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 1 atm. (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) Explain the variation in solubility among the hydrocarbons listed (the first three compounds), based on their molecular structures and intermolecular forces. (c) Compare the solubilities of ${\mathrm{O}}_2,{\mathrm{N}}_2$, and $\mathrm{NO},$ and account for the variations based on molecular structures and intermolecular forces. (d) Account for the much larger values observed for ${\mathrm{H}}_2\mathrm{S}$ and ${\mathrm{SO}}_2$ as compared with the other gases listed. (e) Find several pairs of substances with the same or nearly the same molecular masses (for example, ${\mathrm{C}}_2{\mathrm{H}}_4$ and ${\mathrm{N}}_2$ ), and use intermolecular interactions to explain the differences in their solubilities. $$\overline{\begin{array}{lc}\text{ Gas }& \text{ Solubility }(\mathrm{m}M)\\ {\mathrm{CH}}_4(\text{ methane })& 1.3\\ {\mathrm{C}}_2{\mathrm{H}}_6\text{ (ethane) }& 1.8\\ {\mathrm{C}}_2{\mathrm{H}}_4\text{ (ethylene) }& 4.7\\ {\mathrm{N}}_2& 0.6\\ {\mathrm{O}}_2& 1.2\\ \mathrm{NO}& 1.9\\ {\mathrm{H}}_2\mathrm{S}& 99\\ {\mathrm{SO}}_2& 1476\end{array}}$$

Short answer

a) The volume of CH4(g) contained in 4.0 L of a saturated solution at 25°C is approximately 0.116 L. b) The solubility variation among hydrocarbons is due to the strength of London dispersion forces which increases with larger molecular size, thus CH4 < C2H6 < C2H4 in terms of solubility. c) The solubilities of O2, N2, and NO are influenced by their intermolecular forces, with N2 < O2 < NO in terms of solubility due to a combination of London dispersion forces and dipole-dipole interactions. d) H2S and SO2 exhibit higher solubility values because they can form hydrogen bonds with water molecules, resulting in a stronger interaction compared to the other gases listed. e) An example of a pair with similar molecular masses is C2H4 and N2. C2H4 is more soluble than N2 due to stronger London dispersion forces, which are influenced by its larger electron cloud and more complex molecular structure compared to N2.

Step-by-step solution

a) Calculating the volume of CH4(g)

Firstly, we are given the solubility of CH4(g) as 1.3 mM, which means there are 1.3 millimoles of CH4 per liter of water at 25°C and 1 atm. Since we are given a 4.0 L solution, the number of moles of CH4 can be calculated by multiplying the solubility by the volume of the solution: Number of moles of CH4 = solubility × volume of solution = 1.3 mM × 4.0 L = 5.2 mmol Now, in order to find the volume of CH4(g) under standard conditions of temperature and pressure, we will use the molar volume of an ideal gas at standard conditions, which is 22.4 L/mol. Volume of CH4(g) = (5.2 mmol / 1000) × 22.4 L/mol = 0.11648 L ≈ 0.116 L

b) Solubility variation in hydrocarbons

Methane (CH4), ethane (C2H6), and ethylene (C2H4) are all hydrocarbons. The solubility of these compounds in water primarily depends on London dispersion forces, which are relatively weak intermolecular interactions based on temporary variations in electron density. The strength of these forces increases as the size of the molecules increase, leading to greater solubilities. Larger molecules have more electrons and thus greater London dispersion forces, making them more soluble in water. Therefore, CH4 < C2H6 < C2H4 in terms of solubility.

c) Solubility comparison of O2, N2, and NO

The solubilities of O2, N2, and NO are primarily influenced by the nature and strength of their intermolecular forces. O2 and N2 are both diatomic molecules and have similar non-polar structures, resulting in weak London dispersion forces. However, O2 is more soluble than N2, due to O2's larger electron cloud, leading to stronger dispersion forces. NO has a polar structure, as nitrogen is more electronegative than oxygen. This results in stronger intermolecular interactions (both London dispersion forces and dipole-dipole interactions) between NO and water molecules, making it more soluble than O2 and N2. Therefore, N2 < O2 < NO in terms of solubility.

d) Larger values for H2S and SO2

H2S and SO2 exhibit much higher solubility values compared to the other gases listed because they can form hydrogen bonds with water molecules. This strong intermolecular interaction allows these two molecules to dissolve in water to a much higher extent than the other gases, which only have London dispersion forces and, in some cases, dipole-dipole interactions.

e) Pairs of substances with similar molecular masses

One example of a pair with similar molecular masses is C2H4 (ethylene) and N2 (nitrogen). Although they have similar molecular masses, their solubilities are quite different, with C2H4 being more soluble than N2. This difference can be explained by the stronger London dispersion forces in C2H4, due to its larger electron cloud and more complex molecular structure compared to N2.

Key concepts

Molar Volume of an Ideal Gas

Understanding the concept of molar volume is crucial when dealing with gases. The molar volume of an ideal gas refers to the volume one mole of a gas occupies at defined conditions of temperature and pressure. For instance, at standard temperature and pressure (STP, which is 0 degrees Celsius and 1 atmosphere), the molar volume of any ideal gas is 22.4 liters per mole. This value is derived from the ideal gas law equation, which relates pressure, volume, temperature, and the number of moles of a gas.

In the context of the textbook solution, calculating the volume of methane gas (CH_{4}) starts with the known quantity of moles, obtained from its solubility, and then uses the standard molar volume to find the equivalent volume of the gas at STP conditions.

London Dispersion Forces

London dispersion forces are a type of van der Waals force, which are the weakest of intermolecular forces. Despite their relatively low strength, they are important for nonpolar molecules and are present in all molecular interactions to some degree. These forces arise due to the momentary imbalances in electron distribution within molecules, creating temporary dipoles that induce dipoles in adjacent molecules. As a result, a fleeting attraction is generated.

In the solubility discussion, hydrocarbons such as methane, ethane, and ethylene rely on London dispersion forces to dissolve in water. The larger the hydrocarbon molecules, the more electrons they possess, and the stronger the London dispersion forces are, increasing their solubility in water.

Intermolecular Forces

Intermolecular forces are the forces of attraction or repulsion between neighboring particles (molecules or atoms). They can be dipole-dipole interactions in polar molecules, hydrogen bonds in molecules with N-H, O-H, or F-H groups, and the aforementioned London dispersion forces in nonpolar molecules.

These forces are essential in explaining physical properties like boiling points, melting points, and the solubility of different substances. For example, oxygen (O_{2}), and nitrogen (N_{2}) primarily exhibit London dispersion forces, affecting their solubility in water. NO, however, with its slightly polar character, displays both London dispersion and dipole-dipole interactions, accounting for its higher solubility compared to O_{2} and N_{2}.

Hydrocarbon Solubility

The solubility of hydrocarbons in water largely depends on their molecular size, shape, and the intermolecular forces at play. Since most hydrocarbons are nonpolar, their solubility in polar solvents like water is generally low. However, as the hydrocarbon chain length and complexity increase, so does their solubility due to enhanced London dispersion forces.

Within the step-by-step solution presented, this trend is shown where ethylene (C_{2}H_{4}), being the largest of the three hydrocarbons listed, exhibits the highest solubility in water due to its greater London dispersion forces.

Hydrogen Bonds

Hydrogen bonding is one of the most significant intermolecular forces and occurs when a hydrogen atom is attached to a highly electronegative atom such as nitrogen, oxygen, or fluorine and is in the vicinity of another electronegative atom with a lone pair of electrons.

Hydrogen bonds are much stronger than London dispersion forces or dipole-dipole interactions, leading to higher solubility for substances capable of forming hydrogen bonds with water. In the example provided, hydrogen sulfide (H_{2}S) and sulfur dioxide (SO_{2}) show much higher solubility compared to nonpolar gases, as they are able to form hydrogen bonds with water molecules, drastically differing in their interaction with the solvent compared to other gases.