Problem 119 - Chemistry The Molecular Nature of Matter

Question

We can determine the heat of vaporization by using the Clausius-Clapeyron equation if we know the vapor pressures of a substance at two different temperatures. Determine the heat of vaporization of diethyl ether if the vapor pressure is \(1.0 \mathrm{~mm} \mathrm{Hg}\) at \(-74.3^{\circ} \mathrm{C}\) and \(425 \mathrm{~mm} \mathrm{Hg}\) at \(18.7^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The heat of vaporization of diethyl ether is approximately 26.04 kJ/mol.

Step 1: Convert temperature to Kelvin

Convert the given temperatures from Celsius to Kelvin by adding 273.15 to each Celsius temperature. The Kelvin temperature scale is required when using the Clausius-Clapeyron equation.\[ T_1 = -74.3^{c}C + 273.15 = 198.85 K \] \[ T_2 = 18.7^{c}C + 273.15 = 291.85 K \]

Step 2: Convert vapor pressure to pascals

Convert the vapor pressures from mmHg to pascals, since the Clausius-Clapeyron equation requires pressure in SI units. The conversion factor is 1 mmHg = 133.322 Pa.\[ P_1 = 1.0 \tmmHg \times 133.322 \tfrac{Pa}{mmHg} = 133.322 \tPa \] \[ P_2 = 425 \tmmHg \times 133.322 \tfrac{Pa}{mmHg} = 56661.85 \tPa \]

Step 3: Apply the Clausius-Clapeyron equation

Use the Clausius-Clapeyron equation to determine heat of vaporization. The equation is: \[ \tln\frac{P_2}{P_1} = \frac{\tDelta H_{vap}}{R}\tcbrace(\frac{1}{T_1} - \frac{1}{T_2}) \] where \( \tDelta H_{vap} \) is the heat of vaporization and R is the ideal gas constant (8.314 J/mol·K).

Step 4: Solve for the heat of vaporization

Rearrange the Clausius-Clapeyron equation to solve for \( \tDelta H_{vap} \) and plug in the known values.\[ \tDelta H_{vap} = \frac{R\tln\frac{P_2}{P_1}}{\tfrac{1}{T_1} - \frac{1}{T_2}} \] Substituting the values: \[ \tDelta H_{vap} = \frac{8.314 \tJ/molcdot K\tln\frac{56661.85 \tPa}{133.322 \tPa}}{\tfrac{1}{198.85 \tK} - \frac{1}{291.85 \tK}} \]

Step 5: Calculate the heat of vaporization

Perform the calculations to find the heat of vaporization.\[ \tDelta H_{vap} = \frac{8.314 \tJ/molcdot K \tln(425)}{\tfrac{1}{198.85 \tK} - \frac{1}{291.85 \tK}} \ttapprox 26.04 \tkJ/mol \] This is the amount of energy required to vaporize one mole of diethyl ether.

Key Concepts

Heat of Vaporization

The heat of vaporization, often symbolized as \( \Delta H_{vap} \), is a crucial thermodynamic quantity that represents the amount of energy needed to convert a mole of a substance from a liquid to a gas at constant pressure and temperature.

When a substance undergoes phase change, it either absorbs or releases energy, without changing its temperature. The energy absorbed when a liquid vaporizes is precisely the heat of vaporization. This energy breaks intermolecular forces that hold the liquid molecules together, allowing them to move freely and occupy more space as a gas.

This heat is measurable in units such as Joules per mole (J/mol) and a high heat of vaporization indicates strong intermolecular forces within the liquid. For substances like water, which has a high heat of vaporization, more energy is required to vaporize it compared to substances with weaker intermolecular forces. The Clausius-Clapeyron equation enables us to calculate this heat if the vapor pressures at two temperatures are known, providing insight into the energetic interactions within the substance.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a given temperature in a closed system. It is a critical property that indicates how easily a substance can vaporize.

At any temperature, molecules in a liquid are in constant motion with varying speeds. Some of these molecules can have enough kinetic energy to escape the liquid’s surface and enter the vapor phase. As these molecules accumulate above the liquid, they exert pressure, which increases until the system reaches a dynamic equilibrium where the rate of evaporation equals the rate of condensation.

Factors Affecting Vapor Pressure

Several factors can affect vapor pressure, with temperature being one of the most significant. As temperature increases, more molecules have enough energy to escape, thus increasing vapor pressure. Intermolecular forces also play a pivotal role; weaker forces result in higher vapor pressures.

Understanding vapor pressure is essential for solving thermodynamic problems like boiling point determination and for many practical applications, including distillation and refrigeration processes.

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat, work, temperature, and energy in systems. It lays out a set of universal laws that govern the flow and conversion of energy.

First Law of Thermodynamics

The first law, often known as the law of energy conservation, states that energy cannot be created or destroyed, only transformed from one form to another.

Second Law of Thermodynamics

The second law introduces the concept of entropy and asserts that spontaneous processes in an isolated system always proceed in a way that increases the total entropy.

Using thermodynamics principles, we can analyze processes such as the vaporization of a substance, and predict how changes in temperature, volume, or pressure affect the system. These principles provide the foundation for equations like the Clausius-Clapeyron relation, which describes the quantitative connection between vapor pressure and temperature, informing us about the energetic changes during phase transitions like vaporization.