Problem 139 - General Chemistry

Question

Consider the following three compounds: $$\mathrm{CH}_{3} \mathrm{CHO}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$$ a. Describe the types of intermolecular forces that you expect to see in each. Explain how you arrived at these types. b. The heats of vaporization of these compounds are (in no particular order): \(25.8 \mathrm{~kJ} / \mathrm{mol}, 38.6 \mathrm{~kJ} / \mathrm{mol}\), and \(19.0 \mathrm{~kJ} / \mathrm{mol}\). What is the heat of vaporization of \(\mathrm{CH}_{3} \mathrm{CHO} ?\) c. If the normal boiling point of \(\mathrm{CH}_{3} \mathrm{CHO}\) is \(21^{\circ} \mathrm{C}\), what is its vapor pressure at \(15^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

a) CH3CHO: dipole-dipole, London; CH3CH2CH3: London; CH3CH2OH: H-bonding, dipole-dipole, London. b) 25.8 kJ/mol for CH3CHO. c) Vapor pressure at 15°C is 0.734 atm.

Step 1: Identify Intermolecular Forces

First, we need to identify the intermolecular forces present in each compound. - **CH3CHO (Acetaldehyde):** This molecule contains a carbonyl group, thus it primarily exhibits *dipole-dipole interactions* due to the polarization of the carbon-oxygen double bond. Additionally, *London dispersion forces* are also present. - **CH3CH2CH3 (Propane):** Being a non-polar molecule, it mainly experiences *London dispersion forces*. - **CH3CH2OH (Ethanol):** This molecule can form *hydrogen bonds* because of the O-H group, along with *dipole-dipole interactions* and *London dispersion forces*.

Step 2: Link Intermolecular Forces to Heat of Vaporization

The strength of intermolecular forces affects the heat of vaporization—a strong force requires more energy (higher heat of vaporization). - **Ethanol** has hydrogen bonding, making it likely to have the highest heat of vaporization among the three. - **Acetaldehyde** has dipole-dipole interactions, likely giving it a middle heat of vaporization. - **Propane** with only London dispersion forces should have the lowest heat of vaporization.

Step 3: Match Heats of Vaporization with Compounds

The possible heats of vaporization are 25.8 kJ/mol, 38.6 kJ/mol, and 19.0 kJ/mol. - **Ethanol** should correspond with the highest value, 38.6 kJ/mol.- **Propane** should correspond with the lowest value, 19.0 kJ/mol.- **Acetaldehyde** (\(CH_3CHO\), which has intermolecular forces between the strength exhibited by ethanol and propane, should correspond with 25.8 kJ/mol.

Step 4: Calculate Vapor Pressure Using Clausius-Clapeyron Equation

The Clausius-Clapeyron Equation \(\ln\left(\frac{P_1}{P_2}\right) = \frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\) can be used, where \(\Delta H_{vap} = 25.8\, \text{kJ/mol} = 25800\, \text{J/mol}\), \(R = 8.314\, \text{J/mol K}\), \(T_1 = 21^{\circ}C = 294\, \text{K}\), and \(T_2 = 15^{\circ}C = 288\, \text{K}\). \(P_1\) is the vapor pressure at \(21^{\circ}C\), which is the normal boiling point, so \(P_1 = 1\) atm. Substitute these into the equation, \(\ln\left(\frac{1}{P_2}\right) = \frac{25800}{8.314}\left(\frac{1}{288} - \frac{1}{294}\right)\). Calculating this gives \(\ln\left(\frac{1}{P_2}\right) = 0.309\). Therefore, \(P_2 = e^{-0.309} \approx 0.734\) atm.

Key Concepts

Dipole-Dipole Interactions

Dipole-dipole interactions occur between polar molecules where there is a permanent net dipole moment due to uneven electron distribution. These forces arise because the positive pole of one molecule is attracted to the negative pole of another. The strength of dipole-dipole forces depends on the magnitude of the dipole moment and the distance between molecules. In molecules like acetaldehyde (\(\text{CH}_3\text{CHO}\)), the carbon-oxygen double bond is polar. This results in substantial dipole-dipole interactions as the more electronegative oxygen atom pulls electron density away from the carbon, creating a net dipole.
Understanding dipole-dipole interactions helps predict physical properties such as boiling points and solubility. These interactions are stronger than London dispersion forces but weaker than hydrogen bonds, significantly influencing the heat of vaporization in substances with this force as a key player.

London Dispersion Forces

Also known as van der Waals forces, London dispersion forces are transient attractive forces experienced by all molecules, whether polar or nonpolar. They are the result of temporary dipoles formed when electron distribution around the nucleus is uneven for a brief moment.
London dispersion forces increase with molecular size and weight because larger atoms or molecules have electrons that are further from the nucleus and less tightly bound, making them more polarizable. Therefore, a nonpolar molecule like propane (\(\text{CH}_3\text{CH}_2\text{CH}_3\)) relies solely on London dispersion forces to hold its molecules together, leading to a lower boiling point. The weakest of the intermolecular forces, London dispersion forces significantly contribute to the heat of vaporization, albeit less so compared to stronger forces like hydrogen bonds or dipole-dipole interactions.

Hydrogen Bonds

Hydrogen bonds are a special type of dipole-dipole interaction that occur when a hydrogen atom is covalently bonded to a highly electronegative atom such as oxygen, nitrogen, or fluorine. This setup makes the hydrogen positively polarized, enabling it to interact strongly with an electronegative atom on a neighboring molecule.
Ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)) can engage in hydrogen bonding due to its O-H group. The presence of hydrogen bonds markedly increases both the boiling point and heat of vaporization as more energy is required to break these strong interactions. Hydrogen bonds are among the strongest types of intermolecular forces, leading to marked deviations from expected boiling points in compounds capable of such bonding.

Heat of Vaporization

The heat of vaporization refers to the energy needed to convert a mole of a liquid into vapor at constant temperature. It is directly related to the strength of intermolecular forces present in the liquid.

• Strong intermolecular forces result in a high heat of vaporization because more energy is needed to overcome these forces.
• For ethanol, with substantial hydrogen bonding, the heat of vaporization is high (38.6 kJ/mol).
• Acetaldehyde, with moderate dipole-dipole interactions, yields a mid-range value of 25.8 kJ/mol.
• Propane, relying solely on weak London dispersion forces, has the lowest heat of vaporization at 19.0 kJ/mol.

Understanding these forces and their influence on heat of vaporization is crucial for predicting and explaining boiling points, volatility, and energetics on a molecular level.

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation relates the vapor pressure of a substance to its temperature, providing insight into phase changes. It is useful for calculating vapor pressures at different temperatures, assuming the heat of vaporization is constant.
The general equation is:\[\ln\left(\frac{P_1}{P_2}\right) = \frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]where:

• \(P_1, P_2\) are the vapor pressures at temperatures \(T_1, T_2\) respectively,
• \(\Delta H_{vap}\) is the heat of vaporization,
• \(R\) is the universal gas constant (8.314 J/mol K).

This equation is particularly handy when predicting pressure changes with temperature for a compound like acetaldehyde, as seen when calculating its vapor pressure at 15°C, knowing its normal boiling point vapor pressure is 1 atm at 21°C. Approximations and assumptions involved ensure it primarily applies to ideal conditions.