Question

In a gas-phase equilibrium mixture of \(\mathrm{SbCl}_{5}, \mathrm{SbCl}_{3}\), and \(\mathrm{Cl}_{2}\) at \(500 . \mathrm{K}, P_{\mathrm{SbCl}_{5}}=0.072\) bar and \(P_{\mathrm{SbCl}_{3}}=5.02 \mathrm{mbar}\). Calculate the equilibrium partial pressure of \(\mathrm{Cl}_{2}\), given that \(K=3.5 \times 10^{-4}\) for the reaction \(\mathrm{SbCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{SbCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\).

Short answer

The equilibrium partial pressure of \(Cl_2\) is \(5.02 \times 10^{-3}\) bar.

Step-by-step solution

Translate Given Information

We are given the partial pressures of \(SbCl_5\) and \(SbCl_3\), which are 0.072 bar and 5.02 mbar, respectively. To keep units consistent, convert the pressure of \(SbCl_3\) from mbar to bar: \(1 \text{ mbar} = 1 \times 10^{-3} \text{ bar}\), so \(P_{SbCl_3} = 5.02 \times 10^{-3} \text{ bar}.\) The equilibrium constant \(K_p\) for the reaction is \(3.5 \times 10^{-4}\).

Apply the Equilibrium Expression

For the reaction \(SbCl_{5}(g) \rightleftharpoons SbCl_{3}(g) + Cl_{2}(g)\), the equilibrium constant expression is \(K_p = \frac{P_{SbCl_3} \times P_{Cl_2}}{P_{SbCl_5}}\). We need to solve for \(P_{Cl_2}\) using the given values.

Solve for \(P_{Cl_2}\)

Using \(K_p = 3.5 \times 10^{-4}\) and the given partial pressures, we plug in the known values to get \(3.5 \times 10^{-4} = \frac{5.02 \times 10^{-3} \times P_{Cl_2}}{0.072}\). Next, solve for \(P_{Cl_2}\) by multiplying both sides by 0.072 and then dividing by 5.02 \times 10^{-3}.

Calculate the Partial Pressure of \(Cl_2\)

Perform the calculation to find \(P_{Cl_2}\): \(P_{Cl_2} = \frac{3.5 \times 10^{-4} \times 0.072}{5.02 \times 10^{-3}}\), which simplifies to \(P_{Cl_2} = \frac{2.52 \times 10^{-5}}{5.02 \times 10^{-3}}\), and then to \(P_{Cl_2} = 5.02 \times 10^{-3}\) bar.

Key concepts

Equilibrium Constant

The equilibrium constant is a fundamental concept in chemical equilibrium that quantifies the ratio of the concentration or partial pressure of the products to the reactants at equilibrium. It is a measure of the extent of a reaction, where a high constant indicates a greater amount of products, and a low constant suggests a reaction that favors the reactants.

To calculate the equilibrium constant in gas-phase reactions, we often use partial pressures denoted by the symbol \(K_p\). For the given reaction \(SbCl_{5}(g) \rightleftharpoons SbCl_{3}(g) + Cl_{2}(g)\), the equilibrium constant expression is \[K_p = \frac{P_{SbCl_3} \times P_{Cl_2}}{P_{SbCl_5}}\]. This calculation is crucial as it helps us determine the position of the equilibrium and predict how the system will respond to changes in conditions, such as pressure, temperature, or concentrations.

Partial Pressure

Partial pressure refers to the pressure exerted by a single gas component in a mixture of gases. Each gas in a mixture exerts pressure as if it were alone in the volume, and the total pressure is the sum of these individual pressures. It's a crucial factor in gas-phase equilibrium calculations where reactions involve gases at constant temperature and volume.

In the exercise, the conversion of units for \(SbCl_3\) from mbar to bar ensures the consistency needed for accurate calculations of partial pressures. To find the equilibrium partial pressure of \(Cl_2\), we use the given partial pressures of \(SbCl_5\) and \(SbCl_3\), along with the equilibrium constant. Remember, it's important to double-check your units to avoid calculation errors.

Gas-Phase Reactions

Gas-phase reactions entail the transformation of reactants to products in the gaseous state. These reactions are dynamic systems where the rate of the forward reaction is equal to the rate of the reverse reaction at equilibrium. For gas-phase reactions, understanding how gas laws apply and how to use partial pressures is fundamental.

In our discussion, the gas-phase reaction \(SbCl_{5}(g) \rightleftharpoons SbCl_{3}(g) + Cl_{2}(g)\) is shown to reach equilibrium. Because all the species involved are gases, we use their partial pressures to compute the equilibrium constant. This approach applies universally to gas-phase equilibria, enabling us to predict how changes in conditions like pressure could affect the system.

Le Chatelier's Principle

Le Chatelier's principle provides insight into how a chemical system at equilibrium responds to changes in concentration, pressure, or temperature. According to this principle, if a dynamic system at equilibrium experiences a change, the system adjusts in a way that counteracts the change imposed.

If the pressure is increased, for example, the system shifts to reduce it, favoring the production of fewer moles of gas. In the context of the given reaction \(SbCl_{5}(g) \rightleftharpoons SbCl_{3}(g) + Cl_{2}(g)\), an increase in the pressure of \(SbCl_{5}\) would lead the equilibrium to shift to the right, increasing the production of \(SbCl_{3}\) and \(Cl_{2}\) to reduce the pressure. Understanding this principle helps us manipulate and predict the outcomes of equilibrium systems.