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 \(0.2\) mole of \(\mathrm{COCl}_{2}\) and \(0.1\) mole each of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\). The equilibrium constant \(\left(\mathrm{K}_{\mathrm{c}}\right)\) for reaction: \(\mathrm{CO}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{COCl}_{2}\) is (a) 15 (b) 5 (c) 20 (d) 10

Short answer

The equilibrium constant \( K_c \) for the reaction is 10.

Step-by-step solution

Write the reaction equation

The reaction given is \( \mathrm{CO} + \mathrm{Cl}_2 \rightleftharpoons \mathrm{COCl}_2 \). This is a simple single-step reaction at equilibrium.

Determine the volume of the system

The reaction occurs in a 0.5 litre vessel. This volume will be used to convert moles to concentrations for each species in equilibrium.

Calculate concentrations at equilibrium

To find the concentration, use the formula: \( \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} \). For \( \mathrm{COCl}_2 \), \( \frac{0.2}{0.5} = 0.4 \, \text{M} \); for \( \mathrm{CO} \), \( \frac{0.1}{0.5} = 0.2 \, \text{M} \); and for \( \mathrm{Cl}_2 \), \( \frac{0.1}{0.5} = 0.2 \, \text{M} \).

Write the expression for equilibrium constant \( K_c \)

For the reaction, the expression is \[ K_c = \frac{[\mathrm{COCl}_2]}{[\mathrm{CO}][\mathrm{Cl}_2]} \].

Substitute the equilibrium concentrations

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

Calculate \( K_c \)

Perform the arithmetic: \[ K_c = \frac{0.4}{0.04} = 10 \]. Thus, the equilibrium constant is 10.

Key concepts

Chemical Equilibrium

In a chemical reaction, equilibrium is the state where the rates of the forward and reverse reactions become equal. At this point, the concentrations of the reactants and products remain constant over time. This does not mean that the concentrations are equal, but that their rates of change are balanced. For the reaction:

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

At equilibrium, the mixture contains specific amounts of carbon monoxide (\(\mathrm{CO}\)), chlorine (\(\mathrm{Cl}_2\)), and phosgene (\(\mathrm{COCl}_2\)), which do not change because the reactions occur at the same rate in both directions.
Understanding chemical equilibrium can help predict how a system will respond to changes in conditions such as pressure, temperature, or concentration. For instance, according to Le Chatelier's principle, if the concentration of \(\mathrm{Cl}_2\) increases, the system will shift to reduce this change, potentially forming more \(\mathrm{COCl}_2\). Understanding these dynamics is crucial for manipulating reactions in industrial and laboratory settings.

Reaction Quotient

The reaction quotient \( Q \) is calculated in the same way as the equilibrium constant \( K \), but it uses the concentrations of the reactants and products at any point in time. This comparison helps determine the direction in which a reaction will shift to reach equilibrium. For the equation:

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

If the reaction quotient \( Q \) matches the equilibrium constant \( K_c \), the system is in equilibrium. If \( Q < K_c \), the reaction will shift to the right, forming more products. Conversely, if \( Q > K_c \), the reaction will shift to the left, forming more reactants.
In our context, since we calculated \( K_c = 10 \) at equilibrium, if mass action calculations at any other point yield \( Q = 10 \) with given concentrations, it confirms the system is at equilibrium. This tool is incredibly useful in determining how far a reaction is from equilibrium and in which direction it must shift.

Concentration Calculations

To understand equilibrium, it's essential to accurately calculate the concentrations of each species involved in the reaction. The formula used is:

• \( \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} \)

This allows one to determine how much of each reactant and product is present in a given volume.
For the reaction \( \mathrm{CO} + \mathrm{Cl}_2 \rightleftharpoons \mathrm{COCl}_2 \):

• The concentration of \(\mathrm{COCl}_2\) at equilibrium is \( \frac{0.2}{0.5} = 0.4 \ \text{M} \)
• The concentrations of both \(\mathrm{CO}\) and \(\mathrm{Cl}_2\) are each \( \frac{0.1}{0.5} = 0.2 \ \text{M} \)

These concentration calculations are vital in determining the equilibrium constant \( K_c \), which further predicts how the system behaves at equilibrium. Being precise in concentration calculations ensures accurate predictions and manipulations of reaction conditions in both industrial and practical settings.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on the balanced chemical equation. It provides the quantitative relationship between substances as they undergo chemical transformations.
For the reaction:

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

The stoichiometric coefficients indicate a 1:1:1 molar relationship between \(\mathrm{CO}\), \(\mathrm{Cl}_2\), and \(\mathrm{COCl}_2\). This means one mole of \(\mathrm{CO}\) reacts with one mole of \(\mathrm{Cl}_2\) to produce one mole of \(\mathrm{COCl}_2\). Such balance is critical for using the reaction equation to understand how changes in the amount of one substance affect the others.
Using stoichiometry ensures that chemical reactions can be scaled up or down without altering the essential characteristics of the system, making it a cornerstone concept in chemical education and application. This principle helps determine the amounts of different substances needed or produced in a reaction, ensuring efficient and complete reactions.