Problem 119 - Chemistry

Question

An 8.00 -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 mixture were 1.80 \(\mathrm{atm}\) and 1.60 \(\mathrm{g} / \mathrm{L}\) , respectively. Calculate \(K_{\mathrm{p}}\) for this reaction.

Step-by-Step Solution

Answer

The equilibrium constant \(K_{\mathrm{p}}\) for this reaction is approximately 0.548.

Step 1: Calculate the number of moles of \(\mathrm{SO_{3}}\) initially present

To calculate the initial amount (in moles) of \(\mathrm{SO_{3}}\), we will use its molar mass: Number of moles of \(\mathrm{SO_{3}} = \dfrac{\text{Given mass of SO3}}{\text{Molar mass of SO3}}\) The molar mass of \(\mathrm{SO_{3}}\) is approximately \(80.07 \, \mathrm{g/mol}\) (32.07 g/mol for S and 3 × 16.00 g/mol for O). Number of moles of $\mathrm{SO_{3}} = \dfrac{8.00 \, \mathrm{g}}{80.07 \, \mathrm{g/mol}} \approx 0.1 \, \mathrm{mol}\)

Step 2: Relate the number of moles of \(\mathrm{SO_{2}}\) and \(\mathrm{O_{2}}\) that form at equilibrium to the decrease in \(\mathrm{SO_{3}}\)

Let x be the number of moles of \(\mathrm{SO_{3}}\) that decompose at equilibrium. We can then determine the number of moles of \(\mathrm{SO_{2}}\) and \(\mathrm{O_{2}}\) formed: - Moles of \(\mathrm{SO_{3}}\) at equilibrium = \(0.1 - x\) - Moles of \(\mathrm{SO_{2}}\) at equilibrium = \(x\) - Moles of \(\frac{1}{2}\mathrm{O_{2}}\) at equilibrium = \(\dfrac{1}{2}x\)

Step 3: Determine the volume of the container

We are given the density of the gaseous mixture: \(1.60 \, \mathrm{g/L}\). We can use this information to determine the volume of the container: Total mass of the gaseous mixture = 8.00g Density = \(\dfrac{\text{Mass}}{\text{Volume}}\) Therefore, Volume = $\dfrac{\text{Mass}}{\text{Density}} = \dfrac{8.00 \, \mathrm{g}}{1.60 \, \mathrm{g/L}} = 5.00 \, \mathrm{L}\$

Step 4: Calculate the partial pressures of all the gases at equilibrium

The total pressure is given as \(1.80 \, \mathrm{atm}\). As the container volume is 5.00L, we can use the ideal gas equation to find the total moles of gas at equilibrium: Total moles at equilibrium = \(\frac{\text{Pressure × Volume}}{\text{Gas constant (R) × Temperature}}\) Using the appropriate values for the gas constant (R) and given temperature: \(PV = nRT \Rightarrow n = \frac{PV}{RT}\) For this problem, we will use the value of R = 0.0821 atm L/mol K. Temperature = \(600^{\circ} \mathrm{C} + 273.15 = 873.15 \, \mathrm{K}\) Total moles at equilibrium = \(\dfrac{1.80 \, \mathrm{atm} \times 5.00 \, \mathrm{L}}{0.0821 \, \mathrm{atm \, L/mol \, K} \times 873.15 \, \mathrm{K}} \approx 0.125 \, \mathrm{mol}\) From Step 2, at equilibrium we have: $$0.1 - x + x + \dfrac{1}{2}x = 0.125$$ Solving for x: $$x = 0.025 \, \mathrm{mol}$$ Using these values, we can calculate the partial pressures: $$P_{\mathrm{SO_{3}}} = \dfrac{\text{Moles of } \mathrm{SO_{3}}}{\text{Volume}} = \dfrac{0.1 - 0.025}{5.00} \times 1.80 \, \mathrm{atm} \approx 0.27 \, \mathrm{atm}$$ $$P_{\mathrm{SO_{2}}} = \dfrac{\text{Moles of } \mathrm{SO_{2}}}{\text{Volume}} = \dfrac{0.025}{5.00} \times 1.80 \mathrm{atm} \approx 0.09 \, \mathrm{atm}$$ $$P_{\frac{1}{2}\mathrm{O_{2}}} = \dfrac{\text{Moles of } \frac{1}{2}\mathrm{O_{2}}}{\text{Volume}} = \dfrac{0.5 \times 0.025}{5.00} \times 1.80 \, \mathrm{atm} \approx 0.045 \, \mathrm{atm}$$

Step 5: Calculate \(K_{\mathrm{p}}\)

Now that we have the partial pressures of all the gases involved in the reaction, we can calculate \(K_{\mathrm{p}}\) using the given reaction: $$K_{\mathrm{p}} = \dfrac{P_{\mathrm{SO_{2}}}\times P_{\frac{1}{2}\mathrm{O_{2}}^{\frac{1}{2}}}{P_{\mathrm{SO_{3}}}}$$ Plugging in the values obtained in Step 4: $$K_{\mathrm{p}} = \dfrac{0.09 \, \mathrm{atm} \times \sqrt{0.045 \, \mathrm{atm}}}{0.27 \, \mathrm{atm}} \approx 0.548$$ Therefore, the equilibrium constant \(K_{\mathrm{p}}\) for this reaction is approximately 0.548.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in which the concentration of reactants and products remains constant over time. It occurs when the rates of the forward and reverse reactions are equal. In the context of the reaction \[\mathrm{SO}_3 (g) \rightleftharpoons \mathrm{SO}_2 (g) + \frac{1}{2} \mathrm{O}_2 (g)\], the equilibrium is achieved when the rate at which \(\mathrm{SO}_3\) decomposes equals the rate at which \(\mathrm{SO}_2\) and \(\mathrm{O}_2\) recombine to form \(\mathrm{SO}_3\).
The equilibrium constant, \(K_p\), is a measure of the position of equilibrium for a reaction involving gases. It is expressed in terms of the partial pressures of the gases. A large \(K_p\) value indicates that the reaction heavily favors products, while a small \(K_p\) value suggests that reactants are favored at equilibrium. For the given reaction, the equilibrium constant is calculated using the formula: \[K_p = \frac{P_{\mathrm{SO_2}} \times \left(P_{\mathrm{O_2}}\right)^{1/2}}{P_{\mathrm{SO_3}}}\].
This equation shows how the pressure of each gas contributes to the equilibrium state, emphasizing the dynamic nature of chemical equilibria.

Gas Laws

Gas laws are essential for understanding the behavior of gases in different conditions, particularly in reactions like the decomposition of \(\mathrm{SO}_3\). The ideal gas law, represented by the equation \(PV = nRT\), connects pressure (P), volume (V), and temperature (T) with the amount of gas in moles (n) and the ideal gas constant (R).
These principles are crucial when calculating the total moles of gas present at equilibrium in the given reaction. By rearranging the ideal gas law, we can determine the number of moles from pressure, volume, and temperature data: \[n = \frac{PV}{RT}\].
This relation helps us set up the equilibrium expression correctly by understanding the amounts of each gaseous component involved.
Finally, understanding gas laws ensures accuracy when determining the volume of containers or systems, crucial for calculating equilibrium concentrations or partial pressures of gases.

Stoichiometry

Stoichiometry involves the quantitative relationships between the reactants and products in a chemical reaction. It enables us to use these relationships to calculate various quantities, such as the amount left or formed at equilibrium.
For the reaction \(\mathrm{SO}_3 (g) \rightleftharpoons \mathrm{SO}_2 (g) + \frac{1}{2} \mathrm{O}_2 (g)\), stoichiometry allows us to relate the moles of each compound present at equilibrium.
Initially, certain moles of \(\mathrm{SO}_3\) decompose. By setting up the balance with a change in moles (denoted as \(x\)), we find how many moles of \(\mathrm{SO}_2\) and \(\mathrm{O}_2\) are produced.

• Initial moles of \(\mathrm{SO}_3\): \(0.1\, \text{mol}\)
• Equilibrium moles \(\mathrm{SO}_3\): \(0.1 - x\, \text{mol}\)
• Equilibrium moles \(\mathrm{SO}_2\): \(x\, \text{mol}\)
• Equilibrium moles \(\mathrm{O}_2\): \(\frac{x}{2}\, \text{mol}\)

By solving these relationships, we derive the concentrations required to determine partial pressures and \(K_p\). This demonstrates stoichiometry's pivotal role in understanding and solving equilibrium problems.

Partial Pressure

Partial pressure is the pressure exerted by a single type of gas in a mixture of gases, and it plays a crucial role in calculating equilibrium constants in gaseous reactions.
For the equation of interest, \(\mathrm{SO}_3 (g) \rightleftharpoons \mathrm{SO}_2 (g) + \frac{1}{2} \mathrm{O}_2 (g)\), the partial pressures for \(\mathrm{SO}_3\), \(\mathrm{SO}_2\), and \(\mathrm{O}_2\) must be determined to calculate the equilibrium constant \(K_p\).
The total pressure of the reaction mixture is given, and we use it alongside the calculated moles of each gas at equilibrium to determine each gas's partial pressure:

• \(P_{\mathrm{SO_3}} = \frac{\text{moles of } \mathrm{SO_3}}{\text{total moles}} \times \text{total pressure}\)
• \(P_{\mathrm{SO_2}} = \frac{\text{moles of } \mathrm{SO_2}}{\text{total moles}} \times \text{total pressure}\)
• \(P_{\mathrm{O_2}} = \frac{\text{moles of } \mathrm{O_2}}{\text{total moles}} \times \text{total pressure}\)

By understanding each component's contribution to the total pressure, we can accurately calculate \(K_p\), which reflects the equilibrium condition of the reaction. This concentration understanding is fundamental when working with mixtures of gases.