Problem 109 - A Complete Resource Book in Chemistry for JEE Main

Question

In a \(0.5\) litre capacity vessel, \(\mathrm{CO}\) and \(\mathrm{Cl}\), are mixed to form \(\mathrm{COCl}_{2} .\) At equilibrium, it contains \(0.2\) mole of \(\mathrm{COCl}_{2}\) and \(0.1\) mole each of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\). The equilibrium constant \(\left(\mathrm{K}_{c}\right)\) for reaction \(\mathrm{CO}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{COCl}_{2}\) is (a) 15 (b) 5 (c) 20 (d) 10

Step-by-Step Solution

Answer

The equilibrium constant \( K_c \) is 10.

Step 1: Write the Balanced Equation

The given chemical reaction is \( \mathrm{CO} + \mathrm{Cl}_2 \rightleftharpoons \mathrm{COCl}_2 \).

Step 2: Determine Concentrations

Calculate the molar concentrations of each species using the given number of moles and the vessel volume of 0.5 L. - For \( \mathrm{COCl}_2 \), the concentration is \( \frac{0.2 \, \text{moles}}{0.5 \, \text{L}} = 0.4 \, \text{mol/L} \).- For \( \mathrm{CO} \), the concentration is \( \frac{0.1 \, \text{moles}}{0.5 \, \text{L}} = 0.2 \, \text{mol/L} \).- For \( \mathrm{Cl}_2 \), the concentration is \( \frac{0.1 \, \text{moles}}{0.5 \, \text{L}} = 0.2 \, \text{mol/L} \).

Step 3: Write the Expression for Kc

The equilibrium constant expression \( K_c \) for the reaction \( \mathrm{CO} + \mathrm{Cl}_2 \rightleftharpoons \mathrm{COCl}_2 \) is:\[ K_c = \frac{[\mathrm{COCl}_2]}{[\mathrm{CO}][\mathrm{Cl}_2]} \]

Step 4: Substitute Concentrations into Kc Expression

Substitute the calculated concentrations into the \( K_c \) expression:\[ K_c = \frac{0.4}{0.2 \times 0.2} \]

Step 5: Calculate Kc

Solve the equation:\[ K_c = \frac{0.4}{0.04} = 10 \]

Key Concepts

Chemical Equilibrium

In chemistry, reaching a state where the forward and reverse reactions occur at the same rate without any net change in the concentration of reactants and products is called chemical equilibrium. At this point, all substances involved in the reaction settle in specific constant concentrations. This concept is vital because it explains how reactions reach stability.
The equilibrium state is dynamic, meaning reactions continue to occur, but there's no noticeable change in concentrations over time. For students learning about equilibrium, it's essential to understand these key points:

• Reactions at equilibrium do not stop; they proceed with no net change in concentrations.
• Equilibrium can be established in both closed systems, where no reactants or products can escape, and certain open systems under specific conditions.
• Temperature, pressure, and concentration changes can shift the equilibrium position, described by Le Chatelier's Principle.

Molar Concentration

Molar concentration, often called molarity, measures how much of a substance is present in a given volume of solution. It's expressed as moles per liter (mol/L), symbolized as \(M\). Molar concentration is crucial for calculating many chemical equations, particularly in finding equilibrium constants.
In our equilibrium example, calculating the molar concentration of each substance involved was crucial for finding the equilibrium constant \(K_c\):

• For \(\mathrm{COCl}_2\), we calculated its concentration by dividing the moles of \(\mathrm{COCl}_2\) by the volume of the vessel, resulting in \(0.4\, \text{mol/L}\).
• Similarly, \(\mathrm{CO}\) and \(\mathrm{Cl}_2\) each had concentrations of \(0.2 \text{mol/L}\) calculated in the same manner.

This step is necessary because these concentrations directly substitute into the equilibrium expression to yield the equilibrium constant. Knowing how to calculate molar concentration is essential not only for equilibrium but for many other areas of chemistry.

Balanced Chemical Equation

The essence of a balanced chemical equation is that it respects the Law of Conservation of Mass, ensuring that the number of atoms for each element in the reactants equals the number of atoms for those elements in the products.
Balancing is essential because it provides a clear understanding of the stoichiometry of the reaction, which is the quantitative relationship between reactants and products. The equation we dealt with was:

• \(\mathrm{CO} + \mathrm{Cl}_2 \rightleftharpoons \mathrm{COCl}_2\)

For this reaction, it's balanced because there's one carbon (C) and two chlorine (Cl) atoms present on both sides of the equation. This balance allows us to calculate the equilibrium constant because it informs us about the relation and proportion of moles participating in the reaction. Additionally, a balanced equation is foundational for any quantitative chemical analysis, ensuring accurate calculations and predictions.