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

Question

In a \(0.5\) litre capacity vessel, \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) are mixed to form \(\mathrm{COCl}_{2}\). At equilibrium, it contains \(0.2 \mathrm{~mole}\) of \(\mathrm{COCl}_{2}\) and \(0.1\) mole each of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\). The equilibrium constant \(\left(\mathrm{K}_{0}\right)\) 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 \( K_0 \) is 10.

Step 1: Write the Equilibrium Expression

Start by writing the equilibrium expression for the reaction: \( \mathrm{CO} + \mathrm{Cl}_2 \rightleftharpoons \mathrm{COCl}_2 \). The equilibrium constant \( K_0 \) is defined as \( K_0 = \frac{\text{[Products]}}{\text{[Reactants]}} = \frac{[\mathrm{COCl}_2]}{[\mathrm{CO}][\mathrm{Cl}_2]} \).

Step 2: Calculate Equilibrium Concentrations

Next, calculate the molarity of each substance at equilibrium. The solution is in a \( 0.5 \) litre vessel. For each substance, use the formula: concentration (M) = moles/volume \( (L) \). - \([\mathrm{CO}] = \frac{0.1}{0.5} = 0.2 \text{ M} \)- \([\mathrm{Cl}_2] = \frac{0.1}{0.5} = 0.2 \text{ M} \)- \([\mathrm{COCl}_2] = \frac{0.2}{0.5} = 0.4 \text{ M} \)

Step 3: Substitute Concentrations into Equilibrium Expression

Substitute the calculated concentrations back into the equilibrium expression: \[ K_0 = \frac{0.4}{(0.2)(0.2)} \]

Step 4: Simplify and Solve for \( K_0 \)

Simplify the expression from the previous step: \[ K_0 = \frac{0.4}{0.04} = 10 \].

Step 5: Compare with Given Options

Finally, compare your calculated value of \( K_0 \) with the given options. The calculated \( K_0 \) is 10, which matches option (d): 10.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentration of reactants and products remains constant over time. It occurs in reversible reactions where the forward and reverse reaction rates are equal. This means that, although the individual molecules continue to react, there is no net change in the concentration of reactants and products.
Here are some important points to keep in mind about chemical equilibrium:

• The system is in a dynamic balance, not static. Molecules constantly react, but there is no overall change.
• Equilibrium can be reached from both directions of the reaction — starting with reactants or products.
• It does not mean reactants and products are equal in concentration but that their rates of formation are equal.

A key aspect of chemical equilibrium is understanding the equilibrium constant ( K_0 ). This expression relies on the concentrations of products and reactants at equilibrium, offering a snapshot of the reaction's extent. The equation for K_0 in the given exercise is structured as the concentration of the product divided by the concentrations of the reactants.

Molarity Calculation

Molarity is the measure of the concentration of a solute in a solution, expressed as the number of moles of solute per liter of solution (mol/L). This concept is crucial when dealing with chemical reactions in solution, as it directly relates to how much of a substance is involved in the reaction.
In the problem at hand, calculating molarity involved using the formula:

• Molarity (M) = \( \text{moles of solute} \ \text{/}\ \text{volume of solution in liters} \)

For example, for \(\text{CO}\), which is available in a 0.5 L vessel:

• \( [\text{CO}] = \frac{0.1\ \text{moles}}{0.5\ \text{L}} = 0.2\ \text{M} \)
• The same calculation applies to \(\text{Cl}_2\), resulting in a molarity of 0.2 M.
• For \(\text{COCl}_2\), the molarity is \(\frac{0.2\ \text{moles}}{0.5\ \text{L}} = 0.4\ \text{M} \)

Understanding molarity allows you to correctly input the concentrations into the equilibrium expression, which is essential for determining the equilibrium constant.

Reaction Stoichiometry

Reaction stoichiometry involves the calculation of the relative quantities of reactants and products in a chemical reaction. Reflecting the balanced chemical equation, this concept helps determine how much reactant is needed to produce a certain amount of product.
In the context of the provided exercise, stoichiometry identifies the optimal ratios of molecules. For the reaction \( \text{CO} + \text{Cl}_2 \rightleftharpoons \text{COCl}_2 \), the stoichiometric coefficients are all one, implying a 1:1:1 ratio. This means:

• One mole of CO reacts with one mole of \(\text{Cl}_2\) to form one mole of \(\text{COCl}_2\).
• Whenever you have certain moles of CO and \(\text{Cl}_2\), they should theoretically yield an equal mole amount of the product, \(\text{COCl}_2\).

Stoichiometry ensures the reaction aligns with these ratios at any point, especially at equilibrium, helping predict the amounts of each compound present. It is an essential tool in Reactants-Products calculations, leading to a deeper understanding and description of the chemical process at an equilibrium state.