Problem 109 - Chemistry An Atoms First Approach

Question

An \(8.00-\mathrm{g}\) sample of \(\mathrm{SO}_{3}\) was placed in an evacuated container, where it decomposed at \(600^{\circ} \mathrm{C}\) according to the following reaction: $$\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)$$ At equilibrium the total pressure and the density of the gaseous mixtures were 1.80 atm and \(1.60 \mathrm{g} / \mathrm{L},\) respectively. Calculate \(K_{\mathrm{p}}\) for this reaction.

Step-by-Step Solution

Answer

The short answer to this question is: To calculate the equilibrium constant \(K_p\) for the given reaction, follow these steps: 1. Calculate the initial moles of SO3 2. Calculate the moles of SO3, SO2, and O2 at equilibrium 3. Calculate the total moles of gases at equilibrium 4. Calculate the volume of the container 5. Calculate the partial pressures of the gases at equilibrium 6. Calculate the total pressure at equilibrium 7. Solve for x (the moles of SO3 decomposed) 8. Calculate the equilibrium constant \(K_p\) By following these steps, you can find the equilibrium constant Kp for this reaction.

Step 1: 1. Calculate the Initial Moles of SO3

To find the initial moles of SO3, we will calculate it using the provided mass by dividing it by the molar mass of SO3: \(n_{SO3(initial)} = \frac{8.00g}{(32.07g/mol + 3(16g/mol))}\)

Step 2: 2. Calculate the Moles of SO3 at Equilibrium

Let's assume x moles of SO3 have decomposed at equilibrium. Then, the final moles of SO3 would be: \(n_{SO3(final)} = n_{SO3(initial)} - x\)

Step 3: 3. Calculate the Moles of SO2 and O2 at Equilibrium

As x moles of SO3 have decomposed, the final moles of SO2 and O2 would be: \(n_{SO2} = x\) \(n_{O2} = \frac{1}{2}x\)

Step 4: 4. Calculate the Total Moles of Gases at Equilibrium

Adding the moles of all three gases gives: \(n_{total} = n_{SO_3(final)} + n_{SO_2} + n_{O_2} = (n_{SO_3(initial)} - x) + x + \frac{1}{2}x\)

Step 5: 5. Calculate the Volume of the Container

Using the density, we can calculate the volume of the container: \(V = \frac{mass_{total}}{density} = \frac{8.00 g}{1.60 g/L}\)

Step 6: 6. Calculate the Partial Pressures at Equilibrium

Dividing the moles of each gas by the volume and then multiplying by the gas constant R and the temperature T, we can find the partial pressures of the gases at equilibrium: \(P_{SO3} = \frac{n_{SO3(final)}RT}{V}\) \(P_{SO2} = \frac{n_{SO2}RT}{V}\) \(P_{O2} = \frac{n_{O2}RT}{V}\)

Step 7: 7. Calculate the Total Pressure at Equilibrium

Using the partial pressures, we can find the total pressure at equilibrium: \(P_{total} = P_{SO3} + P_{SO2} + P_{O2}\)

Step 8: 8. Solve for x

Plug in the given total pressure (1.80 atm) into the total pressure equation obtained in step 7 and solve for x.

Step 9: 9. Calculate Kp

Now that we have calculated the partial pressures at equilibrium, we can determine the equilibrium constant Kp: \(K_p = \frac{P_{SO2}P_{O2}^{1/2}}{P_{SO3}}\) By calculating each step of the solution and inputting the values, we'll find the equilibrium constant, Kp, for this reaction.

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K_{p} \) when dealing with gases, is a crucial concept in chemical equilibrium. It provides a measure of the relative concentrations of products and reactants at equilibrium in a reversible chemical reaction. Specifically, \( K_{p} \) is used when the equilibrium concentrations are expressed in terms of partial pressures.

In our reaction, \( 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2\mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \), the equilibrium constant \( K_{p} \) is formulated as:

\[ K_{p} = \frac{P_{\mathrm{SO}_{2}}^{2} \cdot P_{\mathrm{O}_{2}}}{P_{\mathrm{SO}_{3}}^{2}} \]

Understanding \( K_{p} \):

• If \( K_{p} > 1 \), the products are favored at equilibrium.
• If \( K_{p} < 1 \), the reactants are favored.

This constant does not change with changes in pressure or concentration but can change with temperature, thus linking the equilibrium constant to the thermal energy of the system.

Partial Pressure

Partial pressure refers to the pressure that each gas in a mixture would exert if it occupied the entire volume by itself at the same temperature. In a system with multiple gases, knowing the partial pressure helps understand the contribution of each gas to the total pressure.

Using Dalton's Law of Partial Pressures, the total pressure \( P_{total} \) in a mixture of gases can be expressed as the sum of the partial pressures:

\[ P_{total} = P_{\mathrm{SO}_{3}} + P_{\mathrm{SO}_{2}} + P_{\mathrm{O}_{2}} \]

Calculating Partial Pressures:

• Identify the moles of each gas at equilibrium. For instance, the moles of \( \mathrm{SO}_{2} \) and \( \mathrm{O}_{2} \) are derived from the reaction equation based on decomposed \( \mathrm{SO}_{3} \).
• Use the ideal gas law adjusted for partial pressures: \( P_{i} = \frac{n_{i}RT}{V} \), where \( n_{i} \) is the moles of the gas, \( R \) is the gas constant, and \( V \) is the volume.

This concept aids in calculating \( K_{p} \) since it directly involves the partial pressures of the gases at equilibrium.

Decomposition Reaction

A decomposition reaction is a chemical process where one compound breaks down into two or more simpler substances. The reaction in the exercise:

\( \mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \)

This reaction exemplifies decomposition, as \( \mathrm{SO}_{3} \) decomposes into \( \mathrm{SO}_{2} \) and \( \frac{1}{2}\mathrm{O}_{2} \). Decomposition reactions are often endothermic, requiring energy (in the form of heat) to proceed.

In this case, the decomposition of \( \mathrm{SO}_{3} \) occurs in a closed system, allowing us to study the effects of pressure and temperature on the equilibrium position.

Implications for the exercise:

• The balance of forward (decomposition) and reverse reactions affects overall equilibrium.
• By initially knowing the amount of \( \mathrm{SO}_{3} \), we track changes leading to equilibrium.

Gas Laws

The gas laws provide essential tools for understanding the behavior of gases under various conditions of temperature, volume, and pressure.

Key gas laws relevant to the exercise:

• Ideal Gas Law: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature in Kelvin. This law helps relate the partial pressures to moles of gases.
• Dalton's Law: As previously mentioned, it describes the behavior of partial pressures in a mixture of gases.

Application in the exercise:

The argument for determining \( K_{p} \) rests on these gas laws. By calculating moles from the mass of \( \mathrm{SO}_{3} \) and using the ideal gas law, the partial pressures necessary for finding \( K_{p} \) can be derived.