Problem 34 - Chemistry The Central Science In Si Units

Question

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons\) \(\mathrm{PCl}_{5}(g)\). A 7.5-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 \(P_{\mathrm{PCl}_{3}}=12.56 \mathrm{kPa}, P_{\mathrm{Cl}_{2}}=15.91 \mathrm{kPa},\) and \(P_{\mathrm{PCl}_{5}}=131.7 \mathrm{kPa}\) (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

(a) \( K_p = 0.6584 \); (b) favors reactants; (c) \( K_c = 0.000176 \).

Step 1: Understand the Reaction Process

The chemical equation is \( \mathrm{PCl}_{3}(g) + \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \). This is a reversible reaction where two reactants form one product. At equilibrium, each gas has a specific pressure.

Step 2: Calculate the Equilibrium Constant \( K_p \)

The expression for the equilibrium constant in terms of partial pressure \( K_p \) is given by \[ K_p = \frac{P_{\mathrm{PCl}_{5}}}{P_{\mathrm{PCl}_{3}} \times P_{\mathrm{Cl}_{2}}} \]. Substituting the given pressures: \[ K_p = \frac{131.7}{12.56 \times 15.91} = \frac{131.7}{199.9496} \approx 0.6584 \].

Step 3: Determine if Equilibrium Favors Reactants or Products

Since \( K_p = 0.6584 \) and is less than 1, it indicates that the concentration of reactants \( [\mathrm{PCl}_{3}] \) and \( [\mathrm{Cl}_{2}] \) is favored over products \( [\mathrm{PCl}_{5}] \) at equilibrium.

Step 4: Calculate the Equilibrium Constant \( K_c \)

We use the relation between \( K_p \) and \( K_c \): \[ K_p = K_c (RT)^{\Delta n} \], where \( R = 8.314 \) \( \mathrm{J/mol \cdot K} \), \( T = 450 \) \( \mathrm{K} \), and \( \Delta n = -1 \). Rearranging gives \[ K_c = \frac{K_p}{(RT)^{-1}} = K_p \times \frac{1}{RT} \]. Substituting the values: \[ K_c = 0.6584 \times \frac{1}{8.314 \times 450} = 0.6584 \times \frac{1}{3741.3} \approx 0.000176 \].

Key Concepts

Understanding Equilibrium Constant

In chemical reactions, especially reversible ones, we often talk about something called the **equilibrium constant**. It's a measure of how far a reaction goes to produce products before reaching the balance point, known as equilibrium. For a reaction involving gases, like our example with phosphorus trichloride and chlorine forming phosphorus pentachloride, the equilibrium constant can be expressed in terms of partial pressures, noted as \( K_p \).

The formula to calculate \( K_p \) is given by:

• \( K_p = \frac{P_{\mathrm{products}}}{P_{\mathrm{reactants}}} \)

Here, \( P \) stands for the partial pressure of each gas involved. For the reaction in question, it becomes:

• \( K_p = \frac{P_{\mathrm{PCl}_{5}}}{P_{\mathrm{PCl}_{3}} \times P_{\mathrm{Cl}_{2}}} \)

This provides a quantitative way to express which side of the reaction is favored. A \( K_p \) value greater than 1 indicates that products are favored, while a value less than 1 suggests reactants are more prevalent. In our exercise, the \( K_p \) was calculated to be approximately 0.6584, favoring reactants.

Decoding Partial Pressure

Partial pressure is a concept that plays a crucial role in understanding how gases behave in a mix. In a container with a mixture of different gases, partial pressure refers to the pressure exerted by an individual type of gas independently. Imagine each gas in a mix as if it is the only one in the container.

In our scenario, we have partial pressures for three gases: phosphorus trichloride \( (P_{\mathrm{PCl}_{3}}) \), chlorine \( (P_{\mathrm{Cl}_{2}}) \), and phosphorus pentachloride \( (P_{\mathrm{PCl}_{5}}) \). The pressures were given as:

• \( P_{\mathrm{PCl}_{3}} = 12.56 \) kPa
• \( P_{\mathrm{Cl}_{2}} = 15.91 \) kPa
• \( P_{\mathrm{PCl}_{5}} = 131.7 \) kPa

To find the equilibrium constant \( K_p \), we substitute these partial pressures into the equilibrium expression. Knowing partial pressures is key to calculating important quantities in gas reactions and understanding the dynamics within a mixture.

Exploring Reversible Reactions

Reversible reactions are fascinating because, unlike what we might expect, they don't just go from reactants to products until they're used up. Instead, these reactions can proceed in both directions, which means products can turn back into reactants.

Take our given reaction: \( \mathrm{PCl}_{3}(g) + \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \). The double-headed arrow is the key sign here, indicating the reaction can go both ways, reaching a point where the rate of the forward reaction equals the rate of the reverse reaction. This state is known as equilibrium.

At equilibrium, the system remains constant, allowing us to apply the equilibrium constant concepts. Understanding the nature of reversible reactions allows us to explore how changes to the system, like pressure, temperature, or concentration, affect the balance between reactants and products. This is a key aspect of equilibriatic processes in chemistry.