Q.When hydrochloric acid is added to cobalt nitrate solution at room temperature, the following reaction takes place and the reaction mixture becomes blue. On cooling the mixture it becomes pink. On the basis of this information mark the correct answer.
[Co (H2O)6]^3+ (aq) + 4Cl^- (aq) ⇌ [CoCl4]^2- (aq) + 6H2O (l)
(pink) (blue)

The key idea here is Le Chatelier's Principle, which describes how a system at equilibrium responds to changes in conditions like temperature.

1. The given reaction is: $[Co(H_2O)_6]^{3+} (aq) + 4Cl^- (aq) \rightleftharpoons [CoCl_4]^{2-} (aq) + 6H_2O (l)$ (pink) (blue)
2. At room temperature, the mixture is blue, indicating the formation of $[CoCl_4]^{2-}$ is favored.
3. On cooling, the mixture becomes pink, indicating the formation of $[Co(H_2O)_6]^{3+}$ is favored. …

When a system at equilibrium is subjected to a change in conditions, it will adjust itself to counteract that change and re-establish a new equilibrium. This fundamental principle is known as Le Chatelier's principle. One of the most important conditions affecting equilibrium is temperature.

For a reversible reaction, the effect of temperature on the equilibrium position depends on whether the reaction is endothermic or exothermic:

• Endothermic reactions ($\Delta H > 0$): These reactions absorb heat. Increasing the temperature (adding heat) shifts the equilibrium to the right, favoring product formation. Conversely, decreasing the temperature (removing heat) shifts the equilibrium to the left, favoring reactant formation.
• Exothermic reactions ($\Delta H < 0$): These reactions release heat. Increasing the temperature (adding heat) shifts the equilibrium to the left, favoring reactant formation. Conversely, decreasing the temperature (removing heat) shifts the equilibrium to the right, favoring product formation.

We can use this principle to determine the sign of $\Delta H$ for the given reaction.

1. Identify the reaction and the colors of the species:

   The given reaction is:

   $[Co(H_2O)_6]^{3+} (aq) + 4Cl^- (aq) \rightleftharpoons [CoCl_4]^{2-} (aq) + 6H_2O (l)$

   (pink) (blue)

   We are told that the pink species is $[Co(H_2O)_6]^{3+}$ and the blue species is $[CoCl_4]^{2-}$.

2. Analyze the effect of cooling on the equilibrium:

   The problem states: "On cooling the mixture it becomes pink."

   This observation tells us that when the temperature is lowered, the equilibrium shifts to the left, favoring the formation of the pink reactant, $[Co(H_2O)_6]^{3+}$.

3. Apply Le Chatelier's Principle to deduce the nature of the reverse reaction:

   Since cooling (decreasing temperature) shifts the equilibrium to the left, it means that the reverse reaction is favored by cooling. …