Question

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.
${[Co{({H_2}O)_6}]^{3 + }}_{(aq)} + 4C{l^ - }_{(aq)} \rightleftarrows {[CoC{l_4}]^{2 - }}_{(aq)} + 6{H_2}{O_{(l)}}$
(A) $\Delta H > 0$ for the reaction
(B) $\Delta H < 0$ for the reaction
(C) $\Delta H = 0$ for the reaction
(D) The sign of $\Delta H$ can not be predicted using this information

Answer 1

${[Co{({H_2}O)_6}]^{3 + }}_{(aq)}$ solution is pink in colour and ${[CoC{l_4}]^{2 - }}_{(aq)}$ solution is blue in colour. If the reaction is endothermic then $\Delta H > 0$ for the reaction and if the reaction is exothermic then $\Delta H < 0$ for the reaction.

Complete answer:
First we need to know that the solution of ${[Co{({H_2}O)_6}]^{3 + }}_{(aq)}$ is pink in colour. When hydrochloric acid is added into it, ${[CoC{l_4}]^{2 - }}_{(aq)}$ forms and it has a blue colour.
As the information in given in the question that at room temperature, the reaction occurs in the forward direction and formation of ${[CoC{l_4}]^{2 - }}_{(aq)}$ results in the colour change in the solution. When we cool this resulting solution, then we get pink coloured solution once again, this means that ${[Co{({H_2}O)_6}]^{3 + }}_{(aq)}$ is regenerated.
As this reaction is in equilibrium, it is possible that equilibrium shifts with change in temperatures.
Now, let’s talk about the enthalpy of this reaction. At relatively higher temperatures, means if we provide the reaction system with some heat, then reaction occurs in a forward direction. That means that this reaction is endothermic. Both of these things suggest that $\Delta H > 0$ for this reaction.
So, correct answer is (A) $\Delta H > 0$ for the reaction.

Additional Information:
- In the endothermic type of reactions, heat is absorbed by the reaction system and hence the value of $\Delta H$ is always taken in positive sign.
- In exothermic types of reactions, heat is released by the reaction system and hence the value of $\Delta H$ is always taken in negative sign.

Note:
You need to remember the sign of $\Delta H$ value in certain types of reactions (endothermic or exothermic) as this is often confusing. Remember that the $\Delta H$ value of reaction that they have asked about of the forward reaction, do not consider the reverse reaction here as the value will be inverted here.