Why are gases less soluble at higher temperatures?
The solubility of any solute in a solvent is dependent on temperature. When a solid mixes with a liquid, heat is used to break the bonds between the atoms of the solid. Similarly, when a new bond is formed between the solute and solvent, heat is liberated. This formation of new bonds between the solute and solvent is called dissolution. In the above example, the solid is said to be soluble in liquid.
Solubility is dependent on temperature, and can be explained in two ways:
1. Reduced solubility with an increase in temperature.
2. Enhanced solubility with an increase in temperature.
In the first case, the heat that is given out is more than the heat that is taken in while breaking the bonds of the solute. In this case, the reaction that is happening in the solvent is called exothermic. Now, if we increase the temperature of the system, which does not require it at all, it stops the process of dissolution – because of the greater heat that was already given out by the reaction. Hence, in this situation, an increase in temperature does not increase the solubility of the solute.
In the second case, the heat given out by the reaction in the solution is less than the heat used for breaking the solute bonds. The reaction is called endothermic. If we increase the temperature of the system, the breaking of solute bonds to react with the solvent will be more, and this results in greater solubility of the solute. Hence, in this situation, an increase in temperature increases the solubility of the solute.
The above explanation is related to the Le Chatelier’s principle. According to this principle, if a system is at equilibrium, and is applied with some stress, the equilibrium moves towards the side of removing the stress. Therefore, gaseous oxygen + solvent = oxygen solution + heat. When temperature is added to the system, which is releasing the heat already, the gas becomes less soluble, and is separated from the solvent.