Question

Consider the reaction. $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ A solution is made containing initial \(\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]=0.020 \mathrm{M}\). At equilibrium, \(\left[\mathrm{Cl}_{2}\right]=1.2 \times 10^{-2} \mathrm{M}\). Calculate the value of the equilibrium constant. Hint: Use the chemical reaction stoichiometry to calculate the equilibrium concentrations of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) and \(\mathrm{SO}_{2}\).

Short answer

The equilibrium constant (Kc) for the reaction is 0.018.

Step-by-step solution

Write Down the Equilibrium Expression

For the given reaction, the equilibrium constant expression (Kc) is based on the concentrations of the gases at equilibrium. It is given by: \( K_c = \frac{[\text{SO}_2][\text{Cl}_2]}{[\text{SO}_2\text{Cl}_2]} \)

Use Stoichiometry to Find Other Equilibrium Concentrations

Since \( \text{SO}_2\text{Cl}_2 \) dissociates into \( \text{SO}_2 \) and \( \text{Cl}_2 \) in a 1:1:1 ratio, the concentration of \( \text{SO}_2 \) at equilibrium will be the same as \( \text{Cl}_2 \), which is \( 1.2 \times 10^{-2} \text{M} \). The change in concentration of \( \text{SO}_2\text{Cl}_2 \) will also be \( 1.2 \times 10^{-2} \text{M} \), so the equilibrium concentration of \( \text{SO}_2\text{Cl}_2 \) is \( 0.020 \text{M} - 1.2 \times 10^{-2} \text{M} \).

Calculate the Change in Concentration

Subtract the change from the initial concentration to find the equilibrium concentration of \( \text{SO}_2\text{Cl}_2 \):\[ [\text{SO}_2\text{Cl}_2]_{eq} = 0.020 \text{M} - 1.2 \times 10^{-2} \text{M} = 8.0 \times 10^{-3} \text{M} \]

Calculate the Equilibrium Constant (Kc)

Plug in the equilibrium concentrations into the equilibrium expression: \[ K_c = \frac{(1.2 \times 10^{-2} \text{M})(1.2 \times 10^{-2} \text{M})}{8.0 \times 10^{-3} \text{M}} \]Calculate Kc to find its value.

Solve for Kc

Perform the arithmetic: \[ K_c = \frac{(1.2 \times 10^{-2} \text{M})^2}{8.0 \times 10^{-3} \text{M}} = \frac{1.44 \times 10^{-4} \text{M}^2}{8.0 \times 10^{-3} \text{M}} = 0.018 \]

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry, representing a state in which the forward and reverse reactions occur at the same rate, leading to constant concentrations of products and reactants over time. In the context of the given reaction, \(\text{SO}_2\text{Cl}_2(g) \rightleftharpoons \text{SO}_2(g) + \text{Cl}_2(g)\), equilibrium will be achieved when the rate at which \(\text{SO}_2\text{Cl}_2\) dissociates into \(\text{SO}_2\) and \(\text{Cl}_2\) equals the rate at which \(\text{SO}_2\) and \(\text{Cl}_2\) recombine to form \(\text{SO}_2\text{Cl}_2\).

This equilibrium state does not imply that the reactants and products are present in equal amounts, but rather that their ratios remain constant over time. It's a dynamic equilibrium because reactions continue happening on both the forward and reverse sides, but without any net change in the concentrations of each species. Understanding this balance is crucial when calculating the equilibrium constant, which quantifies the relative amounts of reactants and products at equilibrium.

Reaction Stoichiometry

Reaction stoichiometry deals with the quantitative relationship between reactants and products in a chemical reaction. It allows us to predict the amounts of products formed from a given amount of reactants or vice versa. In the step-by-step solution provided, stoichiometry plays a vital role in determining the equilibrium concentrations.

In the reaction \(\text{SO}_2\text{Cl}_2(g) \rightleftharpoons \text{SO}_2(g) + \text{Cl}_2(g)\), the stoichiometry is 1:1:1. This means for every mole of \(\text{SO}_2\text{Cl}_2\) that dissociates, one mole of \(\text{SO}_2\) and one mole of \(\text{Cl}_2\) are produced. By utilizing this ratio, we can determine the change in concentration of the reactants and products as the system approaches equilibrium.

Equilibrium Concentrations

Equilibrium concentrations are the amounts of each reactant and product present when the chemical system is at equilibrium. These concentrations are used to calculate the equilibrium constant, which describes the extent of the reaction. As per the exercise solution, once the concentration of \(\text{Cl}_2\) at equilibrium was given as \(1.2 \times 10^{-2} \text{M}\), the same concentration applied to \(\text{SO}_2\) due to the 1:1 stoichiometric ratio. By knowing the initial concentration of \(\text{SO}_2\text{Cl}_2\) and the change in concentration (which is equal to that of \(\text{Cl}_2\)), we easily calculated the equilibrium concentration of \(\text{SO}_2\text{Cl}_2\) to be \(8.0 \times 10^{-3} \text{M}\).

This demonstrates the importance of finding all equilibrium concentrations to ultimately evaluate the equilibrium constant, an indicator of how far the reaction proceeds toward products or reactants under given conditions.

Kc Expression

The equilibrium constant expression, denoted as \(K_c\), relates the concentrations of the products to the concentrations of the reactants, each raised to the power of their respective coefficients in the balanced equation. For the reaction under study, the \(K_c\) expression is \( K_c = \frac{[\text{SO}_2][\text{Cl}_2]}{[\text{SO}_2\text{Cl}_2]} \).

To calculate \(K_c\), we substitute the equilibrium concentrations of \(\text{SO}_2\), \(\text{Cl}_2\), and \(\text{SO}_2\text{Cl}_2\) into the expression. As indicated in the step-by-step solution, this results in \(K_c = \frac{(1.2 \times 10^{-2} \text{M})^2}{8.0 \times 10^{-3} \text{M}} = 0.018\). By doing this, we quantify the position of equilibrium, with a larger \(K_c\) value suggesting a greater concentration of products relative to reactants at equilibrium.