Problem 38 - Chemistry and Chemical Reactivity

Question

Sulfur dioxide is readily oxidized to sulfur trioxide. $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_{3}(\mathrm{g}) \quad K_{c}=279 $$ If we add \(3.00 \mathrm{g}\) of \(\mathrm{SO}_{2}\) and \(5.00 \mathrm{g}\) of \(\mathrm{O}_{2}\) to a \(1.0-\mathrm{L}\). flask, approximately what quantity of \(\mathrm{SO}_{3}\) will be in the flask once the reactants and the product reach equilibrium? (a) \(2.21 \mathrm{g}\) (c) \(3.61 \mathrm{g}\) (b) \(4.56 \mathrm{g}\) (d) \(8.00 \mathrm{g}\) (Note: The full solution to this problem results in a cubic equation. Do not try to solve it exactly. Decide only which of the answers is most reasonable.)

Step-by-Step Solution

Answer

The quantity of \(\text{SO}_3\) is approximately \(3.61 \text{ g}\).

Step 1: Convert Masses to Moles

First, we need to convert the given masses of \( \text{SO}_2 \) and \( \text{O}_2 \) to moles using their molar masses.- The molar mass of \( \text{SO}_2 \) is approximately \( 64.07 \, \text{g/mol} \). - Moles of \( \text{SO}_2 = \frac{3.00 \, \text{g}}{64.07 \, \text{g/mol}} \approx 0.0468 \, \text{mol} \).- The molar mass of \( \text{O}_2 \) is approximately \( 32.00 \, \text{g/mol} \). - Moles of \( \text{O}_2 = \frac{5.00 \, \text{g}}{32.00 \, \text{g/mol}} \approx 0.156 \, \text{mol} \).

Step 2: Convert to Concentrations

Since the volume of the flask is \( 1.0 \, \text{L} \), the moles directly convert to molar concentrations.- \([\text{SO}_2] = 0.0468 \, \text{M} \)- \([\text{O}_2] = 0.156 \, \text{M} \)

Step 3: Determine Limiting Reactant

The reaction is given as \( 2 \text{SO}_2 + \text{O}_2 \rightarrow 2 \text{SO}_3 \).- We need 2 moles of \( \text{SO}_2 \) for every mole of \( \text{O}_2 \). Here, we have \( 0.0468 \times 2 = 0.0936 \, \text{mol} \) of \( \text{SO}_2 \) needed, but we only have \( 0.0468 \, \text{mol} \).- \( \text{SO}_2 \) is the limiting reactant.

Step 4: Set Up Equilibrium Expression

Using the equilibrium constant \( K_c = 279 \) and the reaction \( 2 \text{SO}_2 + \text{O}_2 \rightarrow 2 \text{SO}_3 \), we write the expression:\[ K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]} = 279 \]Replace concentrations at equilibrium \([\text{SO}_2] = 0.0468 - 2x\), \([\text{O}_2] = 0.156 - x\), and \([\text{SO}_3] = 2x\).

Step 5: Approximation for Solving Cubic Equation

The equation forms a cubic, \([2x]^2 = 279([0.0468-2x]^2)([0.156-x])\). Approximate a reasonable root:- Assume \( x \) is small compared to initial concentrations.- Calculate the change in \( \text{SO}_3 \) approximately to reason through the expected grams formed.

Step 6: Convert Moles of SO₃ to Grams

Assume approximately half of the limiting reactant SO₂ is converted to SO₃ due to high K value indicating more products at equilibrium. Calculate:- Produced moles \(\approx 0.0234 \text{ mol} \).- Moles of \( \text{SO}_3 \) = 0.0468 - small x.- Grams of \( \text{SO}_3 = 0.0234 \times 80.07 \, \text{g/mol} \approx 3.61 \text{ g} \).

Step 7: Conclusion

Considering the high equilibrium constant and the quantities of reactants, sizeable formation of \(\text{SO}_3\) is expected. Therefore, given the options, it is reasonable the answer is approximately \(3.61\text{ g}\).

Key Concepts

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is completely consumed first, preventing further reaction progress. To identify it, you compare the mole ratios of the reactants to the coefficients in the balanced equation.
For example, in the reaction \( 2 \text{SO}_2 + \text{O}_2 \rightarrow 2 \text{SO}_3 \), you need 2 moles of \( \text{SO}_2 \) for every mole of \( \text{O}_2 \).
Here’s how to find the limiting reactant:

• Calculate moles: \( \text{SO}_2 = 0.0468 \text{ mol} \) and \( \text{O}_2 = 0.156 \text{ mol} \).
• Determine needed \( \text{SO}_2 \) for all \( \text{O}_2 \): \( 0.156 \times 2 = 0.312 \text{ mol} \).
• Compare: With only \( 0.0468 \text{ mol} \) \( \text{SO}_2 \) available, it limits the reaction.

This concept ensures that calculations of product formed don’t exceed what's possible.

Equilibrium Constant

The equilibrium constant \( K_c \) is a value that expresses the ratio of concentrations of products to reactants at equilibrium.
It provides insight into the position of the equilibrium in a chemical reaction.
For the reaction \( 2 \text{SO}_2 + \text{O}_2 \rightleftharpoons 2 \text{SO}_3 \), the expression is:

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

A high \( K_c \) means more products are present at equilibrium.
In our example, \( K_c = 279 \), showing a strong tendency to form \( \text{SO}_3 \).
This understanding helps predict that a significant amount of \( \text{SO}_3 \) will be formed when the reactants reach equilibrium.

Mole Conversion

Mole conversion is a fundamental skill for transitioning between different units of measurement in chemistry.
It allows you to convert mass to moles, which is essential for reacting proportions and equilibrium calculations.

• Use the molar mass: \( \text{SO}_2 = 64.07 \text{ g/mol} \) and \( \text{O}_2 = 32.00 \text{ g/mol} \).
• Convert given mass to moles: \( \text{Moles of SO}_2 = \frac{3.00 \text{ g}}{64.07 \text{ g/mol}} \approx 0.0468 \text{ mol} \).
• Similarly, for \( \text{O}_2 \), \( \approx 0.156 \text{ mol} \).

Knowing how to do this helps set up the equation for the chemical balance, determine the limiting reactant, and accurately calculate product formation at equilibrium.