Problem 34 - Chemistry The Central Science

Question

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons\) \(\mathrm{PCl}_{5}(g) . \mathrm{A} 7.5-\mathrm{L}\) gas vessel is charged with a mixture of \(\mathrm{PCl}_{3}(g)\) and \(\mathrm{Cl}_{2}(g),\) which is allowed to equilibrate at 450 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 \(P_{\mathrm{PCl}_{\mathrm{s}}}=1.30 \mathrm{atm}\) (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 \(\mathrm{K}\)

Step-by-Step Solution

Answer

In summary, at 450 K, the value of Kp for the reaction \(PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g)\) is approximately 66.98, indicating that the equilibrium strongly favors the formation of phosphorus pentachloride (\(PCl_5\)). The value of Kc for this reaction at 450 K is approximately 3679.45.

Step 1: Write the equilibrium constant expression for Kp

The expression for Kp is derived from the balanced chemical equation, using the partial pressures of each gas at equilibrium: \(K_p = \dfrac{P_{PCl_5}}{P_{PCl_3} P_{Cl_2}}\)

Step 2: Substitute the given values into the Kp expression

Inserting the provided partial pressures, the equation becomes: \(K_p = \dfrac{1.30}{(0.124)(0.157)}\)

Step 3: Calculate Kp

We will now solve the equation to find Kp: Kp = \(\dfrac{1.30}{(0.124)(0.157)} = 66.98\)

Step 4: Analyze the Kp value for favorability

When Kp > 1, the equilibrium favors the products. In this case, Kp = 66.98, much larger than 1, so we can conclude that the equilibrium strongly favors the formation of phosphorus pentachloride (\(PCl_5\)).

Step 5: The relationship between Kp and Kc

The relationship between the equilibrium constants Kp and Kc is given by: \(K_p = K_c(RT)^{\Delta n}\), where \(\Delta n\) is the change in the number of moles of gaseous species from the reactant side to the product side.

Step 6: Calculate the change in the number of moles

From the balanced chemical equation, we see that 1 mole of \(PCl_3\) and 1 mole of \(Cl_2\) react to form 1 mole of \(PCl_5\). So, \(\Delta n = (1 - 1 - 1) = -1\)

Step 7: Calculate Kc using Kp, R, T, and Δn

Substitute the given values and Kp value into the relationship equation and solve for Kc: \(K_c = \dfrac{K_p}{RT^{Δn}}\) \(K_c = \dfrac{66.98}{(0.08206 \frac{L \cdot atm}{mol \cdot K})(450K)^{-1}}\) \(K_c = \dfrac{66.98}{0.0182}\) Kc = 3679.45 So, the value of Kc for this reaction at 450 K is approximately 3679.45.

Key Concepts

Equilibrium Constant

Understanding the equilibrium constant (\(K\)) is essential when studying chemical equilibrium. It is a numerical measure of the ratio of the concentrations of products to reactants at equilibrium. For gaseous reactions, this constant is often expressed in terms of partial pressures, denoted as \(K_p\). In the example given, we calculate \(K_p\) for the reaction \(\mathrm{PCl}_{3}(g) + \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) based on the equilibrium partial pressures.

The calculation of \(K_p\) involves substituting these partial pressures into the equilibrium constant expression: \(K_p = \frac{P_{PCl_5}}{P_{PCl_3} P_{Cl_2}}\). The resulting high value of \(K_p = 66.98\) suggests that the reaction strongly favors the formation of products, in this case, \(\mathrm{PCl}_{5}(g)\). High values of \(K\) indicate a large concentration of products at equilibrium, whereas low values indicate a larger concentration of reactants.

To further interpret this reaction, the equilibrium constant in terms of concentration, \(K_c\), can be related to \(K_p\) when the gases involved each occupy the same volume. For gases, this relationship includes the temperature (\(T\)) and the ideal gas constant (\(R\)) and considers the change in the number of moles of gas (\(\Delta n\)).

Partial Pressures

Partial pressures play a pivotal role in determining the direction and extent of chemical reactions involving gases. The partial pressure of a gas in a mixture is the pressure that the gas would exert if it alone occupied the entire volume of the mixture at the same temperature.

In our chemical equilibrium example, the partial pressures of \(\mathrm{PCl}_{3}(g)\), \(\mathrm{Cl}_{2}(g)\), and \(\mathrm{PCl}_{5}(g)\) are provided at equilibrium. When calculating the equilibrium constant \(K_p\), these partial pressures are used, reflecting the actual concentrations of gases at that moment. It is the individual partial pressures that are plugged into the equilibrium expression to determine \(K_p\) and they determine the reaction's progress and position of equilibrium.

It’s important to note that if we change the conditions of the reaction, such as the temperature or volume, the partial pressures will change, affecting the value of \(K_p\) and potentially shifting the equilibrium position. This shift can be predicted by applying Le Châtelier's principle.

Le Châtelier's Principle

Le Châtelier's principle is a guiding concept in chemical equilibrium, helping us predict how a system at equilibrium reacts to disturbances. It states that if an external change is applied to a system at equilibrium, such as a change in pressure, temperature, or concentration, the system will adjust itself to counteract the effect of the change and establish a new equilibrium.

For instance, if we were to increase the concentration of \(\mathrm{PCl}_{3}(g)\) in our reaction, the principle predicts that the equilibrium would shift to the right, producing more \(\mathrm{PCl}_{5}(g)\) to relieve the stress of the added reactant. Similarly, increasing the pressure by decreasing the volume of the reaction vessel would shift the equilibrium toward the side with fewer moles of gas, which, in this case, is the product side.

Le Châtelier's principle allows chemists to optimize conditions for the highest yield of desired products. In educational settings, visual aids such as graphs or animations showing the shifts in reaction equilibria can greatly enhance understanding of this principle and how it affects chemical reactions.