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 \(K_{p \cdot}(\mathbf{b})\) By using data in Appendix \(\mathrm{B}\), 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 between liquid water and water vapor is: \( H_2O(l) \rightleftharpoons H_2O(g) \). The expression for the equilibrium constant, Kp, is: \( K_p = P_{H_2O(g)} \). At 30°C, the vapor pressure of water is 4.24 kPa, so Kp is 4.24 kPa. At the normal boiling point of any liquid, the value of Kp is equal to the atmospheric pressure, which is 101.3 kPa.

Step-by-step solution

Write the equilibrium equation for liquid water and water vapor

We start by writing the equilibrium equation between liquid water and water vapor. The reaction can be represented as: \( H_2O(l) \rightleftharpoons H_2O(g) \) Here, \(H_2O(l)\) represents liquid water and \(H_2O(g)\) represents water vapor.

Write the expression for Kp

Since we have the chemical reaction for the equilibrium process, we can now write the expression for the equilibrium constant, Kp. For this reaction, we have: \( K_p = \frac{[H_2O(g)]}{[H_2O(l)]} \) However, as Kp only depends on gaseous species, we need to consider partial pressures of the gases in the expression. Since the liquid water is not in the gaseous phase, its partial pressure has no effect on Kp. Thus, the expression for Kp becomes: \( K_p = P_{H_2O(g)} \)

Find the value of Kp at 30°C

Now that we have the expression for Kp, we can find its value at 30°C using the data in Appendix B. In the appendix, we find the vapor pressure of water at various temperatures. At 30°C, the vapor pressure of water is given as 4.24 kPa. So, the value of Kp is: \( K_p = P_{H_2O(g)} = 4.24 \ kPa \)

Find the value of Kp at the normal boiling point

We are asked to find the value of Kp for any liquid in equilibrium with its vapor at the normal boiling point of the liquid. At normal boiling point, the vapor pressure of the liquid is equal to the atmospheric pressure. For this exercise, we assume that atmospheric pressure is 1 atm (101.3 kPa). Therefore, at the normal boiling point, the value of Kp is: \( K_p = P= 101.3 \ kPa \) So, the value of Kp for any liquid in equilibrium with its vapor at the normal boiling point of the liquid is 101.3 kPa.

Key concepts

Vapor Pressure

Vapor pressure is an essential concept in understanding how liquids transition into gases. It is a measure of the tendency of molecules to escape from a liquid into the vapor phase. This pressure results from the equilibrium established between the molecules escaping into the vapor and those condensing back into the liquid.
At a specific temperature, the vapor pressure is a characteristic property of the liquid. For instance, the vapor pressure of water at 30°C is known to be 4.24 kPa.

• The greater the vapor pressure, the more volatile the liquid.
• Vapor pressure increases with temperature as more molecules gain enough energy to enter the vapor phase.
• When vapor pressure equals atmospheric pressure, the liquid reaches its boiling point.

Understanding vapor pressure helps in predicting how a liquid will behave under various temperatures and is crucial for applications like distillation and weather patterns analysis.

Equilibrium Constant

The equilibrium constant, denoted as \(K_p\) for gaseous reactions, quantifies the ratio of the concentrations of products to reactants at equilibrium. In the context of vapor pressure, \(K_p\) represents the saturation level of the vapor as it equilibrates with its liquid form.
For the equilibrium reaction between liquid water and its vapor, \( H_2O(l) \rightleftharpoons H_2O(g) \), the equilibrium constant \(K_p\) is defined as:

• \(K_p = P_{H_2O(g)}\)

In this case, it focuses solely on the water vapor pressure as the liquid's pressure is negligible.

• At 30°C, the \(K_p\) value corresponds to the vapor pressure of water, which is 4.24 kPa.
• At boiling point, \(K_p\) corresponds to atmospheric pressure, typically 101.3 kPa.

Understanding \(K_p\) is vital in predicting how likely a chemical reaction will reach completion under given conditions and is applied across fields such as chemical engineering and environmental science.

Boiling Point

The boiling point is the temperature at which a liquid's vapor pressure equals the atmospheric pressure surrounding it. This marks the phase change from liquid to gas, termed boiling.

• Water's normal boiling point is at 100°C under 1 atm of pressure (101.3 kPa).
• Liquids with lower boiling points have higher vapor pressures at given temperatures.
• Boiling points vary with altitude due to changes in atmospheric pressure. At higher altitudes, boiling occurs at lower temperatures.

This concept is crucial for processes like cooking, where understanding the boiling behavior of substances can influence the outcome of recipes.
Furthermore, in industrial settings, controlling boiling points is critical in processes such as distillation, where separating components of a liquid mixture depends on their varying boiling points.