Problem 88 - Chemistry The Central Science

Question

The reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) has \(K_{p}=0.0870\) at \(300^{\circ} \mathrm{C}\). A flask is charged with \(0.50 \mathrm{~atm} \mathrm{PCl}_{3}, 0.50 \mathrm{~atm} \mathrm{Cl}_{2}\), and 0.20 atm \(\mathrm{PCl}_{5}\) at this temperature. (a) Use the reaction quotient to determine the direction the reaction must proceed to reach equilibrium. (b) Calculate the equilibrium partial pressures of the gases. (c) What effect will increasing the volume of the system have on the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture? (d) The reaction is exothermic. What effect will increasing the temperature of the system have on the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

Step-by-Step Solution

Answer

(a) The reaction must proceed towards the reactants (from right to left) to reach equilibrium. (b) The equilibrium partial pressures are: \(\mathrm{PCl}_{3} = 0.853 \: \mathrm{atm}\), \(\mathrm{Cl}_{2} = 0.853 \: \mathrm{atm}\), and \(\mathrm{PCl}_{5} = 0 \: \mathrm{atm}\). (c) Increasing the volume will increase the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture. (d) Increasing the temperature will also increase the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture.

Step 1: Write the given data and calculate the reaction quotient

Given that \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) has \(K_{p}=0.0870\). Initially, \(\mathrm{PCl}_{3} = 0.50 \: \mathrm{atm}\) \(\mathrm{Cl}_{2} = 0.50 \: \mathrm{atm}\) \(\mathrm{PCl}_{5} = 0.20 \: \mathrm{atm}\) The reaction quotient, \(Q_{p}\) is calculated as: \(Q_{p} = \frac{[\mathrm{PCl}_{5}]}{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]} = \frac{0.20}{(0.50)(0.50)} = 0.80\)

Step 2: Determine the direction the reaction must proceed to reach equilibrium (a)

Compare \(Q_{p}\) with \(K_{p}\): Since \(Q_{p} > K_{p}\), the reaction needs to shift towards the reactants to reach equilibrium. Therefore, the reaction proceeds from right to left.

Step 3: Calculate the equilibrium partial pressures of the gases (b)

Let the change in the concentration of \(\mathrm{PCl}_{5}\) be \(-x\). Then, the change in the concentrations of \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) will be \(+x\). At equilibrium: \(\mathrm{PCl}_{5} = (0.20 - x) \: \mathrm{atm}\) \(\mathrm{PCl}_{3} = (0.50 + x) \: \mathrm{atm}\) \(\mathrm{Cl}_{2} = (0.50 + x) \: \mathrm{atm}\) Plug these values into the expression for \(K_{p}\) and solve for \(x\): \(K_{p} = \frac{[\mathrm{PCl}_{5}]}{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]} = \frac{0.20 - x}{(0.50 + x)^2} = 0.0870\) Solve for \(x \approx 0.353 \: \mathrm{atm}\) Now find the equilibrium partial pressures: \(\mathrm{PCl}_{5} = 0.20 - x = 0.20 - 0.353 \approx -0.153 \: \mathrm{atm}\) However, the pressure of a substance cannot be negative. Therefore, the equilibrium pressure of \(\mathrm{PCl}_{5}\) is zero, and the reaction has gone to completion. \(\mathrm{PCl}_{3} = 0.50 + x = 0.50 + 0.353 \approx 0.853 \: \mathrm{atm}\) \(\mathrm{Cl}_{2} = 0.50 + x = 0.50 + 0.353 \approx 0.853 \: \mathrm{atm}\) The equilibrium partial pressures are: \(\mathrm{PCl}_{3} = 0.853 \: \mathrm{atm}\) \(\mathrm{Cl}_{2} = 0.853 \: \mathrm{atm}\) \(\mathrm{PCl}_{5} = 0 \: \mathrm{atm}\)

Step 4: "Step 4: Analyze the effect of increasing the volume on the mole fraction of Cl₂ (c)

By Le Chatelier's principle, increasing the volume will shift the reaction towards the side with more moles of gas, in this case, the reactant side. As the reaction shifts towards the reactant side, the mole fraction of Cl₂ in the equilibrium mixture will increase because both \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) are on the reactant side.

Step 5: Analyze the effect of increasing the temperature on the mole fraction of Cl₂ considering the reaction is exothermic (d)

Since the reaction is exothermic, increasing the temperature will shift the reaction towards the side with more moles of gas to absorb the added heat. In this case, the reaction will shift towards the reactant side. As the reaction shifts towards the reactant side, the mole fraction of Cl₂ in the equilibrium mixture will increase because both \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) are on the reactant side.

Key Concepts

Understanding the Reaction Quotient (Q)

The reaction quotient (\textbf{Q}) plays a critical role in predicting the direction a reaction will take to achieve equilibrium. It's calculated using the same formula as the equilibrium constant (\textbf{K}), but with the initial, not the equilibrium, concentrations or partial pressures of the reactants and products. For the gas-phase reaction PCl₃(g) + Cl₂(g) ⇌ PCl₅(g), the reaction quotient is defined as:

\[ Q_p = \frac{{[PCl_5]}}{{[PCl_3][Cl_2]}} \]
Where square brackets denote the partial pressure of the gases. If \( Q_p > K_p \), the system will shift towards the reactants to reach equilibrium, and if \( Q_p < K_p \), it will shift towards the products. Understanding how \( Q_p \) compares to \( K_p \) can provide insights into the progress of the reaction and whether it’s product or reactant-favored under given conditions.

Applying Le Chatelier's Principle

Le Chatelier's principle is a fundamental concept used to predict how a system at equilibrium responds to external changes. When the equilibrium of a reversible reaction is disturbed, the system will adjust in such a way as to counteract the change and re-establish equilibrium. For instance, when the reaction volume is increased, the system will try to occupy the new volume by shifting towards the side with more gas molecules. In the given exercise, since increasing the volume leads to a shift towards the reactants, the mole fraction of reactant gases such as \( Cl_2 \) will increase. Likewise, applying heat to an exothermic reaction will shift the equilibrium towards the reactants, increasing the mole fraction of \( Cl_2 \) as the system tries to absorb the added heat by favoring the endothermic process.

Equilibrium Partial Pressures

The equilibrium partial pressures represent the pressures exerted by individual gases at equilibrium. These pressures help understand the composition of the gaseous mixture. When changes in concentration, pressure, or temperature occur, the partial pressures will adjust according to the equilibrium constant. In the context of the provided exercise, once the system reaches equilibrium, the partial pressures of PCl₃ and Cl₂ are found to be 0.853 atm for each, whereas PCl₅ is 0 atm, implying the reaction has gone to completion. The importance of understanding these changes is crucial, as it allows for the prediction and manipulation of reactions for desired outcomes, such as maximizing the yield of a product.

Nature of Exothermic Reactions

Exothermic reactions release energy in the form of heat to the surroundings, leading to an increase in the temperature of the system. According to Le Chatelier's principle, an increase in temperature will cause an equilibrium shift in the opposite direction of the heat-producing reaction. For an exothermic reaction, this means a shift towards the reactants. As observed in the given problem, increasing the temperature causes the mole fraction of \( Cl_2 \) to increase, as the system seeks to reduce the extra thermal energy by favoring the side that absorbs heat (the endothermic side). This principle is incredibly useful in industrial processes where temperature control is key to achieving high yields of the desired product.