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 \((\mathrm{K})\) 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 \(\mathrm{K}\) is 10. The answer is (d).

Step 1: Write the Equilibrium Expression

The equilibrium expression for the reaction \[ \mathrm{CO} + \mathrm{Cl}_2 \rightleftharpoons \mathrm{COCl}_2 \]is given by:\[ \mathrm{K} = \frac{[\mathrm{COCl}_2]}{[\mathrm{CO}][\mathrm{Cl}_2]} \]

Step 2: Calculate Molarity of Each Component

Calculate the molarity of each component, using the formula\[ \text{Molarity} = \frac{\text{moles}}{\text{volume}} \]The volume of the container is 0.5 litres.For \(\mathrm{COCl}_{2}\):\[ [\mathrm{COCl}_2] = \frac{0.2}{0.5} = 0.4 \, \text{M} \]For \(\mathrm{CO}\):\[ [\mathrm{CO}] = \frac{0.1}{0.5} = 0.2 \, \text{M} \]For \(\mathrm{Cl}_2\):\[ [\mathrm{Cl}_2] = \frac{0.1}{0.5} = 0.2 \, \text{M} \]

Step 3: Substitute the Molarities into Equilibrium Expression

Substitute the molarities calculated in Step 2 into the equilibrium expression from Step 1:\[\mathrm{K} = \frac{0.4}{0.2 \times 0.2}\]

Step 4: Calculate the Equilibrium Constant

Perform the arithmetic to solve for \(\mathrm{K}\):\[\mathrm{K} = \frac{0.4}{0.04} = 10\]

Key Concepts

Chemical Equilibrium

Imagine a balance scale, perfectly level, with equal weights on both sides. Chemical equilibrium is a bit like that. It is the state in which both the reactants and products remain unchanged over time. In a chemical reaction, the forward and backward processes occur at the same rate. This means the amount of each substance stays constant. It doesn’t stop the reactions from happening, but rather achieves a balance. In our example, the reaction of carbon monoxide (\( \mathrm{CO} \)) with chlorine (\( \mathrm{Cl}_2 \)) to form phosgene (\( \mathrm{COCl}_2 \)) reaches a point where the formation of \( \mathrm{COCl}_2 \) and its breakdown happen at the same rate, maintaining steady concentrations.

Reaction Quotient

Similar to a snapshot taken of a reaction mixture at any given time, the reaction quotient (\( Q \)) helps predict the direction a reaction will proceed to reach equilibrium. It is calculated like the equilibrium constant (\( K \)), using initial concentrations of reactants and products. - If \( Q = K \), the system is at equilibrium.- If \( Q < K \), the reaction will proceed forward (producing more products) to reach equilibrium.- If \( Q > K \), the reaction will go in reverse (producing more reactants) to reach equilibrium. In the exercise given, finding \( K \) confirms what \( Q \) would approach when the system reaches equilibrium. Both concepts assure us that reactions adjust to achieve this balance.

Mole Concept

The mole is a basic counting unit in chemistry, analogous to a dozen or a pair, but significantly larger. One mole contains exactly \( 6.022 \times 10^{23} \) entities, be it atoms, molecules, or ions. This immense number is named Avogadro's number. It's crucial for quantifying chemical reactions. To find out how much of each substance reacts or forms, chemists use the mole concept. In the given problem, the numbers of moles for each chemical are directly used to calculate their concentrations. Here, we have:- \( 0.1 \) mole of \( \mathrm{CO} \) and \( \mathrm{Cl}_2 \) each.- \( 0.2 \) mole of \( \mathrm{COCl}_2 \).These values relate to the reaction's stoichiometry, providing a solid base to calculate molarities and understand the reaction dynamics.

Molarity Calculation

Molarity is a way to express the concentration of a solution. It tells us how many moles of a solute are in one liter of solution (\( \text{mol/L} \), M). To find the molarity:- Divide the amount of substance in moles by the solution's volume in liters.- This is highly practical, especially when using the mole concept to convert mass from grams to moles, and then to molarity using the solution volume. In the exercise, the container volume is \( 0.5 \) liters. The molarities are thus:- \( [\mathrm{COCl}_2] = \frac{0.2}{0.5} = 0.4 \, \text{M} \)- \( [\mathrm{CO}] = \frac{0.1}{0.5} = 0.2 \, \text{M} \)- \( [\mathrm{Cl}_2] = \frac{0.1}{0.5} = 0.2 \, \text{M} \)These calculations are crucial for the equilibrium expression, providing the means to solve for \( K \), the equilibrium constant. Understanding molarity helps grasp the relationship between the quantities of substances in a reaction and their concentrations at equilibrium.