# Temperature Effects on the Solubility of Gases

[ "article:topic", "enthalpy of solution", "Solubility of gases", "authorname:delmar", "entropy of solvation", "enthalpy of solvation", "entropy of solution", "showtoc:no" ]

For a gas dissolving in a liquid, the entropy of solution is negative ($$\Delta S _{sol} < 0$$), since the solute in the gas phase will have higher entropy than when dissolved in the solvent. As with all processes under constant pressure and constant temperature, the solution will occur only if $$\Delta G < 0$$. Hence, the only way that Gibbs energy of solution will be negative, i.e,

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

is if the solution process is exothermic, i.e, $$\Delta H_{sol}<0$$. In contrast to dissolving solids in liquids, where $$\Delta_{sol} > 0$$, dissolving a gas is not entropically driven and occurs due to the enthalpy differences from making and breaking intermolecular interactions in the solvent and solution. There are three basic steps involves in dissolving a solute of any state into a solution each with a corresponding enthalpy change.

Steps involved in Dissolving of a solute in a solvent

Dissolution can be viewed as occurring in three steps:

1. Breaking solute-solute attractions (endothermic), i.e., lattice energy in salts ($$\Delta H_1>0$$)
2. Breaking solvent-solvent attractions (endothermic), i.e., hydrogen bonding in water ($$\Delta H_2>0$$).
3. Forming solvent-solute attractions (exothermic), in solvation ($$\Delta H_3<0$$).

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

$\Delta H_{sol} = \Delta H_1 + \Delta H_2 + \Delta H_3$

### Solubility of Gases in Water

For gases that follow the ideal gas equation of state, the enthalpy change associated with Step 1 is $$\Delta H_1 =0$$ (since the lattice energy is zero for gaseous solute). Since $$\Delta H_3 > \Delta H_2$$ in polar solvents like water, the dissolution of most gases is exothermic. That is, when a gas dissolves in a liquid solvent, energy is released as heat, warming both the system (i.e. the solution) and the surroundings.

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

Consequently, the solubility of a gas is dependent on temperature (Figure $$\PageIndex{1}$$). The solubility of gases in liquids decreases with increasing temperature. Attractive intermolecular interactions in the gas phase are essentially zero for most substances. When a gas dissolves, it does so because its molecules interact with solvent molecules. Because heat is released when these new attractive interactions form, dissolving most gases in liquids is an exothermic process ($$ΔH_{soln} < 0$$). 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.

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. Image used with permission (CC BY-SA OpenStax).

The thermodynamic perspective is that at elevated temperatures, the negative entropy term in Equation \ref{eq1} will greater dominate the enthalpic term that is driving the dissolution process and make $$ΔG_{soln}$$ less negative and hence less spontaneous, i.e., weaker.

Example $$\PageIndex{1}$$:

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 N2(g) at 0 °C and STP.

NOTE: Molarity is equivalent to: moles of solute/ L of solution

At STP 1 mol=22L

\begin{align} \text{Molarity of } \ce{N2} &= 21\,ml \left( \dfrac{1L}{1000\,ml}\right) \nonumber \\[5pt] &=0.021\,L \,\ce{N2} \nonumber \\[5pt] &= 0.021\,L \,\ce{N2} \left(\dfrac{1\,mol}{22\,L}\right)\nonumber \\[5pt] &=\dfrac{0 .000954\,mol}{1\, L} \nonumber \\[5pt] &= 9.5 \times 10^{-4}\, M\, \ce{N2} \nonumber \end{align} \nonumber

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

$C = k_H P_{gas} \nonumber$

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 \\[5pt] &= 3.29 \times 10^{-5}\, M\, N_2 \nonumber \end{align} \nonumber

The Survival of Fish

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. Image used with permission (Public Domain; United States Fish and Wildlife Service)

### Solubility of Gases in Organic Solvents

In organic solvents, the solvation enthalpy ($$\Delta H_1$$) are considerably weaker than in water, especially for non-polar gases in nonpolar solvents. In many cases, the $$\Delta H_{sol} \approx 0$$, which means that

$\Delta H_2 \approx \Delta H_3$

and the enthalpy needed to break the solvent-solvent interactions is comparable to the enthalpy released in making solvent-gas interactions. Gases dissolved in organic solvents can actually be more soluble at higher temperatures but not for all gases in all organic solvents (Table $$\PageIndex{1}$$).

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 $$10.78 \times 10^4$$ $$5.05 \times 10^4$$ $$3.02 \times 10^4$$ $$0.086 \times 10^4$$ $$5.695 \times 10^4$$ -

Deviations from the trend in Figure $$\PageIndex{1}$$ occurs when $$\Delta H_{sol} > 0$$ for these systems or alternatively

$\Delta H_2 > \Delta H_3$

A Le Chatelier perspective, like that used above for water, can help understand why. The associated reaction with a $$\Delta H_{sol} > 0$$ is

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

that shows that adding heat to a solution by increasing the temperature will shift in the equilibrium to favor dissolution (i.e, push 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 $$\PageIndex{1}$$).

Exercise $$\PageIndex{1}$$

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 $$\PageIndex{1}$$

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.

Exercise $$\PageIndex{2}$$

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.