Question

The following data present the temperatures at which certain vapor pressures are achieved for dichloromethane \(\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)\) and methyl iodide \(\left(\mathrm{CH}_{3} \mathrm{I}\right)\) : $$ \begin{array}{lcccl} \hline \begin{array}{l} \text { Vapor Pressure } \\ \text { (torr): } \end{array} & \mathbf{1 0 . 0} & \mathbf{4 0 . 0} & \mathbf{1 0 0 . 0} & \mathbf{4 0 0 . 0} \\ \hline T \text { for } \mathrm{CH}_{2} \mathrm{Cl}_{2}\left({ }^{\circ} \mathrm{C}\right): & -43.3 & -22.3 & -6.3 & 24.1 \\ T \text { for } \mathrm{CH}_{3} \mathrm{I}\left({ }^{\circ} \mathrm{C}\right): & -45.8 & -24.2 & -7.0 & 25.3 \\ \hline \end{array} $$ (a) Which of the two substances is expected to have th greater dipole-dipole forces? Which is expected to hav the greater London dispersion forces? Based on your a swers, explain why it is difficult to predict which con pound would be more volatile. (b) Which compour would you expect to have the higher boiling poin Check your answer in a reference book such as the \(C R\) Handbook of Chemistry and Physics. (c) The order volatility of these two substances changes as the ten perature is increased. What quantity must be differer for the two substances in order for this phenomenon occur? (d) Substantiate your answer for part (c) drawing an appropriate graph.

Short answer

In summary, dichloromethane has stronger dipole-dipole forces, while methyl iodide has stronger London dispersion forces, making it difficult to predict which compound is more volatile. The boiling points are 39.6 °C for dichloromethane and 42.5 °C for methyl iodide. The change in volatility order as temperature increases is due to the difference in their enthalpies of vaporization. A graph plotting vapor pressure against temperature with different slopes for both substances demonstrates this phenomenon.

Step-by-step solution

Analyze the data

First, take a look at the provided data in the table. For a given vapor pressure, compare the temperatures for dichloromethane and methyl iodide. In general, if vapor pressure increases with decreasing temperature for a substance, it indicates stronger intermolecular forces. Here's the given data: \(T\) for \(\mathrm{CH}_{2}\mathrm{Cl}_{2}: -43.3, -22.3, -6.3, 24.1 (^{\circ}\mathrm{C}) \(T\) for \(\mathrm{CH}_{3}\mathrm{I}: -45.8, -24.2, -7.0, 25.3 (^{\circ}\mathrm{C})

Determine the type of forces

In order to answer (a), compare the number of electrons in each molecule. Dichloromethane has a total of 20 electrons, while methyl iodide has 22 electrons. Since London dispersion forces increase with the number of electrons, methyl iodide is expected to have greater London dispersion forces. For dipole-dipole forces, analyze the polarity of the two substances. Dichloromethane is more polar than methyl iodide because chlorine is more electronegative than iodine. Therefore, dichloromethane is expected to have greater dipole-dipole forces.

Volatility explanation

Based on our answers in Step 2, it is difficult to predict which compound would be more volatile because dichloromethane has stronger dipole-dipole forces while methyl iodide has stronger London dispersion forces. These two opposing factors make it challenging to determine which substance has overall stronger intermolecular forces and hence would be less volatile.

Predict boiling point

To answer (b), look for any correlation between boiling points and the intermolecular forces analyzed in Step 2. In general, substances with stronger intermolecular forces have higher boiling points. Since dichloromethane has stronger dipole-dipole forces and methyl iodide has stronger London dispersion forces, it is difficult to predict which compound has a higher boiling point. Check in a reference book to verify the answer: Dichloromethane has a boiling point of 39.6 °C and methyl iodide has a boiling point of 42.5 °C.

Identify the reason for change in volatility

For part (c), we need to identify the quantity that must be different for the two substances for the change in volatility as temperature increases. Enthalpy of vaporization is the quantity responsible for different temperature dependencies of the substances, influencing their volatility.

Draw a graph

Lastly, to answer (d), draw a graph that substantiates your answer in Step 5. On the x-axis, plot the temperature, and on the y-axis, plot the vapor pressure. Add data points for both substances and connect them with lines. The graph should show two lines with different slopes, indicating different enthalpies of vaporization for the two substances, which explains the change in volatility order as the temperature increases.

Key concepts

Vapor Pressure

Vapor pressure refers to the pressure exerted by the vapor of a liquid when it is in equilibrium with its liquid or solid phase in a closed container. Simply put, it's a measure of a substance's tendency to evaporate. The concept of vapor pressure is crucial because it helps us understand why different substances evaporate at different rates.

In the exercise, we are provided with a table that displays the temperatures at which certain vapor pressures are achieved for dichloromethane and methyl iodide. The data helps us analyze intermolecular forces by showing how each substance's vapor pressure varies with temperature.

• High vapor pressure at a lower temperature indicates weaker intermolecular forces, making the substance more volatile under those conditions.
• If a substance has stronger intermolecular forces, its vapor pressure will be lower at a given temperature because it is less likely to evaporate.

Understanding vapor pressure is key to predicting how substances behave under changing temperatures, which plays into determining their boiling points and volatility.

Boiling Point

The boiling point of a substance is the temperature at which its vapor pressure equals the external pressure surrounding it, usually the atmospheric pressure. When a liquid reaches its boiling point, it changes to vapor as the molecules have enough energy to break free from the liquid's surface.

Intermolecular forces play a significant role in determining the boiling point. Typically, the stronger the intermolecular forces, the higher the boiling point, as more energy is required to separate the molecules.

• Dichloromethane, with a boiling point of 39.6 °C, has greater dipole-dipole forces, suggesting that these forces don't increase the boiling point as much as the London dispersion forces do for methyl iodide.
• Methyl iodide, with a boiling point of 42.5 °C, exhibits stronger London dispersion forces, which contribute to its slightly higher boiling point than dichloromethane.

Boiling point comparisons offer insight into the relative strengths of intermolecular forces within each compound, which is crucial in understanding their physical properties and behaviors.

Dipole-Dipole Forces

Dipole-dipole forces occur between molecules that have permanent dipoles, meaning molecules with regions of partial positive and negative charges. These forces are a type of van der Waals force and are stronger than London dispersion forces but weaker than hydrogen bonding.

Dichloromethane and methyl iodide have different dipole moments because of the differing electronegativities of their atoms.

• In dichloromethane, the presence of two chlorine atoms, which are more electronegative, creates a strong dipole with significant dipole-dipole forces.
• Methyl iodide, on the other hand, is less polar due to iodine’s lower electronegativity, resulting in weaker dipole-dipole forces compared to dichloromethane.

These differences in dipole moments explain why dichloromethane is expected to have greater dipole-dipole forces. Hence, these forces affect the physical properties like boiling point and solubility of the substance.

London Dispersion Forces

London dispersion forces are a type of transient attractive force among all atoms and molecules, whether polar or non-polar. They are the weakest of the van der Waals forces but still significant, especially in larger atoms and molecules. These forces arise from temporary shifts in electron density that create short-lived dipoles.

The magnitude of London dispersion forces is related to the number of electrons in a molecule and the ease with which these electrons can be polarized.

• Methyl iodide, with more electrons than dichloromethane, experiences stronger London dispersion forces. This increased strength is due to the larger number of electrons and the larger atomic size of iodine compared to chlorine.
• In contrast, dichloromethane has fewer electrons, so its London dispersion forces are weaker.

While typically weaker than dipole-dipole forces, London dispersion forces are nonetheless crucial in determining boiling points and solubility, especially for non-polar substances or those with significant electron clouds.