Dissolving a gas: X (g) ⇌ X (aq)
Dissolving a gas always results in a negative ΔS° (system) because the system becomes more ordered.
Since ΔS° (total) = ΔS° (system) + ΔS° (surroundings). For a gas to be soluble, the above reaction must be thermodynamically spontaneous, i.e. ΔS° (total) must be +ve. Therefore, ΔS° (surroundings) has to be +ve in order to compensate for the –ve value of ΔS° (system). Exothermic reactions have a +ve ΔS° (surroundings). Dissolving a gas always results in a negative ΔS° (system) because the system becomes more ordered.
∴ Gasses dissolve exothermically.
For example: CO2 (g) ⇌ CO2 (aq) ΔH negative
The above equilibrium is driven to the left, the endothermic side, by an increase in temperature. Hence, gases such as carbon dioxide are more soluble in cold water than hot water.
Note: Most gases have a negative enthalpy of solution (ΔH solution). A negative enthalpy of solution means that the solute is less soluble at high temperatures.
Dissolving a solid: X (s) ⇌ X (aq)
Contrary to the expected approach to entropy, dissolving solids does not always result in +ve ΔS° (system); the system becoming more disordered. The solute (e.g. salt) becomes more disordered as it goes from a highly ordered solid to a more random solution, however the solvent (e.g. water) can become more ordered due to the forces of attraction between the solute and the solvent. This is more likely to happen when compounds containing ions of high charge density dissolve in water.
Note: treat ΔH solution as ΔH reaction; therefore, if ΔH solution is –ve it means that the reaction is exothermic. If ΔH solution is +ve then the reaction is endothermic.
Therefore for this reaction X (s) ⇌ X (aq):
ΔH | ΔS° (surroundings) |
+ve (endothermic) | -ve (less disordered) |
-ve (exothermic) | +ve (more disordered) |
Thus, at higher temperatures, the equilibrium: X (s) ⇌ X (aq):
is driven to the left (less soluble) if ΔH solution is exothermic and to the right (more soluble) if ΔHsolution is endothermic.
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