Question

At \(100^{\circ} \mathrm{C}, K_{c}=0.078\) for the reaction $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{so}_{2}(g)+\mathrm{Cl}_{2}(g)$$ In an equilibrium mixture of the three gases, the concentrations of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) and \(\mathrm{SO}_{2}\) are 0.108 \(\mathrm{M}\) and \(0.052 \mathrm{M},\) respectively. What is the partial pressure of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

Short answer

Based on the given information, the partial pressure of Cl2 can be found using the mole fraction and the total pressure of the system. The mole fraction of Cl2 is \(\frac{0.161}{0.321}\). To determine the partial pressure of Cl2, we need to know the total pressure, Ptot, and multiply the mole fraction by Ptot. Some additional information, such as temperature or the volume of the vessel, is needed to solve for the total pressure.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the given reaction is: \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g) + \mathrm{Cl}_{2}(g)\)

Use the Kc formula to find the Cl2 concentration

The equilibrium constant, Kc for the reaction can be written as: \( K_c = \frac{[\mathrm{SO}_2][\mathrm{Cl}_2]}{[\mathrm{SO}_2\mathrm{Cl}_2]} \) We are given Kc, [SO2Cl2], and [SO2]. Now we can solve for the concentration of Cl2: \( 0.078 = \frac{(0.052)[\mathrm{Cl}_2]}{(0.108)} \) Now, let's solve for [Cl2]: \([\mathrm{Cl}_2]=\frac{(0.078)(0.108)}{(0.052)}\) \([\mathrm{Cl}_2]=0.161 \, \mathrm{M}\)

Use the ideal gas law to find the partial pressure of Cl2

The ideal gas law can be given as \(PV=nRT\). We can convert it to the form of concentration using the relation \(n=CV\), where \(C\) is the concentration and \(V\) is the volume. Thus, the ideal gas law becomes \(P = CRT\). However, for our case, we only need the ratio of partial pressures between Cl2 and the other gases (SO2 and SO2Cl2). As they are all in the same container, we can assume that they have the same temperature and volume. Thus, the ratio of partial pressures is the same as the ratio of their concentrations. At equilibrium, \([\mathrm{Cl}_{2}]:[\mathrm{SO}_{2}]:[\mathrm{SO}_{2} \mathrm{Cl}_{2}] = 0.161:0.052:0.108\) Now, let's find the partial pressure of Cl2. Suppose the equilibrium pressure of the system is given by Ptot. Partial Pressure of Cl2 = \( \frac{0.161}{0.161 + 0.052 + 0.108} \cdot P_{tot}\) Now, we need to find the total pressure of the system. We know the total moles of reactants and products at equilibrium. As all the reactants and products are in the gaseous state, we can use the mole fraction to find the partial pressure of Cl2: \(P_{Cl_2} = \mathrm{Mole\:Fraction\:of\:Cl_2} \times P_{tot}\) Mole fraction of Cl2 = \(\frac{0.161}{0.161 + 0.052 + 0.108}=\frac{0.161}{0.321}\) Now we have the mole fraction of Cl2, we need to find the total pressure, Ptot. Some additional information is needed, such as temperature or the volume of the vessel, to solve for the total pressure. Once we have the total pressure, we can solve for the partial pressure of \(Cl_2\) by multiplying its mole fraction by the total pressure.

Key concepts

Equilibrium Constant (Kc)

In chemistry, the equilibrium constant, known as \( K_c \), is a crucial value that helps us understand the balance of a chemical reaction. It is derived from the concentrations of reactants and products at equilibrium. For gaseous reactions, \( K_c \) is determined by the relationship between the concentration of each substance.
To calculate the equilibrium constant, the general formula is used:

• \( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

where \([A]\), \([B]\), \([C]\), \([D]\) are the molar concentrations of the reactants and products, and \(a\), \(b\), \(c\), \(d\) are their respective stoichiometric coefficients.
In our exercise, even though we're only directly calculating the concentration of \( \text{Cl}_2 \) using \( K_c \), understanding this constant allows us to gauge how much of each compound will be present when the system reaches equilibrium. A lower \( K_c \) indicates a reaction favoring reactants, while a higher \( K_c \) means more products are present at equilibrium.

Partial Pressure

Partial pressure is the pressure exerted by a single type of gas in a mixture of gases. In a container with multiple gases, each gas contributes part of the total pressure. To find the partial pressure, we rely on Dalton's Law of Partial Pressures.
For a gas \( A \) in a mixture:

• \( P_A = x_A \times P_{tot} \)

Here, \( P_A \) is the partial pressure, \( x_A \) is the mole fraction, and \( P_{tot} \) is the total pressure of the gas mixture.
Understanding this concept is vital because it helps to isolate the behavior of individual gases within a mixture. When given their concentrations, like in our example, you can find the mole fraction of each gas and subsequently its partial pressure. This tells us exactly how much pressure each type of gas contributes to the total pressure.

Ideal Gas Law

The Ideal Gas Law is an equation of state that describes the behavior of an ideal gas. It's a helpful tool in chemistry that relates the pressure, volume, temperature, and number of moles of a gas together. The formula is:
\[ PV = nRT \]
where:

• \( P \) is the pressure of the gas
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the ideal gas constant, typically 0.0821 L atm/mol K
• \( T \) is the temperature in Kelvin

For our exercise, we used a form of the Ideal Gas Law to relate concentration and pressure, which simplifies to \( P = CRT \) when you rearrange the variables. Knowing this allows us to determine the total pressure of the gaseous system at equilibrium by considering the concentrations of the substances involved. Although exact values weren't given for total pressure, the Ideal Gas Law reminds us of how pressure intricately depends on the variables of gas concentration, volume, and temperature.