Question

Because solid \(p\) -dichlorobenzene, \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2},\) sublimes rather easily, it has been used as a moth repellent. From the data given, estimate the sublimation pressure of \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}(\mathrm{s})\) at \(25^{\circ} \mathrm{C} .\) For \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2} ; \mathrm{mp}=\) \(53.1^{\circ} \mathrm{C} ;\) vapor pressure of \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}(1)\) at \(54.8^{\circ} \mathrm{C}\) is \(10.0 \mathrm{mmHg} ; \Delta H_{\text {fus }}=17.88 \mathrm{kJ} \mathrm{mol}^{-1} ; \Delta H_{\text {vap }}=\) \(72.22 \mathrm{k}] \mathrm{mol}^{-1}\)

Short answer

After calculating, the sublimation pressure of \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}\) at \(25^{\circ}\mathrm{C}\) is found to be approximately \(0.511 kPa\).

Step-by-step solution

Identify given values

From the problem, the given values are: Melting point: \(53.1 °C = 326.25 K\)Vapor pressure at \(54.8°C = 327.95 K\): \(10.0 mmHg = 1.33 kPa\)Heat of fusion, \(\Delta H_{fus} = 17.88 kJ/mol = 17880 J/mol\)Heat of vaporization, \(\Delta H_{vap} = 72.22 kJ/mol = 72220 J/mol\) Temperature we want to find the vapor pressure at: \(25 °C = 298.15 K\)

Calculate the enthalpy of sublimation

The enthalpy of sublimation, \(\Delta H_{sub}\), is the sum of the heats of fusion and vaporization:\[\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap} = 17880 J/mol + 72220 J/mol = 90100 J/mol\]

Use the Clausius-Clapeyron equation

In form \[ln(P_f/P_i) = -\Delta H_{sub}/R · (1/T_f - 1/T_i)\]where \(P_f\) is the final pressure (which we're looking for), \(P_i\) is the initial pressure (1.33 kPa at 327.95 K), \(\Delta H_{sub}\) is the enthalpy of sublimation, and \(R\) is the ideal gas constant (8.314 J/K·mol). \(T_f\) is the final temperature (298.15 K), and \(T_i\) is the initial temperature (327.95 K).Rearrange the equation to solve for \(P_f\):\[P_f = P_i · exp[\Delta H_{sub}/R · (1/T_i - 1/T_f)]\]

Calculate the sublimation pressure

Substitute in the known values:\[P_f = 1.33 kPa · exp[90100 J/mol / (8.314 J/K·mol) · (1/327.95 K - 1/298.15 K)]\]

Key concepts

Clausius-Clapeyron Equation

Understanding the Clausius-Clapeyron equation is pivotal when diving into the field of thermodynamics, specifically while dealing with a substance's phase change. The equation forms a crucial link between temperature and pressure during a phase transition.

The equation's foundation rests on the relationship that a higher pressure is required to sustain a phase change, like sublimation, at a higher temperature. It mathematically expresses how pressure changes with temperature for a phase equilibrium situation. Expressed symbolically in its simplified form, it reads as \[ln\left(\frac{P_f}{P_i}\right) = -\frac{\Delta H}{R} \left(\frac{1}{T_f} - \frac{1}{T_i}\right)\] where \(P_i\) and \(P_f\) are the initial and final pressures, \(T_i\) and \(T_f\) are the initial and final temperatures, \(\Delta H\) is the enthalpy change associated with the phase transition, and \(R\) is the universal gas constant.

For students and educators alike, it's reassuring to recognize that the Clausius-Clapeyron Equation isn't just a theoretical expression. It's a practical tool that finds use in real-world applications like predicting the weather, designing refrigeration systems, and even estimating sublimation pressures, as seen in our exercise.

Enthalpy of Sublimation

Enthalpy of sublimation, often symbolized as \(\Delta H_{sub}\), is the total heat required for a substance to change from solid to gas without passing through the liquid phase. It's a thermodynamic concept that combines two other forms of heat changes: the heat of fusion (solid to liquid) and the heat of vaporization (liquid to gas).

In mathematical terms: \[\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap}\]
Approaching the topic from an instructional angle, breaking down the enthalpy of sublimation into these two components simplifies understanding. Students can grasp that sublimation isn't a mysterious process; rather, it's the sum of two transitions they're already familiar with.

When a compound like p-dichlorobenzene sublimes easily, as noted in our textbook problem, it's valuable to calculate the enthalpy of sublimation to predict how it behaves under different temperature and pressure conditions. This calculation bridges the gap between theory and practice, allowing for accurate predictions of sublimation pressures.

Phase Transition

The term 'phase transition' embodies the transformation of a substance from one state of matter to another. Familiar transitions include melting, freezing, boiling, and condensing, but there's also sublimation which is the direct shift from solid to gas. Each transition corresponds to a change in energy and organization of the molecules within a material.

During a phase transition, substances absorb or release energy, which is where the concepts of enthalpy of fusion, vaporization, and sublimation come into play. It's essential to know that these transitions occur at specific temperatures and pressures, where two phases can coexist in equilibrium. This equilibrium is beautifully captured by the Clausius-Clapeyron equation, guiding students and scientists to predict phase transitions under various conditions.

Tailoring this information for educational platforms, it's crucial to ensure learners understand that temperature doesn't change during a phase transition at constant pressure, and it's the energy input or output that facilitates the phase change.

Vapor Pressure

Vapor pressure is a fundamental concept in thermodynamics, referring to the pressure exerted by a vapor in equilibrium with its solid or liquid phase. It's an intrinsic property of a substance and depends on temperature. The higher the temperature, the more molecules have enough energy to escape from the surface, leading to higher vapor pressure.

Linking back to the classroom context, the concept of vapor pressure helps explain why liquids evaporate faster at higher temperatures, and why some substances, like p-dichlorobenzene mentioned in the exercise, can sublimate quite readily. When attempting to understand how substances behave, vapor pressure provides insight into their volatility and stability under various conditions.

It's enlightening for students to see how vapor pressure influences everyday phenomena, such as the drying of wet clothing or the preservation of food. By connecting theoretical concepts to tangible examples, the educational journey becomes more engaging and meaningful.