What are the effect on solubility of gas in liquid with decreasing temperature and pressure?

Solubility

The definition of solubility is the maximum quantity of solute that can dissolve in a certain quantity of solvent or quantity of solution at a specified temperature or pressure (in the case of gaseous solutes). In CHM1045 we discussed solubility as a yes or no quality. But the reality is that almost every solute is somewhat soluble in every solvent to some measurable degree.

As stated in the definition, temperature and pressure play an important role in determining the degree to which a solute is soluble.

Let's start with temperature:

For Gases, solubility decreases as temperature increases (duh...you have seen water boil, right?) The physical reason for this is that when most gases dissolve in solution, the process is exothermic. This means that heat is released as the gas dissolves. This is very similar to the reason that vapor pressure increases with temperature. Increased temperature causes an increase in kinetic energy. The higher kinetic energy causes more motion in the gas molecules which break intermolecular bonds and escape from solution. Check out the graph below:

As the temperature increases, the solubility of a gas decreases as shown by the downward trend in the graph.

For solid or liquid solutes:

CASE I: Decrease in solubility with temperature:

If the heat given off in the dissolving process is greater than the heat required to break apart the solid, the net dissolving reaction is exothermic (See the solution process). The addition of more heat (increases temperature) inhibits the dissolving reaction since excess heat is already being produced by the reaction. This situation is not very common where an increase in temperature produces a decrease in solubility. But is the case for sodium sulfate and calcium hydroxide.

CASE II: Increase in solubility with temperature:

If the heat given off in the dissolving reaction is less than the heat required to break apart the solid, the net dissolving reaction is endothermic. The addition of more heat facilitates the dissolving reaction by providing energy to break bonds in the solid. This is the most common situation where an increase in temperature produces an increase in solubility for solids.

The use of first-aid instant cold packs is an application of this solubility principle. A salt such as ammonium nitrate is dissolved in water after a sharp blow breaks the containers for each. The dissolving reaction is endothermic - requires heat. Therefore the heat is drawn from the surroundings, the pack feels cold.

The effect of temperature on solubility can be explained on the basis of Le Chatelier's Principle. Le Chatelier's Principle states that if a stress (for example, heat, pressure, concentration of one reactant) is applied to an equilibrium, the system will adjust, if possible, to minimize the effect of the stress. This principle is of value in predicting how much a system will respond to a change in external conditions. Consider the case where the solubility process is endothermic (heat added). An increase in temperature puts a stress on the equilibrium condition and causes it to shift to the right. The stress is relieved because the dissolving process consumes some of the heat. Therefore, the solubility (concentration) increases with an increase in temperature. If the process is exothermic (heat given off). A temperature rise will decrease the solubility by shifting the equilibrium to the left.

Now let's look at pressure:

Solids and liquids show almost no change in solubility with changes in pressure. But gases are very dependent on the pressure of the system. Gases dissolve in liquids to form solutions. This dissolution is an equilibrium process for which an equilibrium constant can be written. For example, the equilibrium between oxygen gas and dissolved oxygen in water is O2(aq) <=> O2(g). The equilibrium constant for this equilibrium is K = p(O2)/c(O2). The form of the equilibrium constant shows that the concentration of a solute gas in a solution is directly proportional to the partial pressure of that gas above the solution. This statement, known as Henry's law, was first proposed in 1800 by J.W. Henry as an empirical law well before the development of our modern ideas of chemical equilibrium.

Henry's Law:

Sg stands for the gas solubility, kH is the Henry's Law constant and Pg is the partial pressure of the gaseous solute.

Table: Molar Henry's Law Constants for Aqueous Solutions at 25oC

Gas	Constant	Constant
	(Pa/(mol/dm3))	(atm/(mol/dm3))
He	282.7e6	2865
O2	74.68e6	756.7
N2	155.0e6	1600
H2	121.2e6	1228
CO2	2.937e6	29.76
NH3	5.69e6	56.9

The inverse of the Henry's law constant, multiplied by the partial pressure of the gas above the solution, is the molar solubility of the gas. Thus oxygen at one atmosphere would have a molar solubility of (1/756.7)mol/dm3 or 1.32 mmol/dm3. Values in this table are calculated from tables of molar thermodynamic properties of pure substances and aqueous solutes

Summary of Factors Affecting Solubility

Normally, solutes become more soluble in a given solvent at higher temperatures. One way to predict that trend is to use Le Chatelier's principle. Because DHsoln is positive for most solutions, the solution formation reaction is usually endothermic. Therefore, when the temperature is increased, the solubility of the solute should also increase. However, there are solutes that do not follow the normal trend of increasing solubility with increasing temperature. One class of solutes that becomes less soluble with increasing temperature is the gasses. Nearly every gas becomes less soluble with increasing temperature.

Another property of gaseous solutes in summarized by Henry's law which predicts that gasses become more soluble when their pressures above a liquid solution are increased. That property of gaseous solutes can be rationalized by using Le Chatelier's principle. Imagine that you have a glass of water inside of a sealed container filler with nitrogen gas. If the size of that container were suddenly halved, the pressure of nitrogen would suddenly double. To decrease the pressure of nitrogen above the solution (as is required by Le Chatelier's principle), more nitrogen gas becomes dissolved in the glass of water.

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As with all processes under constant pressure and constant temperature, dissolving a solution into solution will occur only if \(\Delta G_{total} < 0\).

\[\Delta G_{sol} = \Delta H_{sol} - T\Delta S_{sol} < 0\label{eq1}\]

For dissolving solids in liquids, \(\Delta S_{sol} > 0\), but for dissolving gases solutes, the entropy of solution is negative (\(\Delta S _{sol} < 0\)) since the entropy of the gas phase solute is appreciably greater than the entropy of that solute in solution. Consequently, the only way that \(\Delta G_{sol}<0\) for a dissolving a gas in solution is if the solution process is exothermic (i.e., \(\Delta H_{sol}<0\)). This occurs due to the enthalpy differences from making and breaking intermolecular interactions in the solvent and solution. There are three basic steps involved in dissolving a solute from a condensed state (or a non-ideal gas) into a solution each with a corresponding enthalpy change.

Dissolution can be viewed as occurring in three steps:

Breaking solute-solute attractions (endothermic), i.e., lattice energy in salts (\(\Delta H_{\text{solute-solute}}>0\)).
Breaking solvent-solvent attractions (endothermic), i.e., hydrogen bonding and dipole-dipole interactions in water (\(\Delta H_{\text{solvent-solvent}}>0\).
Forming solvent-solute attractions (exothermic), i.e., solvation energy (\(\Delta H_{\text{solute-solvent}}<0\)).

The enthalpy of solution \(\Delta H_{sol}\) is the sum of these three individual steps.

\[ \Delta H_{sol} = \Delta H_{\text{solute-solute}} + \Delta H_{\text{solvent-solvent}} + \Delta H_{\text{solute-solvent}} \label{eq2}\]

For solids solutes, \(\Delta H_{\text{solute-solute}}\) is just the lattice energy of the solute, but for gases that follow the ideal gas equation of state, the enthalpy change associated with Step 1 is zero.

\[\Delta H_{\text{solute-solute}} (gas) =0 \]

This is because are no intermolecular interactions exist in ideal gases (van der Waals gases will differ as expected).

In polar solvents like water, \(\Delta H_{\text{solute-solvent}} > \Delta H_{\text{solvent-solvent}}\), so the dissolution of most gases is exothermic (i.e., \(\Delta H_{sol} <0\)). Hence, when a gas dissolves in a liquid solvent, thermal energy is released which warms both the system (the solution) and the surroundings.

\[\ce{ solute (gas) + water (l) \rightleftharpoons solute (aq) + water (aq)} + \Delta \label{eq4}\]

where \(Delta\) is thermal enegy. Consequently, the solubility of a gas is dependent on temperature (Figure \(\PageIndex{1}\)). The solubility of gases in liquids decreases with increasing temperature. Conversely, adding heat to the solution provides thermal energy that overcomes the attractive forces between the gas and the solvent molecules, thereby decreasing the solubility of the gas; pushes the reaction in Equation \ref{eq4} to the left. The thermodynamic perspective is that at elevated temperatures, the negative entropy term in Equation \ref{eq1} will dominate the enthalpic term that is driving the dissolution process and make \(ΔG_{soln}\) less negative and hence less spontaneous.

Figure \(\PageIndex{1}\): The solubilities of these gases in water decrease as the temperature increases. All solubilities were measured with a constant pressure of 101.3 kPa (1 atm) of gas above the solutions. (CC BY; OpenStax).

Determine the solubility of \(\ce{N2(g)}\) when combined with \(\ce{H2O}\) at 0.0345 °C the pressure of \(\ce{N2}\) is 1.00 atm, and its solubility is 21.0 ml at STP.

Solution

Begin by determining the molarity (solubility) of \(\ce{N2(g)}\) at 0 °C and STP.

At STP 1 mol=22L

\[\begin{align*} \text{Molarity of } \ce{N2} &= 21\,ml \left( \dfrac{1L}{1000\,ml}\right) \\[4pt] &=0.021\,L \,\ce{N2} \\[4pt] &= 0.021\,L \,\ce{N2} \left(\dfrac{1\,mol}{22\,L}\right)\\[4pt] &=\dfrac{0 .000954\,mol}{1\, L} \\[4pt] &= 9.5 \times 10^{-4}\, M\, \ce{N2} \end{align*}\]

Now that the molarity of \(\ce{N2}\). \(C\) has been attained, the Henry's law constant, \(k\), can be evaluated.

Henry's Law:

\[C = k_H P_{gas} \nonumber\]

where \(C\) is solubility, \(k_H\) is Henry's constant, and \(P_{gas}\) is the partial pressure of the gas being considered.

Rearranging the formula to solve for \(k_H\)

\[\begin{align} k_H&= \dfrac{C}{P_{gas}} \\ &= 9.5 \times 10^{-4}\, M \,N_2/ 1\,atm \nonumber \end{align} \nonumber\]

Now substitute k and the partial pressure of \(\ce{N2}\) into Henry's law:

\[\begin{align} C&= (9.5 \times 10^{-4}\,M\, N_2)(0.0345) \nonumber \\[4pt] &= 3.29 \times 10^{-5}\, M\, N_2 \nonumber \end{align} \nonumber\]

A fish kill can occur with rapid fluctuations in temperature or sustained high temperatures. Generally, cooler water has the potential to hold more oxygen, so a period of sustained high temperatures can lead to decreased dissolved oxygen in a body of water. A short period of hot weather can increase temperatures in the surface layer of water, as the warmer water tends to stay near the surface and be further heated by the air. In this case, the top warmer layer may have more oxygen than the lower, cooler layers because it has constant access to atmospheric oxygen.

There are many causes of fish kill, but oxygen depletion is the most common cause. (Public Domain; United States Fish and Wildlife Service)

Dissolving gases in non-polar organic solvents is a different situation than in polar solvents like water discussed above. For non-polar solvents, both the solvent-solvent interactions (\(\Delta H_{\text{solvent-solvent}}\)) and the solvation enthalpies (\(\Delta H_{\text{solute-solvent}}\)) are considerably weaker than in polar liquids like water due to the absence of strong dipole-dipole intermolecular interactions (or hydrogen bonding). In many cases, the enthalpy needed to break solvent-solvent interactions is comparable to the enthalpy released in making solvent-gas interactions.

\[\Delta H_{\text{solvent-solvent}} \approx \Delta H_{\text{solute-solvent}}\]

which means the \(\Delta H_{sol} \approx 0\) via Equation \ref{eq2}. In some solvent-solute combination

\[\Delta H_{\text{solvent-solvent}} > \Delta H_{\text{solute-solvent}}\]

so that \(\Delta H_{sol} > 0\). In this case, we can write the solution reactions thusly

\[ \Delta + \ce{ solute (gas) + solvent (solvent) \rightleftharpoons solute (sol) + solvent (sol)} \label{eq5}\]

where \(Delta\) is thermal energy. In these cases, gases dissolved in organic solvents can actually be more soluble at higher temperatures. A Le Chatelier perspective, like that used above for water, can help understand why. Increasing the temperature will shift in the equilibrium to favor dissolution (i.e, shift Equation \ref{eq5} to the right). As a result, the solubilities of gases in organic solvents often increase with increasing temperature (Table \(\PageIndex{1}\)) in contrast to the trend observed in water (Figure \(\PageIndex{1}\)).

Table \(\PageIndex{1}\): Mole Fraction Solubility of Hydrogen gas at a Hydrogen Partial Pressure of 101.325 kPa in selects solvents at select temperatures.

Temperature (ºC)	\(\ce{H2}\) in n-hexane	\(\ce{H2}\) in Toluene	\(\ce{H2}\) in Acetonitrile	\(\ce{O_2}\) in Water	\(\ce{O_2}\) in Ethanol	\(\ce{O_2}\) in Decane
25	\(7.13 \times 10^4\)	\(3.15 \times 10^4\)	\(1.78 \times 10^4\)	\(0.249 \times 10^4\)	\(5.59 \times 10^4\)	\(21.78\times 10^4\)
50	\(8.20 \times 10^4\)	\(3.75 \times 10^4\)	\(2.16 \times 10^4\)	\(0.196 \times 10^4\)	\( 5.598 \times 10^4\)	\(21.76\times 10^4\)
100	\(10.78 \times 10^4\)	\(5.05 \times 10^4\)	\(3.02 \times 10^4\)	\(0.086 \times 10^4\)	\(5.695 \times 10^4\)	-

Based off of the three steps outlined above for dissolving a solute a liquid, explain why the solubility of hydrogen is smaller in acetonitrile than in toluene and that is less than in n-hexane.

Figure \(\PageIndex{2}\): Solvents used in Table \(\PageIndex{1}\) Answer

Acetinitrile is a polar organic solvent and has stronger intermolecular bonds than toluene, which is a weakly polar solvent and has stronger intermolecular interactions than n-hexane.

A system of cyclopentane and oxygen gas are at equilibrium with an enthalpy of -1234 kJ; predict whether the solubility of oxygen gas will be greater when heat is added to the system or when a temperature decrease occurs.

Answer

Cyclopentane is an organic solvent. Oxygen gas and cyclopentane in a system at equilibrium, where the entropy is negative, will be be displaced from equilibrium when any type of temperature change is inflicted on the system. Because the dissolution of the gas is endothermic, more heat increases the solubility of the gas.

References

Petrucci, et al. General Chemistry Principles & Modern Applications. 9th ed. Upper Saddle River, NJ: Pearson Prentice Hall, 2007
Garde,Shekhar, Garcia, Angel, Pratt, Lawrence, Hummer, Gerhard. "Temperature Dependence of the Non-polar Solubility of Gases in Water". Biophysical Chemistry. Volume 78. Issues 1-2. 1999.