Question

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons\) \(\mathrm{PCl}_{5}(\mathrm{~g})\). A \(7.5-\mathrm{L}\) gas vessel is charged with a mixture of \(\mathrm{PCl}_{3}(g)\) and \(\mathrm{Cl}_{2}(g)\), which is allowed to equilibrate at 450 \(\mathrm{K}\). At equilibrium the partial pressures of the three gases are \(P_{\mathrm{PCl}_{3}}=0.124 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.157 \mathrm{~atm}\), and \(P_{\mathrm{PCl}_{5}}=1.30 \mathrm{~atm}\). (a) What is the value of \(K_{p}\) at this temperature? (b) Does the equilibrium favor reactants or products? (c) Calculate \(K_{c}\) for this reaction at \(450 \mathrm{~K}\).

Short answer

The value of \(K_p\) at 450 K is 66.43, and the equilibrium favors the formation of products, i.e., \(\mathrm{PCl}_{5}\). The value of \(K_c\) for the given reaction at 450 K is approximately 2433.49.

Step-by-step solution

Write the expression for \(K_p\)

To find the value of \(K_p\), we need to write the equilibrium constant expression. For the reaction, we can write the expression for \(K_p\) as: \(K_p = \frac{P_{\mathrm{PCl}_{5}}}{P_{\mathrm{PCl}_{3}} \cdot P_{\mathrm{Cl}_{2}}}\)

Calculate \(K_p\)

To calculate \(K_p\), we will substitute the given partial pressures for each gas at equilibrium in the expression for \(K_p\). We are given: \(P_{\mathrm{PCl}_{3}} = 0.124\) atm, \(P_{\mathrm{Cl}_{2}} = 0.157\) atm, \(P_{\mathrm{PCl}_{5}} = 1.30\) atm Substitute these values in the \(K_p\) expression: \(K_p = \frac{1.30}{0.124 \cdot 0.157} = 66.43\) The value of \(K_p\) at 450 K is 66.43.

Analyze the equilibrium position

To analyze whether the equilibrium favors reactants or products, we look at the value of \(K_p\). If \(K_p > 1\), the reaction favors the formation of products, and if \(K_p < 1\), the reaction favors the formation of reactants. Since \(K_p = 66.43 > 1\), the equilibrium favors the formation of products, i.e., \(\mathrm{PCl}_{5}\).

Write the expression for \(K_c\)

In order to calculate \(K_c\), we first need to write the expression for \(K_c\). For the given reaction, we can write the expression as: \(K_c = \frac{[\mathrm{PCl}_{5}]}{[\mathrm{PCl}_{3}] \cdot [\mathrm{Cl}_{2}]}\)

Convert partial pressures to molar concentrations

To calculate \(K_c\), we need to find the concentrations of the reactants and products in the reaction. We can convert the partial pressures of the gases to molar concentrations using the ideal gas law: \(PV = nRT\) Here, \(P\) represents the partial pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. We have: \(V = 7.5 \mathrm{~L}\) , \(R = 0.0821 \mathrm{\frac{L \cdot atm}{mol \cdot K}}\) , \(T = 450 \mathrm{~K}\) We can convert the partial pressures to molar concentrations using the formula: \([\text{gas}] = \frac{P}{RT}\)

Calculate molar concentrations of gases

Convert the partial pressures of the gases to their molar concentrations using the formula: \(\frac{P_{\mathrm{PCl}_5}}{RT} = \frac{1.30}{ (0.0821)(450)} = 0.0355 \mathrm{M}\) \(\frac{P_{\mathrm{PCl}_3}}{RT} = \frac{0.124}{ (0.0821)(450)} = 0.00338 \mathrm{M}\) \(\frac{P_{\mathrm{Cl}_2}}{RT} =\frac{0.157}{ (0.0821)(450)} = 0.00428 \mathrm{M}\) The molar concentrations are \([\mathrm{PCl}_5] = 0.0355\) M, \([\mathrm{PCl}_3] = 0.00338\) M, and \([\mathrm{Cl}_2] = 0.00428\) M.

Calculate \(K_c\)

Now, substitute the molar concentrations calculated in Step 6 into the \(K_c\) expression: \(K_c = \frac{0.0355}{0.00338 \cdot 0.00428} = 2433.49\) The value of \(K_c\) for the given reaction at 450 K is approximately 2433.49.

Key concepts

Equilibrium Constant (Kp and Kc)

When studying chemical reactions, we encounter a point called chemical equilibrium, where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant over time. Quantitatively, this point can be expressed using equilibrium constants, symbolized as Kp for partial pressures and Kc for molar concentrations.

The equilibrium constant, Kp, is calculated using the formula \[K_p = \frac{P_{\text{products}}}{P_{\text{reactants}}}\],where P stands for the partial pressures of the gases involved. It's essential to raise each partial pressure to the power of its coefficient in the balanced equation. In contrast, the equilibrium constant Kc is calculated with the formula \[K_c = \frac{[\text{products}]}{[\text{reactants}]}\],where square brackets indicate the concentration of substances. For gases, the partial pressures can be converted into concentrations using the ideal gas law, hence establishing a relationship between Kp and Kc: \[K_p = K_c(RT)^{\Delta n}\].Here, R is the ideal gas constant, T is temperature in Kelvin, and Δn is the change in moles of gas in the reaction.

Partial Pressure

In gas-phase chemical reactions, partial pressure is a crucial concept. It represents the pressure that each gas in a mixture would exert if it were alone in the container. According to Dalton's Law of Partial Pressures, the total pressure in a mixture of gases is the sum of the individual partial pressures. When calculating equilibrium constants like Kp, partial pressures need to be used for gases.

The partial pressure can be determined using the relationship provided by the ideal gas law, expressed as: \(P = \frac{nRT}{V}\),where P is the pressure, n is the number of moles of gas, R is the gas constant, T is the temperature, and V is the volume. Understanding how to convert between moles and partial pressures, as shown in the steps for solving the exercise, enables a deeper comprehension of how reaction conditions affect gas behavior at equilibrium.

Reaction Quotient

The reaction quotient, Q, is a number that describes the relative amount of products and reactants at any point in time during a reaction, not just at equilibrium. It uses the same mathematical formula as the equilibrium constants: \[Q_c = \frac{[\text{products}]}{[\text{reactants}]}\],for concentrations, and \[Q_p = \frac{P_{\text{products}}}{P_{\text{reactants}}}\],for partial pressures. The comparison of Q to the equilibrium constant K can predict which direction a reaction will shift to reach equilibrium. If Q is less than K, the reaction will proceed forward, favoring the formation of products. If Q is greater than K, the reaction will shift to favor reactants.

Le Chatelier's Principle

Le Chatelier's principle helps us predict how a reaction will respond to changes in temperature, pressure, concentration, or volume. Simply put, if an external stress is applied to a system at equilibrium, the system adjusts in a way that counteracts the change and a new equilibrium is established. For instance, increasing the concentration of reactants will shift the equilibrium position toward more products. Conversely, an increase in temperature for an exothermic reaction will shift the balance in favor of the reactants.

This principle explains the shifts in equilibrium we observe in various exercises, reinforcing the importance of understanding equilibrium constants and reaction quotients. Mastery of Le Chatelier's principle allows students to reason through the impact of real-world factors on chemical processes.