Problem 111 - Chemistry The Central Science In Si Units

Question

The following table presents the solubilities of several gases in water at \(25^{\circ} \mathrm{C}\) under a total pressure of gas and water vapor of \(101.3 \mathrm{kPa}\). (a) What volume of \(\mathrm{CH}_{4}(g)\) under standard conditions of temperature and pressure is contained in \(4.0 \mathrm{~L}\) of a saturated solution at \(25^{\circ} \mathrm{C} ?\) (b) The solubilities (in water) of the hydrocarbons are as follows: methane \(<\) ethane \(<\) ethylene. Is this because ethylene is the most polar molecule? (c) What intermolecular interactions can these hydrocarbons have with water? (d) Draw the Lewis dot structures for the three hydrocarbons. Which of these hydrocarbons possess \(\pi\) bonds? Based on their solubilities, would you say \(\pi\) bonds are more or less polarizable than \(\sigma\) bonds? (e) Explain why \(\mathrm{NO}\) is more soluble in water than either \(\mathrm{N}_{2}\) or \(\mathrm{O}_{2}\). (f) \(\mathrm{H}_{2} \mathrm{~S}\) is more water-soluble than almost all the other gases in table. What intermolecular forces is \(\mathrm{H}_{2} \mathrm{~S}\) likely to have with water? \((\mathbf{g}) \mathrm{SO}_{2}\) is by far the most water-soluble gas in table. What intermolecular forces is \(\mathrm{SO}_{2}\) likely to have with water? $$ \begin{array}{lc} \hline \text { Gas } & \text { Solubility (mM) } \\ \hline \mathrm{CH}_{4} \text { (methane) } & 1.3 \\ \mathrm{C}_{2} \mathrm{H}_{6} \text { (ethane) } & 1.8 \\ \mathrm{C}_{2} \mathrm{H}_{4} \text { (ethylene) } & 4.7 \\ \mathrm{~N}_{2} & 0.6 \\ \mathrm{O}_{2} & 1.2 \\ \mathrm{NO} & 1.9 \\ \mathrm{H}_{2} \mathrm{~S} & 99 \\ \mathrm{SO}_{2} & 1476 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

(a) 0.12 L of CH4 at STP; (b) Not solely due to polarity; (c) Dispersion forces; (d) Ethylene has π bonds, more polarizable; (e) Polar nature of NO; (f) Dipole interactions and hydrogen bonding; (g) Strong dipole interactions.

Step 1: Calculate Moles of Methane in Solution

First, we calculate the moles of methane dissolved in the solution. Since the solubility of \( \mathrm{CH}_{4} \) is 1.3 mM, this means there are 1.3 mmol per liter. For 4.0 L of solution, the moles of methane are given by:\[ \text{Moles of } \mathrm{CH}_{4} = 1.3 \times 10^{-3} \text{ mol/L} \times 4.0 \text{ L} = 0.0052 \text{ mol}\]

Step 2: Convert Moles to Volume under STP Conditions

At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 L. Therefore, the volume of 0.0052 moles of methane gas is:\[ \text{Volume } = 0.0052 \text{ mol} \times 22.4 \text{ L/mol} = 0.11648 \text{ L} \approx 0.12 \text{ L}\]

Step 3: Analyze Solubility Based on Molecular Structure

Ethylene is more soluble than ethane and methane, indicating that polarity may indeed play a role, but it's also due to the fact it has unsaturation, which likely enhances its interaction with water. Thus, ethylene's higher solubility is not solely due to its polarity.

Step 4: Discuss Hydrocarbon-Water Interactions

These hydrocarbons, being non-polar, primarily interact with water through weak van der Waals (dispersion) forces. Methane and ethane have tetrahedral geometries, whereas ethylene being planar might have more strategic orientations for interaction.

Step 5: Draw Lewis Structures and Analyze Bond Types

### Lewis Structures:- Methane (\( \mathrm{CH}_{4} \)): \(\text{Tetrahedral with only } \sigma \text{ bonds.}\).- Ethane (\( \mathrm{C}_{2}\mathrm{H}_{6} \)): \(\text{Like two linked \( \mathrm{CH}_{3}\) units, only } \sigma \text{ bonds.}\)- Ethylene (\( \mathrm{C}_{2}\mathrm{H}_{4} \)): \(\text{Contains a C=C } \pi \text{ bond.}\)Ethylene possesses \(\pi\) bonds. In general, \(\pi\) bonds are more polarizable than \(\sigma\) bonds, which increases ethylene's solubility.

Step 6: Explain Enhanced Solubility of NO

NO is more soluble than \(\mathrm{N}_2\) or \(\mathrm{O}_2\) due to its polar nature and presence of an unpaired electron, enabling better interaction with polar water molecules.

Step 7: Discuss Intermolecular Forces of H2S with Water

\(\mathrm{H}_{2}\mathrm{S}\) can engage in dipole-dipole interactions and hydrogen bonding with water due to its bent molecular shape and sulfur being significantly electronegative, providing a polar interaction possibility.

Step 8: Discuss Intermolecular Forces of SO2 with Water

\(\mathrm{SO}_{2}\) is polar with a significant dipole moment and can engage in strong dipole-dipole interactions with water, as well as potential hydrogen bonding due to the lone pairs on sulfur.

Key Concepts

Intermolecular Forces

When discussing gas solubility in water, it is essential to consider the various types of intermolecular forces that come into play. Intermolecular forces are the forces of attraction between molecules and they significantly influence how gases dissolve in water.

There are several types of intermolecular forces:

• London Dispersion Forces: These are weak forces resulting from temporary shifts in electron density within a molecule. All molecules, regardless of their polarity, experience these forces.
• Dipole-Dipole Interactions: These occur between molecules with permanent dipole moments, meaning that one part of the molecule is partially negative and another part is partially positive. Water, being a highly polar molecule, strongly engages in dipole-dipole interactions.
• Hydrogen Bonds: A special type of dipole-dipole interaction, hydrogen bonds occur when hydrogen is directly bonded to a highly electronegative atom, such as oxygen, nitrogen, or fluorine. These bonds are stronger than regular dipole-dipole interactions and significantly enhance solubility.

For gases such as SO₂ and H₂S, the presence of dipole moments and ability to form hydrogen bonds with water make them much more soluble than gases like methane, which primarily experience dispersion forces.

Lewis Structures

Lewis structures are a vital tool for visualizing the valence electron configurations of molecules. They help us understand bonding patterns and predict molecular geometry, which is pivotal in determining how molecules interact with one another.

For the hydrocarbons mentioned:

• Methane (C 2 4): A simple Lewis structure with a central carbon atom bonded to four hydrogen atoms, forming a perfect tetrahedral geometry with only 1 bonds.
• Ethane (C 2 3): Similar to methane but with two carbon atoms bonded by a single 1 bond, making each carbon appear as a C 1 unit linked to two hydrogens.
• Ethylene (C 2 4): Features a double bond between the two carbon atoms, introducing a 3 bond in addition to 1 bonds. This double bond makes the molecule planar, allowing more strategic positioning for interactions with other molecules.

The presence of 3 bonds in ethylene makes it more polarizable and thus more soluble in water compared to methane and ethane.

Polarizability

Polarizability refers to the ability of a molecule's electron cloud to be distorted by external electric fields. Highly polarizable molecules have electron clouds that are easily distorted, allowing them to induce more temporary dipoles.

Regarding hydrocarbons, 3 bonds are generally more polarizable than 1 bonds because of the electron density above and below the 3 bond plane. Ethylene, with its 3 bond, can create inducible dipoles more readily. This increased polarizability aids in its higher solubility in water.

Polarizability enhances the London dispersion forces experienced by molecules. So, when comparing the solubilities of methane, ethane, and ethylene, the greater polarizability due to the presence of a 3 bond in ethylene results in its higher solubility in water. This is not just due to increased interaction with water's permanent dipoles but also because of stronger induced dipole interactions.

Standard Temperature and Pressure (STP)

Understanding standard temperature and pressure (STP) is crucial for carrying out calculations involving gases. STP is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm (101.3 kPa). At these conditions, one mole of any ideal gas occupies 22.4 liters, which is a figure commonly used in gas law calculations.

When calculating the volume of gases at STP, such as methane in the exercise, this 22.4 L/mol conversion factor is used. For instance, if you determine that 0.0052 moles of methane are dissolved, then at STP, these moles will occupy approximately 0.12 L of space. This standardized benchmark is critical as it allows for the comparison of gas volumes under consistent conditions.

It is important to note that real gases may deviate from this volume under actual conditions due to intermolecular forces or non-ideal behavior. However, STP offers a simplified approach to understanding and comparing gas volumes in a variety of chemical contexts.