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 \(K_{P}\). (b) By using data in Appendix B, give the value of \(K_{p}\) for this reaction at \(30^{\circ} \mathrm{C}\). (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

The equilibrium equation for liquid water and water vapor is \(\text{H}_{2}\text{O} (\text{liquid}) \rightleftharpoons \text{H}_{2}\text{O} (\text{gas})\), and the expression for \(K_P\) is \(K_P = k[\text{H}_2\text{O}(\text{gas})]\). At \(30^\circ \text{C}\), the vapor pressure of water is \(31.82 \text{ mmHg}\), and although we cannot find the exact value of \(K_P\), it is directly proportional to the vapor pressure. At the normal boiling point, the value of \(K_P\) for any liquid in equilibrium with its vapor is directly proportional to the atmospheric pressure, which is typically \(760 \text{ mmHg}\).

Step-by-step solution

Write the equilibrium equation

For liquid water in equilibrium with water vapor, we have the following equation: \[ \text{H}_{2}\text{O} (\text{liquid}) \rightleftharpoons \text{H}_{2}\text{O} (\text{gas}) \]

Express Kp for the equilibrium

For this equilibrium, we can write the expression for \(K_P\) as follows: \[ K_P = \dfrac{\mathrm{[H_2O(g)]}}{\mathrm{[H_2O(l)]}} \] Keep in mind that concentrations of pure liquids do not change during the reaction, so the concentration of liquid water \([\text{H}_2\text{O}(\text{liquid})]\) is constant. Therefore, the expression for \(K_P\) simplifies to: \[ K_P = k[\text{H}_2\text{O}(\text{gas})] \] where \(k\) is the proportionality constant.

Find the value of Kp at 30°C using data

According to Appendix B, the vapor pressure of water at \(30^\circ \text{C}\) is \(31.82 \text{ mmHg} \). Recall that the expression for \(K_P\) is: \[ K_P = k[\text{H}_2\text{O}(\text{gas})] \] Because the concentration of water vapor is directly proportional to its pressure, we get: \[ K_P = k \times (31.82 \text{ mmHg}) \] However, we do not have the exact value of the proportionality constant \(k\). Even though we cannot calculate the exact value of \(K_P\), we now know that it is directly proportional to the vapor pressure at the given temperature.

Finding Kp for any liquid at its normal boiling point

At the normal boiling point of a liquid, the vapor pressure of the liquid equals the atmospheric pressure (normally \(1 \text{atm} \) or \(760 \: \text{mmHg} \)). Now, the expression for \(K_P\) stands as: \[ K_P = k[\text{liquid} (\text{gas})] \] At the normal boiling point, the vapor pressure of the liquid in equilibrium with its vapor equals the atmospheric pressure, so we have: \[ K_P = k \times (760 \text{ mmHg}) \] Similar to the previous step, we do not have the exact value of the proportionality constant \(k\). However, we can deduce that the value of \(K_P\) for any liquid in equilibrium with its vapor at its normal boiling point is directly proportional to the atmospheric pressure (assuming a standard atmospheric pressure of \(760 \: \text{mmHg} \)).

Key concepts

Equilibrium Constant

The equilibrium constant, represented by the symbol \( K \) or \( K_{eq} \), is a crucial concept in chemistry that indicates the extent of a chemical reaction at equilibrium. It is a ratio of the concentration of the products to the concentration of the reactants, each raised to the power of their stoichiometric coefficients. In the case of vapor pressure equilibrium for water, the equilibrium constant is defined in terms of the pressure of the vapor, as only the gas phase contributes to the pressure.

When the system is at equilibrium, the rate at which the water molecules evaporate from the liquid phase equals the rate at which they condense from the vapor phase. This dynamic balance does not imply that the chemical reaction has stopped, but rather that the forward and reverse reactions are proceeding at the same rate. The equilibrium constant is a direct reflection of the vapor pressure at a specific temperature and is unique for each substance.

Vapor Pressure of Water

Vapor pressure is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. The vapor pressure of water increases as the temperature increases because the molecules have more energy to escape from the liquid surface into the vapor phase.

Factors Affecting Vapor Pressure

• Temperature: Higher temperatures lead to higher vapor pressure.
• Intermolecular Forces: Stronger intermolecular forces result in lower vapor pressure.

Understanding the concept of vapor pressure is vital for predicting how a substance will behave under various temperature conditions. For water, the vapor pressure at any given temperature can be found in published tables and is important for calculating the equilibrium constant.

Normal Boiling Point

The normal boiling point of a liquid is the temperature at which the vapor pressure of the liquid equals the external pressure, typically the atmospheric pressure at sea level (1 atm or 760 mmHg). At this point, the liquid turns into vapor (boils).

Each liquid has its own unique normal boiling point which is a characteristic property of the substance. For instance, the normal boiling point of water is \(100^{\text{o}}C\) at 1 atm. This boiling point changes if the external pressure is different, such as at higher altitudes where atmospheric pressure is lower and water boils at temperatures lower than \(100^{\text{o}}C\). The concept of normal boiling point is essential for understanding the phase transitions of substances and the conditions under which they occur.

Equilibrium Expression

The equilibrium expression, sometimes known as the equilibrium constant expression, relates the concentrations (or partial pressures) of the reactants and products in a reversible reaction at equilibrium. It is mathematically represented based on the chemical equation of the reaction.

For the equilibrium involving liquid water and water vapor, the expression for the equilibrium constant in terms of pressure (\( K_{P} \)) becomes particularly simple because the concentration of the liquid water remains constant. The expression therefore only includes the concentration (or pressure) of the water vapor. This illustrates that for pure substances in a single phase, such as liquid water, their concentrations do not appear in the equilibrium expression, simplifying the calculation of the equilibrium constant.