Problem 7 - Chemistry: Principles and Reactions

Question

Chloroform, \(\mathrm{CHCl}_{3}\), was once used as an anesthetic. In spy movies it is the liquid put in handkerchiefs to render victims unconscious. Its vapor pressure is \(197 \mathrm{~mm} \mathrm{Hg}\) at \(23^{\circ} \mathrm{C}\) and \(448 \mathrm{~mm} \mathrm{Hg}\) at \(45^{\circ} \mathrm{C}\). Estimate its (a) heat of vaporization. (b) normal boiling point.

Step-by-Step Solution

Answer

Question: Calculate the heat of vaporization and normal boiling point of chloroform (CHCl3) given its vapor pressures at two different temperatures, 23°C at 197 mm Hg and 45°C at 448 mm Hg. Answer: The heat of vaporization of chloroform is approximately 29.57 kJ/mol and the normal boiling point of chloroform is approximately 61.35°C.

Step 1: Convert temperatures to Kelvin

To work with the temperatures in the Clausius-Clapeyron equation, we need to convert them to Kelvin. Add 273.15 to the Celsius temperatures: \(T_1 = 23^{\circ} \mathrm{C} + 273.15 = 296.15 \mathrm{K}\) \(T_2 = 45^{\circ} \mathrm{C} + 273.15 = 318.15 \mathrm{K}\)

Step 2: Determine the heat of vaporization (L)

Rearrange the Clausius-Clapeyron equation to solve for the heat of vaporization: \(L = -R(\frac{\ln(\frac{P_2}{P_1})}{\frac{1}{T_2} - \frac{1}{T_1}})\) Input the given values for vapor pressures, temperatures, and the gas constant \(R = 8.314 \mathrm{J/(mol \cdot K)}\): \(L = -8.314 \frac{\ln(\frac{448}{197})}{\frac{1}{318.15} - \frac{1}{296.15}}\) Calculate the heat of vaporization: \(L \approx 29570 \mathrm{J/mol}\) (a) The heat of vaporization of chloroform is approximately \(29.57 \mathrm{kJ/mol}\).

Step 3: Determine the normal boiling point

The normal boiling point is the temperature at which the vapor pressure of a liquid equals the atmospheric pressure. In this case, we will use 760 mm Hg as the atmospheric pressure. Rearrange the Clausius-Clapeyron equation to solve for the boiling point temperature: \(T_2 = \frac{1}{\frac{1}{T_1} -\frac{R}{L}\ln(\frac{P_2}{P_1})}\) Use the previously calculated heat of vaporization, \(L \approx 29570 \mathrm{J/mol}\), and plug in the values: \(T_2 = \frac{1}{\frac{1}{296.15} -\frac{8.314}{29570}\ln(\frac{760}{197})}\) Calculate the boiling point temperature in Kelvin: \(T_2 \approx 334.5 K\) Now convert the temperature to Celsius: \(T_2 = 334.5 - 273.15 = 61.35^{\circ} \mathrm{C}\) (b) The normal boiling point of chloroform is approximately \(61.35^{\circ} \mathrm{C}\).

Key Concepts

Heat of Vaporization

The heat of vaporization is an essential concept in thermodynamics, representing the amount of energy required to transform a given quantity of a substance from a liquid into a gas at constant pressure. In simpler terms, it measures how much energy you need to turn a liquid into a vapor. This concept is crucial to understand, as it provides insights into the strength of molecular forces within a substance.

The heat of vaporization is usually expressed in terms of kilojoules per mole (kJ/mol), which signifies how much energy is necessary to vaporize one mole of a liquid.

• If a substance has a high heat of vaporization, it requires a lot of energy to change from liquid to gas because the intermolecular forces are strong.
• Conversely, a lower heat of vaporization means that the intermolecular forces are weaker, and less energy is needed for the phase change.

In the case of chloroform, when we calculated its heat of vaporization using the Clausius-Clapeyron equation, we found it to be approximately 29.57 kJ/mol. This figure reflects the amount of energy required to break the intermolecular forces in liquid chloroform to transform it into vapor at a given temperature range.

Normal Boiling Point

The normal boiling point of a liquid is the temperature at which its vapor pressure equals the atmospheric pressure, usually 760 mm Hg (also known as standard atmospheric pressure). It's important to note that the normal boiling point is specific to each substance and is regarded as a characteristic property.

Understanding a substance's normal boiling point helps in determining its behavior and interactions under different temperature conditions. Here are a few points to consider:

• A higher normal boiling point usually indicates strong intermolecular forces within the liquid, while a lower boiling point suggests weaker forces.
• Knowing the normal boiling point is essential in applications like cooking, where pressure variations play a crucial role in the boiling process.

For chloroform, the Clausius-Clapeyron equation allowed us to estimate its normal boiling point to be approximately 61.35°C. This temperature is where chloroform's vapor pressure matches atmospheric pressure, making it possible for the liquid to transition to a gaseous state.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid phase at a given temperature. It plays a significant role in understanding the phase changes and volatility of substances. When a liquid's molecules gain enough energy, they transition into a gas, increasing vapor pressure.

Here are some helpful details about vapor pressure:

• A higher vapor pressure at a given temperature indicates a more volatile substance, meaning it evaporates more easily.
• Substances with lower vapor pressures are less volatile, showing stronger intermolecular attractions within the liquid phase.

For chloroform, the vapor pressure changes with temperature, as evident from the values of 197 mm Hg at 23°C and 448 mm Hg at 45°C in the problem. These vapor pressure values help us analyze how easily chloroform transitions from a liquid to a gaseous state as the temperature rises, reflecting its volatility characteristics. The Clausius-Clapeyron equation utilizes these values to indirectly provide a pathway to compute the heat of vaporization and boiling points, showcasing the significance of vapor pressure in such calculations.