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

Short answer

The equilibrium constant \(K_p\) for this reaction is \(1.80\).

Step-by-step solution

1. Calculate the initial moles of \(\mathrm{SO}_{3}\)

To begin, we must first convert the initial mass of \(\mathrm{SO}_{3}\) to moles. We can do this using the molar mass of \(\mathrm{SO_3}\), which is \(32 + 3 \times 16 = 80\,\text{g/mol}\). Since the initial mass is \(8.00\,\text{g}\), Initial moles of \(\mathrm{SO_3} =\frac{8.00\,\text{g}}{80\,\text{g/mol}} = 0.10\,\text{mol}\).

2. Find the moles of the gaseous mixture at equilibrium

We are given the density of the gaseous mixture at equilibrium, which is \(1.60\,\text{g/L}\). We can write the equation for the density as follows: Density = \( \frac{\text{Mass of the mixture}}{\text{Volume of the mixture}}\) Before proceeding, let's consider that: - The total mass of the gaseous mixture is \(8.00\,\text{g}\) (since none of the gases are lost) - The total moles of the mixture at equilibrium = \(n_{\mathrm{SO_2}} + n_{\mathrm{O_2}} + n_{\mathrm{SO_3}}\) - The partial pressures of the gases are in proportion to their moles Now, let's use the density to find the moles of the gaseous mixture at equilibrium: \(\frac{8.00\,\text{g}}{\text{Volume of the mixture}}=1.60\,\text{g/L}\) \(\text{Volume of the mixture} = \frac{8.00\,\text{g}}{1.60\,\text{g/L}} = 5.00\,\text{L}\) Assuming ideal gas behavior, we can use the equation \(PV=nRT\) to find the total moles of the gaseous mixture at equilibrium. Given, total pressure \(P = 1.80\,\text{atm}\), volume \(V = 5.00\,\text{L}\), and temperature \(T = 600^{\circ}\text{C}\), we need to convert the temperature to Kelvin. \(T = 600 + 273.15 = 873.15\,\text{K}\). Also, \(R = 0.0821\,\text{atm}\text{L/mol}\text{K}\) Now, we can solve for \(n_{\text{total}}\): \(n_{\text{total}}=\frac{PV}{RT} = \frac{1.80\,\text{atm} \cdot 5.00\,\text{L}}{0.0821\,\text{atm}\text{L/mol}\text{K} \cdot 873.15\,\text{K}}=0.135\,\text{mol}\)

3. Calculate the moles of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) at equilibrium

Let \(x\) represent the moles of \(\mathrm{SO_3}\) that decompose. According to the stoichiometry of the reaction given, we can say \(x\) moles of \(\mathrm{SO_2}\) and \(x/2\) moles of \(\mathrm{O_2}\) are produced at equilibrium. Then, we can write the equilibrium expressions in terms of moles as: \(n_{\mathrm{SO_3}} = 0.10 - x\) \(n_{\mathrm{SO_2}} = x\) \(n_{\mathrm{O_2}} =\frac{x}{2}\) Since we found that the total moles at equilibrium are \(0.135\,\text{mol}\), we can write the equation: \(n_{\mathrm{SO_3}} + n_{\mathrm{SO_2}} + n_{\mathrm{O_2}} = 0.135\,\text{mol}\) Plugging the expressions for the moles, we get: \((0.10 - x) + x + \frac{x}{2} = 0.135\) Solving for \(x\): \(0.10 + \frac{x}{2} = 0.135\) \(x = 0.07\,\text{mol}\) Now we can determine the moles of each species at equilibrium: \(n_{\mathrm{SO_3}} = 0.10 - 0.07 = 0.03\,\text{mol}\) \(n_{\mathrm{SO_2}} = 0.07\,\text{mol}\) \(n_{\mathrm{O_2}} =\frac{0.07}{2} = 0.035\,\text{mol}\)

4. Calculate the partial pressures of the gases at equilibrium

We will now use the mole fractions and the total pressure to determine the partial pressures of the gas species. \(P_{\mathrm{SO_3}} = \frac{n_{\mathrm{SO_3}}}{n_{\text{total}}} \times P\) \(P_{\mathrm{SO_3}} = \frac{0.03\,\text{mol}}{0.135\,\text{mol}} \times 1.80\,\text{atm} = 0.40\,\text{atm}\) Similarly, \(P_{\mathrm{SO_2}} = \frac{0.07\,\text{mol}}{0.135\,\text{mol}} \times 1.80\,\text{atm} = 0.93\,\text{atm}\) \(P_{\mathrm{O_2}} = \frac{0.035\,\text{mol}}{0.135\,\text{mol}} \times 1.80\,\text{atm} = 0.47\,\text{atm}\)

5. Calculate the equilibrium constant \(K_{\mathrm{p}}\)

Finally, we can use the partial pressures to calculate the equilibrium constant \(K_p\) for the reaction. The expression for \(K_p\) is as follows: \(K_p=\frac{P_{\mathrm{SO_2}} \times \sqrt{P_{\mathrm{O_2}}}}{P_{\mathrm{SO_3}}}\) By plugging in the values, we get: \(K_p = \frac{0.93\,\text{atm} \times \sqrt{0.47\,\text{atm}}}{0.40\,\text{atm}} = 1.80\) So, the equilibrium constant \(K_p\) for this reaction is \(1.80\).

Key concepts

Chemical Equilibrium

In chemical equilibrium, a reaction proceeds in both directions at an equal rate, making the concentrations of reactants and products constant over time. This dynamic state does not imply that the reactant and product are equal in concentration, but rather that their rates of formation are the same.

• A system at equilibrium is balanced, meaning that changes in conditions, such as pressure or temperature, can shift the equilibrium to favor one side of the reaction.
• To determine whether a reaction has reached equilibrium, changes such as concentrations or pressures can be monitored over time until they stabilize.
• The equilibrium state is represented by the equilibrium constant, expressed in terms of concentrations ( K _c ) or partial pressures ( K _p ).

These constants provide information about the position of equilibrium, with higher K _p values favoring product formation.

Partial Pressure

Partial pressure is an important concept when dealing with gases at equilibrium. It refers to the pressure exerted by an individual gas within a mixture. According to Dalton’s Law of Partial Pressures, the total pressure is the sum of the partial pressures of all gases present.

• Each gas in a mixture behaves as though it occupies the entire volume alone, contributing to the total pressure by its proportion of moles to the total moles of gas.
• Partial pressure is essential in calculating equilibrium constants for reactions involving gases, as K_p uses these values.
• In our example, each gas, such as \(\text{SO}_3\), \(\text{SO}_2\), and \(\text{O}_2\), contributes to the total pressure according to its mole fraction and influences the equilibrium dynamics.

By calculating partial pressures, we can establish the behavior of gases in equilibrium systems.

Ideal Gas Law

The ideal gas law is a crucial equation used to relate the pressure, volume, and temperature of a gas to the amount of substance. It is expressed as \(PV=nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.

• The ideal gas law assumes gas particles are in constant, random motion and that there are no intermolecular forces affecting them — an idealistic scenario used for simplification in calculations.
• By using the ideal gas law, we can find the total moles of gas present at equilibrium, which is essential to calculate equilibrium constants accurately.
• In this exercise, understanding the ideal gas law allows for the determination of moles from given conditions, bridging the gap between measurable quantities and molecular scale processes.

Applying the ideal gas law simplifies the understanding of gas behavior in chemical reactions at equilibrium.

Reaction Stoichiometry

Reaction stoichiometry involves using balanced chemical equations to determine the relationships between reactants and products in terms of moles. This concept is foundational in chemistry as it allows for the quantitative study of chemical reactions.

• Stoichiometric coefficients in a chemical equation indicate the ratio of moles in which the species react.
• In our reaction, \(\text{SO}_3\) decomposes to form \(\text{SO}_2\) and \(\frac{1}{2} \text{O}_2\), highlighting that every mole of decomposed \(\text{SO}_3\) produces one mole of \(\text{SO}_2\) and half a mole of \(\text{O}_2\).
• Stoichiometry allows us to calculate the changes in moles as a reaction progresses, which is fundamental when assessing the moles and pressures at equilibrium.

By utilizing stoichiometric principles, one can understand how reactants are converted into products, leading to calculated decisions in a chemical equilibrium.