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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 \(\mathrm{K} .\) 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
At 450 K, the value of \(K_{p}\) for the given reaction is approximately 0.646, which means the equilibrium favors reactants. The value of \(K_{c}\) at this temperature is approximately 2.41.

Step by step solution

01

Calculate \(K_{p}\)

The expression for \(K_{p}\) is given by: \[K_p = \frac{P_{\mathrm{PCl}_{5}}}{ P_{\mathrm{PCl}_{3}} \cdot P_{\mathrm{Cl}_{2}} }\] Using the given equilibrium partial pressures, we can calculate \(K_{p}\): \[K_p = \frac{131.7}{12.56 \cdot 15.91} \approx 0.646 \]
02

Determine if equilibrium favors reactants or products

A \(K_{p}\) value greater than 1 indicates that the equilibrium favors products, while a value less than 1 indicates it favors reactants. In this case, we found \(K_{p}\) to be approximately 0.646, which means that the equilibrium favors reactants.
03

Calculate \(K_{c}\)

We can use the relationship between \(K_{p}\) and \(K_{c}\) to find the value of \(K_{c}\). The relationship is given by: \[K_p = K_c(RT)^{\Delta n}\] Where \(\Delta n\) is the change in the number of moles of gas in the reaction, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. In this case, \(\Delta n = 1 - (1 + 1) = -1\), and at 450 K, \[K_p = K_c(8.314 \times 450)^{-1}\] Solving for \(K_c\), we get: \[K_c = K_p \times (8.314 \times 450) \approx 2.41\] So, at 450 K, \(K_{c} \approx 2.41\).

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, denoted as either \(K_p\) or \(K_c\), is a crucial concept in understanding chemical equilibrium. It tells us about the ratio of concentrations of products to reactants when a reaction is in equilibrium.
For gas reactions, we often use \(K_p\), which is based on partial pressures. For reactions in solution, \(K_c\) is used, based on concentrations.
  • A high equilibrium constant means that, at equilibrium, the products are favored.
  • A low equilibrium constant indicates that the reactants are favored.
If \(K_p\) is less than 1, it suggests that there are more reactants than products at equilibrium. This means the starting materials are more abundant. Conversely, if \(K_p\) is greater than 1, the equilibrium favors the formation of products.
Understanding \(K\) values is key to predicting how chemical reactions proceed in different scenarios.
Partial Pressure
Partial pressure is an important concept when dealing with gases in chemical equilibrium. It refers to the pressure exerted by a single type of gas in a mixture of gases. In a sealed vessel, each gas contributes to the total pressure according to its proportion in the mixture.
  • To find the partial pressure of a gas, multiply the mole fraction of the gas by the total pressure of the mixture.
  • In equilibrium calculations, partial pressures are used to express the concentration of gases.
In the given example reaction, the partial pressures of \(\text{PCl}_3\), \(\text{Cl}_2\), and \(\text{PCl}_5\) were crucial to calculate \(K_p\). Knowing the partial pressures helps us understand how much of each gas is present when the reaction reaches equilibrium, which is fundamental for determining the direction of the reaction's progression.
Gas Reactions
Gas reactions involve the transformation of gaseous reactants to products. They are subject to the gas laws, which describe how gases behave under various conditions such as pressure, volume, and temperature.
  • Temperature changes can shift the equilibrium position because gases are compressible and their behavior is influenced by thermal energy.
  • Pressure also affects gas reactions significantly. When pressure increases, the system shifts to favor the side with fewer gas molecules.
In the given reaction of phosphorus trichloride and chlorine gases forming phosphorus pentachloride, the gas reactions are balanced by considering these factors. The calculations of \(K_p\) depend heavily on the pressures involved and give insight into the extent of the reaction when these gases are mixed. Understanding gas reactions is pivotal in predicting how changes in conditions will influence chemical equilibrium.

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Most popular questions from this chapter

The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each Hb can bind \(4 \mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which \(50 \%\) of the protein is saturated. Fetal hemoglobin has a \(\mathrm{P} 50\) value of \(2.53 \mathrm{kPa},\) and adult hemoglobin has a P50 value of \(3.57 \mathrm{kPa}\). Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction \(4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \rightleftharpoons\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\) in a fetus, compared to \(K_{c}\) for the same reaction in an adult.

If \(K_{c}=1\) for the equilibrium \(3 \mathrm{~A}(g) \rightleftharpoons 2 \mathrm{~B}(g)\), what is the relationship between [A] and [B] at equilibrium?

At \(900^{\circ} \mathrm{C}, K_{c}=0.0108\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ A mixture of \(\mathrm{CaCO}_{3}, \mathrm{CaO},\) and \(\mathrm{CO}_{2}\) is placed in a \(10.0-\mathrm{L}\) vessel at \(900^{\circ} \mathrm{C}\). For the following mixtures, will the amount of \(\mathrm{CaCO}_{3}\) increase, decrease, or remain the same as the system approaches equilibrium? (a) \(15.0 \mathrm{~g} \mathrm{CaCO}_{3}, 15.0 \mathrm{~g} \mathrm{CaO},\) and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) (b) \(2.50 \mathrm{~g} \mathrm{CaCO}_{3}, 25.0 \mathrm{~g} \mathrm{CaO},\) and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) (a) \(30.5 \mathrm{~g} \mathrm{CaCO}_{3}, 25.5 \mathrm{~g} \mathrm{CaO},\) and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\)

A flask is charged with \(152.0 \mathrm{kPa}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(101.3 \mathrm{kPa}\) \(\mathrm{NO}_{2}(g)\) at \(25^{\circ} \mathrm{C},\) and the following equilibrium is achieved: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is \(51.9 \mathrm{kPa}\). (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?(\mathbf{b})\) Calculate the value of \(K_{p}\) for the reaction. (c) Calculate \(K_{c}\) for the reaction.

Consider the following equilibrium, for which \(K_{p}=7.62\) at \(480^{\circ} \mathrm{C}:\) $$2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g)$$ (a) What is the value of \(K_{p}\) for the reaction \(4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) ?\) (b) What is the value of \(K_{p}\) for the reaction \(\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?\) (c) What is the value of \(K_{c}\) for the reaction in part (b)?

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