Question

Under what pressure must an equimolar mixture of \(\mathrm{Cl}_{2}\) and \(\mathrm{PCl}_{3}\) be place at \(250^{\circ} \mathrm{C}\) in order to obtain \(75 \%\) conversion of \(\mathrm{PCl}_{3}\) into \(\mathrm{PCl}_{5}\) ? Given: \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons\) \(\mathrm{PCl}_{5}(\mathrm{~g}) ; K_{\mathrm{p}}=2 \mathrm{~atm}^{-1}\) (a) \(12 \mathrm{~atm}\) (b) \(6 \mathrm{~atm}\) (c) 15 atm (d) \(30 \mathrm{~atm}\)

Short answer

The pressure required for a 75% conversion of \(\mathrm{PCl}_{3}\) into \(\mathrm{PCl}_{5}\) is 6 atm.

Step-by-step solution

Understand the Chemical Equilibrium

Begin by analyzing the chemical reaction \(\mathrm{PCl}_{3}(\mathrm{g}) + \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{5}(\mathrm{g})\). Note that the reaction is at equilibrium and we have been given that \(K_{\mathrm{p}} = 2 \mathrm{~atm}^{-1}\). This indicates that the reaction quotient at equilibrium, involving partial pressures, equals 2 atm^-1.

Set Up the Equilibrium Expression

Express the equilibrium constant in terms of the partial pressures of the reacting gases as \(K_{\mathrm{p}} = \frac{P_{\mathrm{PCl}_5}}{P_{\mathrm{PCl}_3} \cdot P_{\mathrm{Cl}_2}}\). Since at equilibrium \(K_{\mathrm{p}} = 2 \(\textrm{atm}^{-1}\)\), this relationship must be upheld by the partial pressures of the gases.

Determine the Initial Moles and Change

Since we start with an equimolar mixture, let's assume that initially we have 1 mole of \(\mathrm{PCl}_{3}\) and 1 mole of \(\mathrm{Cl}_{2}\) with no \(\mathrm{PCl}_{5}\) present. Upon 75% conversion of \(\mathrm{PCl}_{3}\) into \(\mathrm{PCl}_{5}\), 0.75 moles of \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) react. This leaves 0.25 moles of \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\), and produces 0.75 moles of \(\mathrm{PCl}_{5}\).

Express the Partial Pressures in Terms of Moles

Let the total pressure in the system be \(P\). The partial pressure of each gas can be expressed in terms of its mole fraction and the total pressure. For \(\mathrm{PCl}_{5}\), the partial pressure is \(0.75 \cdot P\); for \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) it is \(0.25 \cdot P\) each.

Substitute the Partial Pressures into the Equilibrium Expression

We now substitute the expressions of partial pressures into the equilibrium constant equation: \(K_{\mathrm{p}} = \frac{0.75 \cdot P}{(0.25 \cdot P)(0.25 \cdot P)} = \frac{0.75}{0.0625} \cdot \frac{1}{P}\). Since we know \(K_{\mathrm{p}} = 2 \(\textrm{atm}^{-1}\)\), we can set the expression equal to 2 and solve for \(P\).

Solve for the Total Pressure

Setting up the equation, we have \(2 = \frac{0.75}{0.0625} \cdot \frac{1}{P} \). Solving for \(P\) gives us: \(P = \frac{0.75}{2 \cdot 0.0625} = 6 \mathrm{~atm}\). Hence, the pressure required to achieve 75% conversion of \(\mathrm{PCl}_{3}\) into \(\mathrm{PCl}_{5}\) is 6 atm.

Key concepts

Equilibrium Constant Expression

Understanding the concept of an equilibrium constant expression is fundamental in solving chemical equilibrium problems. This expression quantifies the ratio of concentrations or partial pressures of products to reactants at equilibrium. In gaseous systems, the equilibrium expression in terms of partial pressures is denoted as \( K_p \). For the reaction \( PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g) \), the equilibrium constant expression is \( K_p = \frac{P_{PCl_5}}{P_{PCl_3} \cdot P_{Cl_2}} \). When given an equilibrium constant value, such as \( K_p = 2 \text{ atm}^{-1} \), you're provided with vital information that relates the pressures of the gases in a very specific proportion when the system is at equilibrium.

Partial Pressures in Chemical Equilibrium

The concept of partial pressures is integral to gas-phase reactions at equilibrium. Each gas in a mixture exerts a certain pressure, proportional to its amount and the total pressure, called its partial pressure. This is important for understanding and calculating the conditions of chemical equilibrium. For example, if a mixture of gases at equilibrium contains \(1 \text{ mole of PCl}_3 \), \(1 \text{ mole of Cl}_2 \), and \(x \text{ moles of PCl}_5 \), the partial pressure exerted by each gas can be expressed as a fraction of the total pressure, and individual partial pressures are essential in determining the direction of the reaction's shift when the system is perturbed.

Calculating Equilibrium Pressure

To calculate the equilibrium pressure for a reaction, the mole fractions of the gases involved along with the equilibrium constant expression are used. In a scenario where we know the initial moles and the extent of the reaction, one can deduce the mole fractions and thereby calculate the partial pressures. In the given exercise, we calculated the total pressure needed to achieve a 75% conversion of \( PCl_3 \) to \( PCl_5 \), taking into account the stoichiometry and the given equilibrium constant, \( K_p \). By systematically substituting the values into the equilibrium expression and solving for the total pressure, we find that a pressure of 6 atm is required to meet the specified conversion.

Le Chatelier's Principle

Le Chatelier's Principle predicts the response of a system in equilibrium to external changes. If a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust to counteract the imposed change and a new equilibrium will be established. When considering a change in pressure for gas-phase reactions, the principle tells us that increasing the pressure of the system will favor the direction that produces fewer moles of gas, and conversely, decreasing the pressure will favor the direction that produces more moles of gas. This principle is especially useful when predicting the behavior of a reaction under different conditions and can help determine the necessary conditions to achieve a desired conversion in a chemical process.