Problem 8 - Chemistry and Chemical Reactivity

Question

An equilibrium mixture of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at a high temperature contains the gases at the following concentrations: \(\left[\mathrm{SO}_{2}\right]=3.77 \times 10^{-3} \mathrm{mol} / \mathrm{L}\) \(\left[\mathrm{O}_{2}\right]=4.30 \times 10^{-3} \mathrm{mol} / \mathrm{L},\) and \(\left[\mathrm{SO}_{3}\right]=4.13 \times\) \(10^{-3} \mathrm{mol} / \mathrm{L} .\) Calculate the equilibrium constant, \(K_{c}\) for the reaction. $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) $$

Step-by-Step Solution

Answer

The equilibrium constant \(K_c\) for the reaction is approximately 278.8.

Step 1: Write the expression for the equilibrium constant

The reaction provided is \(2\, \text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\, \text{SO}_3(\text{g})\). The equilibrium constant, \(K_c\), is given by the ratio of the concentrations of products to reactants raised to the power of their stoichiometric coefficients. Thus, the \(K_c\) expression becomes: \[K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 \cdot [\text{O}_2]}\]

Step 2: Substitute the concentrations into the equilibrium expression

Using the provided concentrations, substitute the values into the \(K_c\) expression: - \([\text{SO}_2] = 3.77 \times 10^{-3}\, \text{mol/L}\)- \([\text{O}_2] = 4.30 \times 10^{-3}\, \text{mol/L}\)- \([\text{SO}_3] = 4.13 \times 10^{-3}\, \text{mol/L}\)Substituting these concentrations, we have: \[K_c = \frac{(4.13 \times 10^{-3})^2}{(3.77 \times 10^{-3})^2 \cdot (4.30 \times 10^{-3})}\]

Step 3: Calculate \(K_c\) value

First, calculate \((4.13 \times 10^{-3})^2 = 1.706 \times 10^{-5}\), then \((3.77 \times 10^{-3})^2 = 1.4229 \times 10^{-5}\). Now, substitute these values back into the expression for \(K_c\): \[K_c = \frac{1.706 \times 10^{-5}}{1.4229 \times 10^{-5} \times 4.30 \times 10^{-3}}\]Calculate the denominator: \[1.4229 \times 10^{-5} \times 4.30 \times 10^{-3} = 6.11847 \times 10^{-8}\]Finally, divide to get \(K_c\): \[K_c \approx \frac{1.706 \times 10^{-5}}{6.11847 \times 10^{-8}} = 278.8\]

Step 4: Finalize the answer

Thus, after calculating the operations in the expression, the equilibrium constant \(K_c\) for the reaction at the given temperatures and concentrations is approximately \(278.8\).

Key Concepts

Chemical Equilibrium

In a chemical reaction, equilibrium is the state where the rate of the forward reaction equals the rate of the reverse reaction. This means that the concentrations of reactants and products remain constant over time, giving the impression that the reaction has "stopped." However, reactions at equilibrium are dynamic, meaning they continuously occur but with no net change in the concentration of reactants and products.
This balance of reactions can be visually represented in the reaction of sulfur dioxide ( SO_2 ), oxygen ( O_2 ), and sulfur trioxide ( SO_3 ). When the system reaches equilibrium, the amounts of each gaseous substance stabilize under specific conditions. The equilibrium state is described by the equilibrium constant, which helps determine the ratio of concentrations at equilibrium. This can differ with changes in temperature or pressure, altering the balance and potentially shifting the reaction towards more products or reactants.

Chemical Reaction

A chemical reaction involves the transformation of reactants into products through the breaking and forming of chemical bonds. In the case of the SO_2 and O_2 reacting to form SO_3, the chemical reaction can be written as:\[2 \, \mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2 \, \mathrm{SO}_3(\mathrm{g})\]

• The symbol "\(\rightleftharpoons\)" indicates that the reaction is reversible, meaning the reactants can convert to products and vice versa.
• This balanced equation reflects the stoichiometry of the reactants and products, showing the exact proportions in which they react or are produced.

In this reaction, two moles of SO_2 react with one mole of O_2 to form two moles of SO_3.

Reaction Quotient

The reaction quotient, Q_c, is a measure used to determine the direction in which a reaction is likely to proceed given the current concentrations of reactants and products. It is calculated using the same expression as the equilibrium constant but with the initial concentrations instead of those at equilibrium.
The reaction quotient for the reaction between SO_2, O_2, and SO_3 can be expressed as:\[Q_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 \cdot [\text{O}_2]}\]

• If \(Q_c = K_c\), the system is at equilibrium.
• If \(Q_c < K_c\), the forward reaction is favored, and the system will generate more products.
• If \(Q_c > K_c\), the reverse reaction is favored, and the system will produce more reactants.

So, the reaction quotient helps predict whether a reaction will move towards equilibrium by forming more products or producing more of the reactants.

Concentration

Concentration refers to the amount of a substance within a certain volume of a solution or mixture. It is typically expressed in moles per liter (mol/L), also known as molarity. In the context of chemical equilibrium, the concentration of reactants and products at equilibrium plays a crucial role in defining the system's equilibrium state and thus, the equilibrium constant, K_c.
For the reaction of SO_2, O_2, and SO_3:

• The concentration of SO_2 is given as \(3.77 \times 10^{-3} \text{ mol/L}\).
• The concentration of O_2 is \(4.30 \times 10^{-3} \text{ mol/L}\).
• The concentration of SO_3 is \(4.13 \times 10^{-3} \text{ mol/L}\).

These concentrations are substituted into the equilibrium expression to evaluate the equilibrium constant and understand the reaction's dynamics at given conditions.

Stoichiometry

Stoichiometry is the study of the quantitative relationships between the reactants and products in a chemical reaction. It allows us to predict how much product will form from given quantities of reactants and is crucial for calculating the equilibrium constant. In stoichiometry, coefficients in a balanced chemical equation represent the mole ratio of the reactants and products.
For the chemical reaction at hand:

• The ratio is 2:1:2, meaning two moles of SO_2 reacted with one mole of O_2 to produce two moles of SO_3.
• The stoichiometric coefficients are used as exponents in the equilibrium expression:\[K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 \cdot [\text{O}_2]}\]

By understanding stoichiometry, you gain insight into the precise balance necessary for carrying out chemical reactions efficiently, helping predict how changes in the amounts of substances can shift the equilibrium position.