Problem 90 - Chemistry The Central Science In Si Units

Question

The reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) has \(K_{p}=\) \(8.59 \times 10^{-4}\) at \(300^{\circ} \mathrm{C}\). A flask is charged with \(50.7 \mathrm{kPa}\) \(\mathrm{PCl}_{3}, 50.7 \mathrm{kPa} \mathrm{Cl}_{2}\), and \(20.3 \mathrm{kPa} \mathrm{PCl}_{5}\) 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. \((\mathbf{c})\) 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 shifts left. (b) Equilibrium pressures: \( P_{\mathrm{PCl}_3} = P_{\mathrm{Cl}_2} \approx 50.671 \), \( P_{\mathrm{PCl}_5} \approx 20.271 \). (c) Increasing volume increases \( \mathrm{Cl}_2 \) mole fraction. (d) Increasing temperature increases \( \mathrm{Cl}_2 \) mole fraction.

Step 1: Determine the Reaction Quotient

The reaction quotient, \( Q_p \), is determined using the initial partial pressures of the gases. For the reaction \( \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \), \( Q_p \) is given by:\[ Q_p = \frac{P_{\mathrm{PCl}_5}}{P_{\mathrm{PCl}_3} \cdot P_{\mathrm{Cl}_2}} = \frac{20.3}{50.7 \times 50.7} \]Calculate \( Q_p \):\[ Q_p = \frac{20.3}{2570.49} \approx 0.0079 \]

Step 2: Compare Reaction Quotient with Equilibrium Constant

With \( Q_p = 0.0079 \) and \( K_p = 8.59 \times 10^{-4} \), compare the two values. Since \( Q_p > K_p \), the reaction will proceed in the direction that reduces \( Q_p \), which means it will proceed to the left (towards reactants) to reach equilibrium.

Step 3: Set Up the Ice Table for Equilibrium Calculation

We use the initial concentrations to set up the ICE (Initial, Change, Equilibrium) table:Initial: \( P_{\mathrm{PCl}_3} = 50.7 \), \( P_{\mathrm{Cl}_2} = 50.7 \), \( P_{\mathrm{PCl}_5} = 20.3 \)Change: \( -x \), \( -x \), \( +x \)Equilibrium: \( 50.7-x \), \( 50.7-x \), \( 20.3+x \)

Step 4: Express Equilibrium Constant in Terms of x

The equilibrium constant K is given by:\[ K_p = \frac{P_{\mathrm{PCl}_5}}{P_{\mathrm{PCl}_3} \cdot P_{\mathrm{Cl}_2}} = \frac{20.3 + x}{(50.7-x)(50.7-x)} \]Set \( K_p = 8.59 \times 10^{-4} \) and solve for \( x \).

Step 5: Solve for x and Calculate Equilibrium Pressures

Solve the quadratic equation to find \( x \). Assuming small \( x \), simplify as:\[ 8.59 \times 10^{-4} = \frac{20.3 + x}{(50.7^2)} \]Find \( x\) and compute:\[ x \approx 0.029 \]Equilibrium:\[ P_{\mathrm{PCl}_3} = P_{\mathrm{Cl}_2} \approx 50.671 \]\[ P_{\mathrm{PCl}_5} \approx 20.271 \]

Step 6: Effect of Volume on Mole Fraction of \( mathrm{Cl_2} \)

Increasing volume decreases total pressure, shifting equilibrium to the side with more moles of gas, here the left side. The mole fraction of \( \mathrm{Cl}_2 \) increases as the system shifts toward the reactant side with more moles.

Step 7: Effect of Temperature on Mole Fraction of \( mathrm{Cl_2} \)

Since the reaction is exothermic, increasing temperature shifts equilibrium to the left (towards reactants), increasing the mole fraction of \( \mathrm{Cl}_2 \).

Key Concepts

Reaction Quotient

The reaction quotient, often represented as \(Q\), is a valuable tool in predicting the direction a chemical reaction will move to reach equilibrium. For the reaction \( \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \), we calculate \(Q_p\) using the initial partial pressures:

• \( P_{\mathrm{PCl}_5} = 20.3 \ \mathrm{kPa} \)
• \( P_{\mathrm{PCl}_3} = 50.7 \ \mathrm{kPa} \)
• \( P_{\mathrm{Cl}_2} = 50.7 \ \mathrm{kPa} \)

We use the formula:\[Q_p = \frac{P_{\mathrm{PCl}_5}}{P_{\mathrm{PCl}_3} \cdot P_{\mathrm{Cl}_2}} = \frac{20.3}{50.7 \times 50.7}\approx 0.0079\]By comparing \(Q_p\) with the equilibrium constant \(K_p\), one can determine which direction the reaction will proceed.

Equilibrium Constant

The equilibrium constant \(K\) is specific to a particular reaction at a given temperature. It indicates the ratio of the concentrations of products to reactants at equilibrium. For gas-phase reactions, we use \(K_p\), where pressure is involved, such as in:\[K_p = \frac{P_{\mathrm{PCl}_5}}{P_{\mathrm{PCl}_3} \cdot P_{\mathrm{Cl}_2}}\]In the given reaction, \(K_p = 8.59 \times 10^{-4}\), which suggests the position of equilibrium favors the reactants.To find equilibrium pressures, adjustments to initial pressures are considered through an ICE table (Initial, Change, Equilibrium).

• If \(Q_p \) (initial condition) is greater than \(K_p\), the reaction will shift to the left, towards reactants, until \(Q_p = K_p\).
• This movement continues until equilibrium is re-established, allowing for the calculation of equilibrium pressures using changes denoted by \(x\).

Le Chatelier's Principle

Le Chatelier's Principle describes how a system at equilibrium responds to disturbances, aiming to re-establish balance. Consider the effects of volume changes. As the volume of a system increases, the pressure decreases, prompting the reaction to shift towards the side with more moles of gas to regain equilibrium.

• In our reaction \( \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \), an increase in volume shifts equilibrium to the left, increasing \( \mathrm{Cl}_2 \)'s mole fraction, as there are more moles on the reactant side.

This principle aids in predicting the response to changes in pressure or temperature.Le Chatelier's Principle is versatile, applying not only to gas volumes but also temperature changes, as seen in exothermic reactions.

Exothermic Reaction

An exothermic reaction releases heat, causing the equilibrium to shift in response to temperature changes. Increasing the temperature of such reactions shifts the equilibrium towards the reactants. This is because additional heat seems like a 'product', and the system attempts to restore balance by favoring the endothermic reaction (opposite direction).

• For the exothermic reaction \( \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \), raising the temperature will increase the mole fraction of \( \mathrm{Cl}_2 \).

This shift decreases the formation of \( \mathrm{PCl}_5 \), directing the reaction towards the reactants, which absorb the extra heat. Understanding these dynamics is crucial for controlling chemical synthesis and understanding reaction behavior in varying conditions.