Problem 99 - General Chemistry

Question

The vapor pressure of benzene is \(100.0 \mathrm{mmHg}\) at \(26.1^{\circ} \mathrm{C}\) and \(400.0 \mathrm{mmHg}\) at \(60.6^{\circ} \mathrm{C}\). What is the boiling point of benzene at \(760.0 \mathrm{mmHg}\) ?

Step-by-Step Solution

Answer

The boiling point of benzene at 760.0 mmHg is approximately 43.29°C.

Step 1: Understand the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is used to relate the pressure and temperature of a substance at two different states. The equation is given by \[ \ln \frac{P_2}{P_1} = \frac{\Delta H_{vap}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \], where \(P_1\) and \(P_2\) are the pressures, \(T_1\) and \(T_2\) are the temperatures in Kelvin, \(\Delta H_{vap}\) is the molar enthalpy of vaporization, and \(R\) is the universal gas constant (8.314 J/mol*K).

Step 2: Convert Temperatures to Kelvin

First, convert the given temperatures from Celsius to Kelvin, as the Clausius-Clapeyron equation requires temperatures in Kelvin. Use the conversion: \[ T(K) = T(^{\circ}C) + 273.15 \]Convert the temperatures: \(26.1^{\circ}C\) to Kelvin: \[ T_1 = 26.1 + 273.15 = 299.25 \, K \] \(60.6^{\circ}C\) to Kelvin: \[ T_2 = 60.6 + 273.15 = 333.75 \, K \]

Step 3: Calculate the Enthalpy of Vaporization

Rearrange the Clausius-Clapeyron equation to solve for \(\Delta H_{vap}\): \[ \Delta H_{vap} = \frac{R \cdot \ln \frac{P_2}{P_1}}{\left( \frac{1}{T_1} - \frac{1}{T_2} \right)} \] Substitute the known values: \(P_1 = 100.0 \mathrm{mmHg}\), \(P_2 = 400.0 \mathrm{mmHg}\), \(T_1 = 299.25 \, K\), \(T_2 = 333.75 \, K\). Calculate: \[ \Delta H_{vap} = \frac{8.314 \cdot \ln \frac{400.0}{100.0}}{\left( \frac{1}{299.25} - \frac{1}{333.75} \right)} \approx 31755.7 \, J/mol \]

Step 4: Rearrange Equation for Boiling Point Calculation

Use the Clausius-Clapeyron equation to find the boiling point \(T_b\) when \(P = 760.0 \mathrm{mmHg}\): \[ \ln \frac{760.0}{100.0} = \frac{31755.7}{8.314} \left( \frac{1}{299.25} - \frac{1}{T_b} \right) \]

Step 5: Solve for Boiling Point in Kelvin

Substitute known values into the rearranged equation to solve for \(T_b\): Calculate \( \ln \frac{760.0}{100.0} = \ln 7.6 \approx 2.028 \). Substitute and solve: \[ 2.028 = \frac{31755.7}{8.314} \left( \frac{1}{299.25} - \frac{1}{T_b} \right) \]Simplify and solve for \( \frac{1}{T_b} \): \[ \frac{1}{T_b} = \frac{1}{299.25} - \frac{2.028 \, \times \, 8.314}{31755.7} \approx \frac{1}{299.25} - 0.0005317 \approx 0.003161 \, K^{-1} \]Finally, solve for \(T_b\): \[ T_b = \frac{1}{0.003161} \approx 316.44 \, K \]

Step 6: Convert Boiling Point Back to Celsius

Convert the boiling point from Kelvin back to Celsius: \[ T(^{\circ}C) = T(K) - 273.15 \] Substitute the value found:\[ T_b = 316.44 - 273.15 \approx 43.29^{\circ}C \]

Key Concepts

Vapor Pressure

Vapor pressure is an essential concept when discussing liquids and their tendency to evaporate. It describes the pressure exerted by the vapor phase of a liquid in equilibrium with its liquid phase at a specific temperature. This equilibrium occurs because at any temperature, molecules at the surface of a liquid gain enough energy to escape into the vapor phase, and some molecules from the vapor phase return to the liquid phase.

The vapor pressure of a liquid depends on:

• The nature of the liquid: Different liquids have different volatilities, which influences their vapor pressures.
• Temperature: Generally, an increase in temperature leads to an increase in vapor pressure as more molecules have the necessary energy to escape the liquid surface.

Understanding vapor pressure is crucial when using the Clausius-Clapeyron equation, which relates vapor pressures at different temperatures to calculate unknown variables.

Enthalpy of Vaporization

Enthalpy of vaporization, \( \Delta H_{vap} \), is the amount of heat energy required to transform a liquid into a gas at its boiling point, under standard pressure. This parameter is essential for understanding energy changes during phase transitions.

Key characteristics include:

• Measured in energy per amount of substance, typically Joules per mole (J/mol).
• A higher enthalpy of vaporization indicates a stronger intermolecular force in the liquid, requiring more energy to overcome.

The Clausius-Clapeyron equation is particularly useful for calculating this value when the vapor pressures and temperatures at two states are known. It helps in predicting how vapor pressure changes with temperature, reflecting the energy required for phase transition.

Boiling Point

The boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure exerted on the liquid. At this point, the liquid begins to boil and turns into vapor.

Key factors influencing the boiling point are:

• External pressure: Lower pressure environments, like at high altitudes, lower the boiling point, while higher pressures increase it.
• Nature of the liquid: Strong intermolecular forces result in higher boiling points as more energy is required to break these forces.

In exercises where the Clausius-Clapeyron equation is applied, calculating the boiling point under new pressure conditions can be achieved by rearranging the equation, highlighting the importance of understanding both temperature and pressure dependencies.

Temperature Conversion

Temperature conversion is crucial for solving science problems, particularly when dealing with equations that require temperatures in a specific unit system. For the Clausius-Clapeyron equation, temperatures must be in Kelvin.

Conversion formula:

• From Celsius to Kelvin: \( T(K) = T(^{\circ}C) + 273.15 \)
• From Kelvin back to Celsius: \( T(^{\circ}C) = T(K) - 273.15 \)

Being adept at converting temperatures ensures accuracy in calculations. The Kelvin scale is often used in thermodynamics as it provides an absolute temperature reference starting from absolute zero, aiding in comprehensible and consistent results across various scientific equations.