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In Section 11.5 we defined the vapor pressure of a liquid in terms of an equilibrium. (a) Write the equation representing the equilibrium between liquid water and water vapor and the corresponding expression for \(K_{p} .\) (b) By using data in Appendix \(\mathrm{B}\), give the value of \(K_{p}\) for this reaction at \(30^{\circ} \mathrm{C} .(\mathrm{c})\) What is the value of \(K_{p}\) for any liquid in equilibrium with its vapor at the normal boiling point of the liquid?

Short Answer

Expert verified
The equilibrium between liquid water and water vapor is represented by the equation \(H_{2}O(l) \rightleftharpoons H_{2}O(g)\). The expression for the equilibrium constant, Kp, is \(K_{p} = P(H_{2}O(g))\). Using data from Appendix B, the value of Kp for this reaction at 30 degrees Celsius is \(K_{p} = 4.24 \: kPa\). The value of Kp for any liquid in equilibrium with its vapor at the normal boiling point of the liquid is \(K_{p} = 101.3 \: kPa\).

Step by step solution

01

Write the Equation for the Equilibrium

Liquid water and water vapor are in equilibrium as per the following reaction: \[H_{2}O(l) \rightleftharpoons H_{2}O(g)\]
02

Write the Expression for the Equilibrium Constant (Kp)

The equilibrium constant Kp is expressed in terms of the partial pressures of the components in the reaction. In this case, the liquid water has no contribution to Kp since it has no partial pressure. Therefore, the expression for Kp becomes: \[K_{p} = \frac{P(H_{2}O(g))}{1}\] which simplifies to: \[K_{p} = P(H_{2}O(g))\]
03

Find the Value of Kp at 30 Degrees Celsius

To find the value of Kp at 30 degrees Celsius, we need the vapor pressure of water at this temperature. Using data from Appendix B, we find that: \[P(H_{2}O(g)) = 4.24 \: kPa\] Therefore, using the Kp expression from Step 2, we get: \[K_{p} = 4.24 \: kPa\]
04

Determine the Value of Kp at the Normal Boiling Point

At the normal boiling point of a liquid, the vapor pressure is equal to 1 atm or 101.3 kPa. Since Kp is equal to the vapor pressure of the substance in equilibrium with its liquid, the value of Kp for any liquid at its normal boiling point is: \[K_{p} = 101.3 \: kPa\]

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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, often symbolized as \( K_p \), is a way to express the balance between reactants and products in a chemical reaction. It is specific to reactions involving gases, where partial pressures are considered. This means that the concentrations of gases, in terms of their pressures, are used to determine the value of \( K_p \).
At equilibrium, the rates of the forward and reverse reactions are equal, leading to a steady state where the composition of the mixture does not change over time.

\( K_p \) depends on the temperature of the reaction. For gaseous equilibria, it's calculated using the formula:
  • \( K_p = \frac{{(P_{products})}}{{(P_{reactants})}} \)
In the specific context of water vapor equilibrium, only the vapor (gas) phase is considered, leading us to an expression that depends solely on the vapor pressure of the substance.
Liquid-Vapor Equilibrium
Liquid-vapor equilibrium refers to the state where both liquid and vapor phases of a substance coexist at equilibrium. This occurs because molecules escape from the liquid surface to the gas phase and condense from the gas back to the liquid at the same rate.

This equilibrium is characterized by the vapor pressure, which is the pressure exerted by the vapor above the liquid when equilibrium is reached. At a given temperature, the vapor pressure is a constant and is unique to each substance.
  • In the equilibrium equation \( H_2O(l) \rightleftharpoons H_2O(g) \), water transitions between its liquid and gas forms while maintaining equilibrium.
  • When the system is at equilibrium, the number of water molecules in the gas returning to liquid equals the number of liquid molecules becoming vapor.
Understanding this equilibrium helps explain why something can go from liquid to vapor but stay in balance without evaporating completely.
Partial Pressure
Partial pressure is the pressure contributed by a single type of gas in a mixture of gases. It's an important concept when evaluating gaseous reactions and dynamic equilibria, as each type of gas contributes to the total pressure in proportion to its concentration.

In chemical equilibrium involving gases, each gas has a partial pressure, summing up to the total pressure, based on its mole fraction. For example:
  • For water in a closed container reaching an equilibrium \( H_2O(l) \rightleftharpoons H_2O(g) \), only the vapor phase contributes to the equilibrium constant \( K_p \).
  • The partial pressure \( P(H_2O) \) indicates how much pressure is exerted by water vapor alone.
At equilibrium at a given temperature, the partial pressure of water equals its vapor pressure, showing why \( K_p \) equals this vapor pressure in such scenarios.

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Most popular questions from this chapter

At \(100^{\circ} \mathrm{C}\) the equilibrium constant for the reaction \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) has the value \(K_{c}=\) \(2.19 \times 10^{-10} .\) Are the following mixtures of \(\mathrm{COCl}_{2}, \mathrm{CO}\) and \(\mathrm{Cl}_{2}\) at \(100^{\circ} \mathrm{C}\) at equilibrium? If not, indicate the direction that the reaction must proceed to achieve equilibrium. (a) \(\quad\left[\mathrm{COCl}_{2}\right]=2.00 \times 10^{-3} \mathrm{M}, \quad[\mathrm{CO}]=3.3 \times 10^{-6} \mathrm{M},\) \(\left[\mathrm{Cl}_{2}\right]=6.62 \times 10^{-6} \mathrm{M} ;\) (b) \(\left[\mathrm{COCl}_{2}\right]=4.50 \times 10^{-2} \mathrm{M}\) \([\mathrm{CO}]=1.1 \times 10^{-7} \mathrm{M},\left[\mathrm{Cl}_{2}\right]=2.25 \times 10^{-6} \mathrm{M} ;(\mathrm{c})\left[\mathrm{COCl}_{2}\right]=\) \(0.0100 M,[\mathrm{CO}]=\left[\mathrm{Cl}_{2}\right]=1.48 \times 10^{-6} \mathrm{M}\)

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 \(P_{\mathrm{PCl}_{3}}=0.124 \mathrm{~atm}\) \(P_{\mathrm{Cl}_{2}}=0.157 \mathrm{~atm},\) 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 \(450 \mathrm{~K}\).

A certain chemical reaction has \(K_{c}=1.5 \times 10^{6}\). Does this mean that at equilibrium there are \(1.5 \times 10^{6}\) times as many product molecules as reactant molecules? Explain.

Consider the following equilibrium: \(2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) \quad K_{c}=1.08 \times 10^{7}\) at \(700^{\circ} \mathrm{C}\) (a) Calculate \(K_{p}\). (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly \(\mathrm{H}_{2} \mathrm{~S} ?\) (c) Calculate the values of \(\mathrm{K}_{c}\) and \(K_{p}\) if you rewrote the balanced chemical equation with \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2}(g)\) instead of \(2 \mathrm{~mol}\).

As shown in Table 15.2 , the equilibrium constant for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is \(K_{p}=4.34 \times 10^{-3}\) at \(300{ }^{\circ} \mathrm{C}\). Pure \(\mathrm{NH}_{3}\) is placed in a 1.00 - \(\mathrm{L}\) flask and allowed to reach equilibrium at this temperature. There are \(1.05 \mathrm{~g} \mathrm{NH}_{3}\) in the equilibrium mixture. (a) What are the masses of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) in the equilibrium mixture? (b) What was the initial mass of ammonia placed in the vessel? (c) What is the total pressure in the vessel?

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