Chemistry – Chemical reactions - Reversible reactions and equilibrium | e-Consult

The reaction between anhydrous cobalt(II) chloride (CoCl₂) and water is an exothermic process, forming hydrated cobalt(II) chloride. The equation is: CoCl₂(s) + xH₂O(l) ⇌ [Co(H₂O)₆]²⁺(aq) + ... (the exact stoichiometry of the hydrated form is complex and not crucial for this question).

Increasing the temperature will shift the equilibrium towards the formation of hydrated cobalt(II) chloride. This is because the reaction is exothermic. Adding heat to the system is equivalent to adding a product, and the equilibrium will shift to favour the reactants (anhydrous CoCl₂ and water) to consume the excess heat and re-establish equilibrium. This results in more anhydrous CoCl₂ reacting with water to form hydrated CoCl₂.

Decreasing the temperature will shift the equilibrium towards the formation of anhydrous cobalt(II) chloride. This is because the reaction is exothermic. Removing heat from the system is equivalent to removing a product, and the equilibrium will shift to favour the products (hydrated CoCl₂) to replace the lost heat. This results in some hydrated CoCl₂ losing water molecules and becoming anhydrous CoCl₂.

Pressure has a negligible effect on this reaction. The reaction involves the dissolution of a solid in a liquid, and the number of moles of solid and liquid are approximately equal. Therefore, changes in pressure will not significantly alter the equilibrium position. However, if the reaction were to involve a change in the number of moles of gas, then pressure would have a more significant effect. Since this reaction does not involve a change in the number of moles of gas, pressure is not a significant factor.

Therefore, temperature is the primary factor influencing the hydration of cobalt(II) chloride, and Le Chatelier's principle explains how changing the temperature can drive the reaction towards either the anhydrous or hydrated state. Pressure has a negligible effect in this case.