Problem 22 - Chemistry and Chemical Reactivity

Question

Vapor pressure data are given here for octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\). $$\begin{array}{cc} \text { Temperature }\left(^{\circ} \mathrm{C}\right) & \text { Vapor Pressure }(\mathrm{mm} \mathrm{Hg}) \\ \hline 25 & 13.6 \\ 50 . & 45.3 \\ 75 & 127.2 \\ 100 . & 310.8 \\ \hline \end{array}$$ Use the Clausius-Clapeyron equation to calculate the molar enthalpy of vaporization of octane and its normal boiling point.

Step-by-Step Solution

Answer

The molar enthalpy of vaporization is 38,674 J/mol and the normal boiling point is 125°C.

Step 1: Understand the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation relates the vapor pressure and temperature of a substance and is given by: \(-\frac{d( ext{ln} P)}{dT} = \frac{\Delta H_{vap}}{R T^2}\), where \( \Delta H_{vap} \) is the molar enthalpy of vaporization, \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(P\) is the vapor pressure.

Step 2: Convert Temperatures to Kelvin

To use the Clausius-Clapeyron equation, temperatures must be in Kelvin. Convert the temperatures: \(T_1 = 25 + 273.15 = 298.15\, K\), \(T_2 = 75 + 273.15 = 348.15\, K\).

Step 3: Use the Clausius-Clapeyron Equation

Use the pressures corresponding to the temperatures converted to Kelvin. The equation can be written in a linear form: \(\ln \left(\frac{P_2}{P_1}\right) = \frac{\Delta H_{vap}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\). Use \(P_1 = 13.6\, mmHg\) at \(T_1 = 298.15\, K\) and \(P_2 = 127.2\, mmHg\) at \(T_2 = 348.15\, K\).

Step 4: Calculate the Slope

Using the values from Step 3, calculate the slope in the equation:\[\ln \left(\frac{127.2}{13.6}\right) = \frac{\Delta H_{vap}}{8.314} \left(\frac{1}{298.15} - \frac{1}{348.15}\right)\]Calculate \(\ln(9.3529)\) and \(\left(\frac{1}{298.15} - \frac{1}{348.15}\right)\).

Step 5: Solve for the Enthalpy of Vaporization

Compute:\[\ln(9.3529) = 2.234\] \[\frac{1}{T_1} - \frac{1}{T_2} = -0.0004812\] Rearrange the equation to solve for \(\Delta H_{vap}\):\[\Delta H_{vap} = \frac{2.234 \times 8.314}{-0.0004812} \approx 38,674\, J/mol\] or \(38.674\, kJ/mol\).

Step 6: Determine the Normal Boiling Point

The normal boiling point occurs when vapor pressure equals 760 mmHg. Solve for \(T\) using the Clausius-Clapeyron equation: \[\ln\left(\frac{760}{310.8}\right) = \frac{38,674}{8.314}\left(\frac{1}{373.15} - \frac{1}{T_b}\right)\] Calculate to find \(T_b\), the temperature at which \(P = 760\, mmHg\):\[T_b \approx 398.5\, K\], which corresponds to 125.35°C.

Key Concepts

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid at a given temperature. Understanding this concept is key to grasping how and when a liquid turns into vapor. The higher the vapor pressure, the more volatile the substance. This means it evaporates more quickly.

Factors influencing vapor pressure include:

• Temperature: As temperature increases, so does vapor pressure. More molecules have enough energy to escape from the liquid.
• Intermolecular Forces: Stronger forces lead to lower vapor pressure as molecules require more energy to vaporize.

In the exercise, octane's vapor pressures at various temperatures help calculate its molar enthalpy of vaporization using the Clausius-Clapeyron equation.

Molar Enthalpy of Vaporization

The molar enthalpy of vaporization (\(\Delta H_{vap}\)) is the energy needed to convert one mole of a liquid into vapor at constant temperature and pressure. It reflects how much energy is required to overcome intermolecular forces to turn a liquid into gas.

This value is expressed in joules per mole (J/mol) or kilojoules per mole (kJ/mol). In the exercise, using the Clausius-Clapeyron equation, the calculated \(\Delta H_{vap}\) for octane was approximately 38.674 kJ/mol, indicating the energy required for vaporization.

It's important to know:

• Higher \(\Delta H_{vap}\) means stronger intermolecular forces.
• This quantity is specific to each substance, which helps explain differences in boiling points across different liquids.

Kelvin Temperature Conversion

Temperature plays a crucial role in thermodynamic calculations, often requiring conversion to the Kelvin scale, the SI unit for temperature. Kelvin eliminates negative numbers, which simplifies math in calculations. To convert Celsius to Kelvin, add 273.15.

For example:

• 25°C converts to 298.15 K
• 75°C converts to 348.15 K

This conversion is essential for the Clausius-Clapeyron equation, ensuring temperatures are correctly used in the formula. Kelvin is favored because it starts at absolute zero, providing a direct measure of thermal energy.

Ideal Gas Constant

The ideal gas constant (\(R\)) is a fundamental constant in chemistry, appearing in equations like the Clausius-Clapeyron and the ideal gas law. It serves as a bridge between macroscopic and microscopic properties of gases.

The value of the ideal gas constant is 8.314 J/(mol·K), and it supports calculations involving:

• Temperature changes
• Pressure measurements

In the context of the exercise, \(R\) helps relate the rate of change in vapor pressure with temperature changes, assisting in finding \(\Delta H_{vap}\) for octane. Understanding \(R\) is crucial for mastering concepts in thermodynamics and physical chemistry.