Why does heating transform the aquacomples of slightly acidified cobalt(II) chloride solution to the chlorido complex?

Question

There is a famous demonstration where, if you mix a $\ce{CoCl2}$-solution with the right amount of $\ce{HCl}$ you will end up with the pink $\ce{[Co(H2O)6]^2+}$ complex but when heating it to boiling temperatures it will change to blue as the $\ce{[CoCl4]^2-}$ complex is formed. This reaction is reversible as the mixture is cooled again. Now literature always describes that this works because that shift towards $\ce{[CoCl4]^2-}$ is endothermic, but why is this the case?

What causes one to be more stable here than the other one but being substituted when heated?

I have some ideas but I can't really compare them:

• First of all there should actually be an entropic advantage towards the chlorido-side, as 6 water ligands leave, meaning that the exothermic step here is against the natural direction of the entropy(?).

• Second, perhaps the chlorido ligand is the better leaving group. I know usually $\ce{H2O}$ is a good leaving group in organic molecules but then its usually a $\ce{H2O+}$. So perhaps chloride is the better leaving group here.

• Maybe there is a kinetic effect, that it is harder to displace 6 ligands than just four ligands

• Or, as much more water is present it just shifts towards the water side because of the huge excess of water?

• Chloride is a weaker ligand than water. But on the other hand the iron-fluorido complex is quite stable in water and towards other ligands.

• So could it be due to the different ligand fields? The splitting in the tetrahedron is much smaller than in the octahedron. Although I don't know how this actually affects the stability.

So those were the ideas I had but I can't really find a common thing among them besides the excess of water which often causes an equlibrium to shift.

By the way $\ce{[Cu(H2O)6]^2+}$ and $\ce{[CuCl4]^2-}$ do the exact same thing.

Does anyone have an idea why this reaction is endothermic?

Answer 1

You may know the formula for Gibbs Free Energy $\Delta G$. This is taught as: $$\Delta G = \Delta H - T\Delta S\tag{1}$$

In this case, we would be observing $\Delta G$ for equilibrium $(2)$.

$$\ce{[Co(H2O)6]^2+ + 4 Cl- <=> [CoCl4]^2- + 6 H2O}\tag{2}$$

Within the temperature range between $\pu{25^\circ C}$ and $\pu{100^\circ C}$, we should not expect the enthalpic term $\Delta H$ to change significantly. In fact, it is usually correct at first or even second approximation to consider it constant. If we do, the only thing in $\Delta G$ that changes with temperature in equation $(1)$ is the $-T\Delta S$ term. It basically implies that a higher temperature will increase the entropic contribution.

You already noted that the product side is favoured entropically since five free-moving particles are turned into seven in the course of the reaction. Higher temperatures should thus favour the product side.

One more thing: since we have established that $\Delta S$ is positive, we can immediately see that $\Delta H$ must be positive, too; i.e. the reaction must be endothermic. This is because $T$ is positive by definition and therefore $-T\Delta S$ is always negative. However, the only way to tip an equilibrium to the opposite side is to change the sign of $\Delta G$ — which can only work if a positive $\Delta H$ can offset a negative $-T\Delta S$.