Q.

[!FORMULA] $[Co(H_2O)_6]^{2+}(aq)\ (\text{pink}) + 4Cl^-(aq) \rightleftharpoons [CoCl_4]^{2-}(aq)\ (\text{blue}) + 6H_2O(l)$

In the above reaction at equilibrium, the reaction mixture is blue in colour at room temperature. On cooling this mixture, it becomes pink in colour. On the basis of this information, which one of the following is true?

Pressure only affects an equilibrium that has gaseous components, and only when the total moles of gaseous reactants differ from the total moles of gaseous products ($\Delta n_g\neq0$). Increasing the pressure (decreasing the volume) shifts the equilibrium toward the side with fewer moles of gas; decreasing the pressure (increasing the volume) shifts it toward the side with more moles of gas. For ammonia synthesis, $N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)$ (4 mol gas $\rightarrow$ 2 mol gas), compressing the system favours $NH_3$ formation; expanding it favours decomposition back to $N_2$ and $H_2$. When $\Delta n_g=0$, as in $H_2(g)+I_2(g)\rightleftharpoons2HI(g)$, pressure changes have no effect at all on the equilibrium position.

Temperature shifts equilibrium in the direction that absorbs the applied change -- raising the temperature always favours whichever direction is endothermic, and lowering the temperature always favours whichever direction is exothermic. For ammonia synthesis, the forward (synthesis) direction is exothermic, so raising the temperature decomposes some $NH_3$ back to $N_2$ and $H_2$ (the endothermic, heat-absorbing direction), while lowering the temperature favours more $NH_3$ formation. This same logic reads a colour change directly as a sign of $\Delta H$: for $[Co(H_2O)_6]^{2+}(\text{pink})+4Cl^-\rightleftharpoons[CoCl_4]^{2-}(\text{blue})+6H_2O$, if the mixture is blue (forward/product-favoured) at room temperature but turns pink (reverse/reactant-favoured) on cooling, then cooling is favouring the reverse reaction -- so heating must favour the forward reaction, meaning the forward reaction is endothermic, $\Delta H>0$. Comparing a reaction's equilibrium constant at two temperatures gives the same conclusion quantitatively: if $K_1$ (room temperature) is greater than $K_2$ (at 700 K), then K is falling as temperature rises -- exactly the signature of an exothermic forward reaction (a lower K at higher T means the endothermic, i.e. reverse, direction is being favoured as T rises, so the forward direction must be exothermic). …