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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 \(450 \mathrm{~K}\).

Short Answer

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

Step by step solution

01

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.
02

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: \[ K_p = \frac{131.7}{12.56 \times 15.91} = \frac{131.7}{199.9496} \approx 0.6584 \].
03

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.
04

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: \[ K_c = 0.6584 \times \frac{1}{8.314 \times 450} = 0.6584 \times \frac{1}{3741.3} \approx 0.000176 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}(g)+\mathrm{Cl}(g) $$ At \(25^{\circ} \mathrm{C},\) the rate constants for the forward and reverse reactions are \(1.4 \times 10^{-28} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) and \(9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?(\mathbf{b})\) Are reactants or products more plentiful at equilibrium?

At \(2000^{\circ} \mathrm{C}\), the equilibrium constant for the reaction \(2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)\) is \(K_{c}=2.4 \times 10^{3} .\) If the initial concentration of \(\mathrm{NO}\) is \(0.250 \mathrm{M},\) what are the equilibrium concentrations of \(\mathrm{NO}, \mathrm{N}_{2}\), and \(\mathrm{O}_{2}\) ?

When \(1.50 \mathrm{~mol} \mathrm{CO}_{2}\) and \(1.50 \mathrm{~mol} \mathrm{H}_{2}\) are placed in a 3.00-L container at \(395^{\circ} \mathrm{C}\), the following reaction occurs: \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\). If \(K_{c}=0.802,\) what are the concentrations of each substance in the equilibrium mixture?

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.15\(), K_{p}\) for the reaction $$ 2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) $$ is about \(1 \times 10^{-11}\). Assuming that the exhaust gas (total pressure \(101.3 \mathrm{kPa}\) ) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2},\) and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the \(\mathrm{CO}_{2}\) reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the \(\mathrm{CO}_{2}\) reaction? Recall that at a fixed pressure and temperature, volume \(\%=\mathrm{mol} \%\).

Le Châtelier noted that many industrial processes of his time could be improved by an understanding of chemical equilibria. For example, the reaction of iron oxide with carbon monoxide was used to produce elemental iron and \(\mathrm{CO}_{2}\) according to the reaction $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \rightleftharpoons 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g) $$ Even in Le Châtelier's time, it was noted that a great deal of CO was wasted, expelled through the chimneys over the furnaces. Le Châtelier wrote, “Because this incomplete reaction was thought to be due to an insufficiently prolonged contact between carbon monoxide and the iron ore [oxide], the dimensions of the furnaces have been increased. In England, they have been made as high as \(30 \mathrm{~m}\). But the proportion of carbon monoxide escaping has not diminished, thus demonstrating, by an experiment costing several hundred thousand francs, that the reduction of iron oxide by carbon monoxide is a limited reaction. Acquaintance with the laws of chemical equilibrium would have permitted the same conclusion to be reached more rapidly and far more economically." What does this anecdote tell us about the equilibrium constant for this reaction?

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