \[ K_c = \frac{[\text{NO}]^2 [\text{Cl}_2]}{[\text{NOCl}]^2} \]. Substitute the equilibrium concentrations to get \[ K_c = \frac{(0.06)^2 \times (0.03)}{(0.44)^2} = \frac{0.000108}{0.1936} \approx 5.6 \times 10^{-4} \].

Key Concepts

Understanding Chemical Equilibrium

In a chemical reaction, chemical equilibrium is reached when the rates of the forward and reverse reactions are equal. Consider it like a busy seesaw where people are moving back and forth at an equal rate. At this point, there's no net change in the concentrations of the reactants and products. This doesn't mean the reactions stop; instead, they continue happening at the same speed in both directions.

For the reaction \(2 \text{NOCl} (g) \rightleftharpoons 2 \text{NO} (g) + \text{Cl}_2 (g)\), the forward reaction is where NOCl dissociates, while the reverse reaction involves the combination of NO and \(\text{Cl}_2\) to form NOCl again. Equilibrium is dynamic, meaning while the concentrations don't change, the molecules constantly react.

• Dynamic Nature: At equilibrium, reactions haven't "stopped"—they're just balanced.
• Concentration Stability: The concentrations of reactants and products remain constant.

Grasping Reaction Stoichiometry

Stoichiometry involves understanding the relationship between the quantities of reactants and products in a chemical reaction. It is the math behind chemical reactions, calculated from a balanced chemical equation.

For the reaction given, each time 2 moles of NOCl decompose, it produces 2 moles of NO and 1 mole of \(\text{Cl}_2\). This ratio remains constant and helps us determine how much of one substance reacts with or is produced from another. In our problem, the change in NOCl concentration determined the changes in the concentrations of NO and \(\text{Cl}_2\):

• NOCl Decomposition: 0.06 M change in NOCl leads to a 0.06 M increase in NO and 0.03 M increase in \(\text{Cl}_2\).
• Proportionality: Each proportion is determined by the coefficients in the balanced equation.

Mastering Molar Concentration

Molar concentration, or molarity, is a measure of the concentration of a solute in a solution. It tells us how much of a substance is present in a certain volume, usually expressed in moles per liter (M).

In the exercise, the molarity is used to reflect the concentration of NOCl initially and at equilibrium. Calculating molarity is done using the formula: \[\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\]Knowing these concentrations helps us understand how much of our starting material has reacted and allows for the determination of the equilibrium constant:

• Initial Molarity: With 1.25 moles of NOCl in a 2.50 L container, the starting molarity is 0.50 M.
• Equilibrium Molarity: Given 1.10 moles remain, calculate as \( \frac{1.10}{2.50} = 0.44 \text{ M} \).

Understanding molarity is crucial for predicting how reactions proceed and how they balance at equilibrium.