Question

At \(20^{\circ} \mathrm{C},\) the vapor pressure of water is \(0.0231 \mathrm{~atm} .\) Calculate \(K_{P}\) and \(K_{\mathrm{c}}\) for the process: $$\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}_{2} \mathrm{O}(g)$$

Short answer

\( K_P = 0.0231 \mathrm{~atm} \), \( K_c = 0.0009636 \mathrm{~mol/L} \).

Step-by-step solution

Understanding the Reaction System

The given process is the phase transition of water from liquid to gas. At equilibrium, the vapor pressure of water is a measure of the tendency of its molecules to escape into the gaseous phase.

Expressing Equilibrium Constant in Terms of Vapor Pressure

The equilibrium constant for a gaseous reaction in terms of partial pressures is denoted as \( K_P \). For the reaction \( \mathrm{H}_2\mathrm{O}(l) \rightleftarrows \mathrm{H}_2\mathrm{O}(g) \), the vapor pressure of water directly represents the equilibrium constant \( K_P \). Thus, \( K_P = P_{\mathrm{H}_2\mathrm{O}} = 0.0231 \mathrm{~atm} \).

Relating Kc to Kp

The equilibrium constant in terms of concentration \( K_c \) and in terms of pressure \( K_P \) are related by the equation \( K_P = K_c(RT)^{\Delta n} \), where \( \Delta n \) is the change in moles of gas, \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \), and \( T = 293 \, \text{K} \) is the absolute temperature. Here, \( \Delta n = 1 - 0 = 1 \).

Solving for Kc

Rearrange the formula \( K_P = K_c(RT)^{\Delta n} \) to find \( K_c \): \[ K_c = \frac{K_P}{(RT)^{\Delta n}} = \frac{0.0231}{(0.0821 \times 293)^{1}} \].

Calculating Kc Value

Calculate the value of \( K_c \) using the expression derived: \( K_c = 0.0231 / (0.0821 \times 293) = 0.0009636 \mathrm{~mol/L} \).

Key concepts

Vapor Pressure

The vapor pressure of a liquid is the pressure exerted by its vapor when the liquid and vapor are in equilibrium. In simple terms, it is the measure of the tendency of liquid molecules to escape into the gas phase.
For a given substance, the vapor pressure primarily depends on temperature. As temperature increases, vapor molecules have more energy to overcome intermolecular forces, resulting in higher vapor pressure.
Water, for instance, has a vapor pressure of 0.0231 atm at 20°C. This indicates that at this temperature, the water molecules in the gas phase exert this level of pressure when in equilibrium with its liquid form.

Phase Transition

Phase transition refers to the process of changing from one state of matter to another, such as from liquid to gas. For water transitioning to vapor, energy in the form of heat is absorbed, allowing water molecules to break free from the liquid state.
This change is observable when you boil water, as it moves from a liquid to a gas. However, it can also occur at temperatures below boiling, as seen in evaporation.
In chemical equilibria, phase transitions are significant because they represent reactions where the equilibrium constant can be expressed in terms of vapor pressure.

Kp and Kc

In the context of chemical reactions involving gases, equilibrium constants can be expressed in two forms:

• Kp: This is the equilibrium constant expressed in terms of partial pressures of gaseous components. It is particularly useful for reactions where gases are involved since pressure is a convenient measurable quantity.


• Kc: This is the equilibrium constant expressed in terms of molar concentrations, often used for reactions in solutions.

For the reaction \( \mathrm{H}_2\mathrm{O}(l) \rightleftarrows \mathrm{H}_2\mathrm{O}(g) \), Kp is equal to the vapor pressure of water, 0.0231 atm. To convert Kp to Kc, you use the relation \( K_P = K_c(RT)^{\Delta n} \), where R is the gas constant and T is the temperature in Kelvin.

Gas Laws

The behavior of gases is described by gas laws, which provide a mathematical relationship between pressure, volume, temperature, and moles.
In the context of equilibrium, these laws guide us in understanding how changes in conditions affect a gaseous system. The ideal gas law \( PV = nRT \) relates pressure (P), volume (V), temperature (T), and the number of moles (n) with the gas constant (R).
For calculating equilibrium constants like Kp, these laws simplify the process by relating pressure to concentrations and temperatures. In reactions where the number of gas moles changes, as denoted by \( \Delta n \), the formula \( K_P = K_c(RT)^{\Delta n} \) is derived from applying gas laws to the concept of equilibrium.