Question

The equilibrium constant for the reaction: $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g) $$ is \(2.8 \times 10^{2}\) at a certain temperature. If \(\left[\mathrm{SO}_{2}\right]=0.0124 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.031 M,\) what is \(\left[\mathrm{SO}_{3}\right] ?\)

Short answer

The concentration of \(\mathrm{SO}_3\) is 0.1155 M.

Step-by-step solution

Write the Equilibrium Expression

For the given reaction \(2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g)\), the equilibrium expression for the equilibrium constant \(K_c\) is defined as \(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}\).

Substitute the Known Values

We are given \(K_c = 2.8 \times 10^2\), \([\text{SO}_2] = 0.0124 \text{ M}\), and \([\text{O}_2] = 0.031 \text{ M}\). Substitute these values into the equilibrium expression: \(2.8 \times 10^2 = \frac{[\text{SO}_3]^2}{(0.0124)^2 \times 0.031}\).

Rearrange the Equation

To find \([\text{SO}_3]\), rearrange the equation to solve for \([\text{SO}_3]^2\): \([\text{SO}_3]^2 = 2.8 \times 10^2 \times (0.0124)^2 \times 0.031\).

Calculate the Right-Hand Side

Calculate the value of the expression on the right-hand side: \([\text{SO}_3]^2 = 2.8 \times 10^2 \times 0.0124^2 \times 0.031 = 0.0133472\).

Solve for [SO3]

Find \([\text{SO}_3]\) by taking the square root of the calculated value: \([\text{SO}_3] = \sqrt{0.0133472} = 0.1155\ \text{M}\).

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_c\), plays a crucial role in understanding chemical equilibrium. It describes the ratio of the concentration of products to reactants at equilibrium for a reversible chemical reaction. This specific value is constant at a given temperature.
In the reaction \(2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g)\), the equilibrium constant \(K_c = 2.8 \times 10^2\) is calculated based on the formula:

• \(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}\)

This equation highlights the relationship between the concentrations of the substances involved in the reaction. It's important to understand that the equilibrium constant is dimensionless and only applicable when concentrations are expressed in molarity (M). Each concentration must be raised to the power of its stoichiometric coefficient from the balanced equation, ensuring the correct calculation of \(K_c\).

Concentration Calculation

Concentration calculations are key in determining the amount of substances present at equilibrium. Given the values and the equilibrium constant, you can calculate unknown concentrations using algebra.
For our example, you are tasked with finding \([\text{SO}_3]\) when given \([\text{SO}_2] = 0.0124 \text{ M}\) and \([\text{O}_2] = 0.031 \text{ M}\). Using the equilibrium expression:

• \(2.8 \times 10^2 = \frac{[\text{SO}_3]^2}{(0.0124)^2 \times 0.031}\)

We rearrange the equation to solve for \([\text{SO}_3]^2\):

• \([\text{SO}_3]^2 = 2.8 \times 10^2 \times (0.0124)^2 \times 0.031\)

After calculating the right side, find the square root to obtain \([\text{SO}_3]\). This step is crucial because it relies on accurate arithmetic manipulation of the equilibrium equation components.

Reaction Quotient

The reaction quotient \(Q\) compares the ratio of product and reactant concentrations at any point in time, not just at equilibrium.
While similar to the equilibrium constant expression, \(Q\) helps you determine the direction in which a reaction will proceed to reach equilibrium.

• If \(Q < K_c\), the reaction will proceed forward, forming more products.
• If \(Q > K_c\), the reaction will proceed in reverse, forming more reactants.
• If \(Q = K_c\), the system is at equilibrium.

In our example problem, knowing that \(Q\) reaches \(K_c\) allows us to proceed with calculating equilibrium concentrations. This tool is invaluable when you mix reactants and products to predict reaction behavior before reaching equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle is a guiding rule to predict how a change in conditions can affect the position of equilibrium. It states that if a dynamic equilibrium system is disturbed, the system will adjust itself to counteract the effect of the disturbance, maintaining balance.
Here are some ways Le Chatelier's Principle can affect the equilibrium:

• Changing concentration: Adding more reactant or product will shift the equilibrium to use up the added substances. For example, adding more \(\text{SO}_2\) would push the reaction right, forming more \(\text{SO}_3\).
• Changing temperature: Raising the temperature of an exothermic reaction will shift the equilibrium left. Lowering it shifts it right.
• Changing pressure: For reactions involving gases, increasing pressure by reducing volume will shift the equilibrium toward the side with fewer gas molecules.

Understanding Le Chatelier’s Principle helps predict the effects of environmental changes on chemical reactions and is vital for chemical equilibrium analysis.