Solubility of solid to liquid

 

As with all processes under constant pressure and constant temperature, dissolving a solution into solution will occur only if ΔGtotal<0.

ΔGsol=ΔHsol−TΔSsol<0

For dissolving solids in liquids, ΔSsol>0, but for dissolving gases solutes, the entropy of solution is negative (ΔSsol<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 ΔGsol<0 for a dissolving a gas in solution is if the solution process is exothermic (i.e., ΔHsol<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.

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 (ΔHsolute-solute>0).

2. Breaking solvent-solvent attractions (endothermic), i.e., hydrogen bonding and dipole-dipole interactions in water (ΔHsolvent-solvent>0

3. Forming solvent-solute attractions (exothermic), i.e., solvation energy (ΔHsolute-solvent<0

The enthalpy of solution ΔHsol is the sum of these three individual steps.

ΔHsol=ΔHsolute-solute+ΔHsolvent-solvent+ΔHsolute-solvent

For solids solutes, ΔHsolute-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.

ΔHsolute-solute(gas)=0

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

Solubility of Gases in Polar Solvents

In polar solvents like water, ΔHsolute-solvent> 0 the dissolution of most gases is exothermic (i.e., ΔHsol<0). Hence, when a gas dissolves in a liquid solvent, thermal energy is released which warms both the system (the solution) and the surroundings.

solute(gas)+water(l)⇌solute(aq)+water(aq)+Δ .....4

where Delta is thermal enegy. Consequently, the solubility of a gas is dependent on temperature . 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 4 to the left. The thermodynamic perspective is that at elevated temperatures, the negative entropy will dominate the enthalpic term that is driving the dissolution process and make ΔGsoln less negative and hence less spontaneous.

 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.

Example 11

Determine the solubility of N2(g)N2(g) when combined with H2OH2O at 0.0345 °C the pressure of N2N2 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.

At STP 1 mol=22L

Molarity of N2=21ml(1L1000ml)=0.021LN2=0.021LN2(1mol22L)=0.000954mol1L=9.5×10−4MN2Molarity of N2=21ml(1L1000ml)=0.021LN2=0.021LN2(1mol22L)=0.000954mol1L=9.5×10−4MN2

Now that the molarity of N2. CC has been attained, the Henry's law constant, kk, can be evaluated.

Henry's Law:

C=kHPgasC=kHPgas

where CC is solubility, kHkH is Henry's constant, and PgasPgas is the partial pressure of the gas being considered.

Rearranging the formula to solve for kHkH

kH=CPgas=9.5×10−4MN2/1atm(5)(5)kH=CPgas=9.5×10−4MN2/1atm

Now substitute k and the partial pressure of N2N2 into Henry's law:

C=(9.5×10−4MN2)(0.0345)=3.29×10−5MN2C=(9.5×10−4MN2)(0.0345)=3.29×10−5MN2

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. (Public Domain; United States Fish and Wildlife Service)

Solubility of Gases in Organic Solvents

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 (ΔHsolvent-solvent) and the solvation enthalpies (ΔHsolute-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.

ΔHsolvent-solvent≈ΔHsolute-solvent

which means the ΔHsol≈0 . In some solvent-solute combination

ΔHsolvent-solvent>ΔHsolute-solvent

so that ΔHsol>0. In this case, we can write the solution reactions thusly

Δ+solute(gas)+solvent(solvent)⇌solute(sol)+solvent(sol)

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 8 to the right). As a result, the solubilities of gases in organic solvents often increase with increasing temperature (