Ch.15 - Chemical Equilibrium

Problem 100c

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Chapter 15, Problem 100c

In Section 11.5, we defined the vapor pressure of a liquid in terms of an equilibrium. (c) What is the value of Kp for any liquid in equilibrium with its vapor at the normal boiling point of the liquid?

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Understand that the normal boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure, typically 1 atm.

Recognize that at equilibrium, the rate of evaporation of the liquid equals the rate of condensation of the vapor.

Recall that the equilibrium constant for the vaporization process, Kp, is defined in terms of the partial pressures of the vapor phase.

At the normal boiling point, the vapor pressure of the liquid is equal to the external pressure, which is 1 atm.

Therefore, the value of Kp at the normal boiling point is equal to the vapor pressure of the liquid, which is 1 atm.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a given temperature. It reflects the tendency of particles to escape from the liquid phase into the vapor phase. At the normal boiling point, the vapor pressure of a liquid equals the external atmospheric pressure, allowing the liquid to transition to gas.

Equilibrium Constant (Kp)

The equilibrium constant, Kp, is a dimensionless number that expresses the ratio of the partial pressures of the products to the reactants at equilibrium for a given reaction. For a liquid in equilibrium with its vapor, Kp can be defined as the ratio of the vapor pressure of the liquid to the pressure of the liquid, which is typically considered to be 1 at the boiling point.

Normal Boiling Point

The normal boiling point of a liquid is the temperature at which its vapor pressure equals the standard atmospheric pressure (1 atm). At this point, the liquid can transition to gas throughout the entire liquid volume, not just at the surface. This concept is crucial for understanding the conditions under which Kp is defined for a liquid in equilibrium with its vapor.

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Open Question

At 800 K, the equilibrium constant for the reaction A₂(g) ⇌ 2 A(g) is K_c = 3.1 × 10^-4. (d) If the temperature is raised to 1000 K, will the reverse rate constant k_r increase or decrease? Will the change in k_r be larger or smaller than the change in k_f?

Textbook Question

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 Kp.

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Textbook Question

In Section 11.5, we defined the vapor pressure of a liquid in terms of an equilibrium. (b) By using data in Appendix B, give the value of Kp for this reaction at 30 C.

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Open Question

Is water dimer formation endothermic or exothermic, given that the Kp for water dimer formation in the gas phase is 0.050 at 300 K and 0.020 at 350 K?

Open Question

The protein hemoglobin (Hb) transports O2 in mammalian blood. Each Hb can bind 4 O2 molecules. The equilibrium constant for the O2 binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the P50 value, defined as the partial pressure of oxygen at which 50% of the protein is saturated. Fetal hemoglobin has a P50 value of 19 torr, and adult hemoglobin has a P50 value of 26.8 torr. Use these data to estimate how much larger Kc is for the aqueous reaction 4 O2(aq) + Hb(aq) ⇌ Hb(O2)4(aq) in a fetus, compared to Kc for the same reaction in an adult.