Problem 44 - Chemistry The Molecular Science

Question

Chemists carried out a study of the high temperature reaction of sulfur dioxide with oxygen in which a sealed reactor initially contained \(0.0076-\mathrm{M} \mathrm{SO}_{2}, 0.0036-\mathrm{M} \mathrm{O}_{2}\), and no \(\mathrm{SO}_{3}\). After equilibrium was achieved, the \(\mathrm{SO}_{2}\) concentration decreased to \(0.0032 \mathrm{M}\). Calculate \(K_{\mathrm{c}}\) at this temperature for $$ 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) $$

Step-by-Step Solution

Answer

The equilibrium constant, \( K_c \), is approximately 1.350.

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the reaction is given by: \[ 2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g) \] This equation shows that 2 moles of sulfur dioxide react with 1 mole of oxygen to produce 2 moles of sulfur trioxide.

Step 2: Set up Initial Concentrations

Initially, the concentration of SO\(_2\) is 0.0076 M, O\(_2\) is 0.0036 M, and SO\(_3\) is 0 M because no SO\(_3\) is present at the start. Write these initial concentrations in a table for clarity: \[ \begin{array}{c|c|c|c} & \text{2 SO}_2 & \text{O}_2 & \text{2 SO}_3 \\hline\text{Initial} & 0.0076 & 0.0036 & 0 \\end{array} \]

Step 3: Determine Change in Concentrations

We know the equilibrium concentration of SO\(_2\) is 0.0032 M. The change in SO\(_2\) concentration is: \[ \Delta [\text{SO}_2] = 0.0032 - 0.0076 = -0.0044 \text{ M} \] Since 2 moles of SO\(_2\) produce 2 moles of SO\(_3\), the change for SO\(_3\) is similar but positive:\[ \Delta [\text{SO}_3] = +0.0044 \text{ M} \] For O\(_2\), the change can be determined using the stoichiometry of the reaction: \[ \Delta [\text{O}_2] = -0.0022 \text{ M} \] (as 1 mole of O\(_2\) reacts for every 2 moles of SO\(_2\)).

Step 4: Calculate Equilibrium Concentrations

Using the initial concentrations and the changes, calculate equilibrium concentrations: \[ \text{[SO}_2] = 0.0076 - 0.0044 = 0.0032 \text{ M} \] \[ \text{[O}_2] = 0.0036 - 0.0022 = 0.0014 \text{ M} \] \[ \text{[SO}_3] = 0 + 0.0044 = 0.0044 \text{ M} \]

Step 5: Write the Expression for the Equilibrium Constant

The expression for the equilibrium constant \( K_c \) for the given reaction is: \[ K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} \] Insert the equilibrium concentrations into this equation.

Step 6: Calculate Kc

Substitute the equilibrium concentrations into the equation:\[ K_c = \frac{(0.0044)^2}{(0.0032)^2 \times 0.0014} \] Calculate:\[ K_c = \frac{0.00001936}{0.00001024 \times 0.0014} \approx \frac{0.00001936}{0.000014336} \approx 1.350 \]

Key Concepts

Equilibrium Constant

When discussing chemical reactions, an important concept to understand is the equilibrium constant, denoted as \( K_c \). This constant provides insight into the relative amounts of reactants and products present once a reaction reaches equilibrium.

For the reaction \( 2 \, \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \, \text{SO}_3(g) \), the equilibrium constant is derived from the concentrations of the compounds at equilibrium. The expression is given by:

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

In this equation, the concentrations of sulfur trioxide, sulfur dioxide, and oxygen are used. Each concentration is raised to the power of its stoichiometric coefficient in the balanced chemical equation.

The value of \( K_c \) indicates whether the equilibrium position favors the formation of products or reactants. A \( K_c \) greater than 1 suggests a higher concentration of products, while a \( K_c \) less than 1 indicates that reactants are predominant at equilibrium.

Stoichiometry

Stoichiometry is a key concept in understanding chemical equations, involving the quantitative relationship between reactants and products. It is derived from the balanced chemical equation and is crucial in predicting how much reactant is consumed or how much product is formed during a reaction.

In our reaction example:

• \( 2 \, \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \, \text{SO}_3(g) \)

The coefficients in the equation tell us that two moles of sulfur dioxide react with one mole of oxygen to form two moles of sulfur trioxide.These coefficients guide calculations of changes in concentrations during chemical reactions. If the concentration of \( \text{SO}_2 \) decreases by \( 0.0044 \, M \), we know from stoichiometry that the concentration of \( \text{SO}_3 \) increases by the same amount, due to their 1:1 ratio in the reaction equation.Understanding stoichiometry is essential for calculating each substance involved at equilibrium, making it a fundamental tool in solving chemical equilibrium problems.

Reaction Rates

Reaction rates tell us how fast a reaction occurs. For equilibrium reactions, like the one discussed, these rates ultimately determine how quickly equilibrium is achieved.

In the case of:

• \( 2 \, \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \, \text{SO}_3(g) \)

The forward reaction rate is initially higher because reactants are abundant. Over time, as reactants are consumed and products are formed, the rate of the forward reaction decreases while the reverse reaction's rate increases.

Equilibrium is reached when the rates of the forward and reverse reactions become equal. At this point, the concentrations of reactants and products remain constant, though molecules continue to convert back and forth. Understanding reaction rates helps us grasp how quickly reactions progress and why altering conditions such as concentration or temperature can shift equilibrium.

Chemical Kinetics

Chemical kinetics involves the study of reaction rates and the steps that occur during chemical reactions. It provides insight into the speed of each elemental step and helps identify factors affecting these rates, such as temperature and concentration.

The equilibrium reaction \( 2 \, \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \, \text{SO}_3(g) \) serves as a great example where kinetics plays a crucial role in examining how quickly the reaction reaches equilibrium.

Here are a few factors affecting chemical kinetics:

• Temperature: As temperature increases, reaction rates usually increase due to greater molecular movement, leading to more effective collisions.
• Concentration: Higher concentrations of reactants can increase the frequency of collisions, thus increasing the reaction rate.
• Presence of Catalysts: Catalysts speed up reactions without being consumed, by lowering the activation energy required for the reaction to occur.

A firm understanding of chemical kinetics allows chemists to manipulate conditions to optimize the speed at which reactions reach equilibrium, providing control over industrial and laboratory processes.