Question

The third law of thermodynamics states that the entropyof a perfect crystal at 0 K is zero. In Appendix 4,${\mathit{F}}^{\mathbf{-}}\left(aq\right)$, role="math" localid="1658167389005" $\mathit{O}{\mathit{H}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$, and${\mathit{S}}^{\mathbf{2}\mathbf{-}}\left(aq\right)$ all have negative standard entropy values. How can role="math" localid="1658167506187" ${\mathit{S}}^{\mathbf{o}}$values be less than zero?

Short answer

It is important to note that these species do not exist in the solid phase, but rather in the aqueous phase, where they coexist with water molecules. When these ions are held in water, there is a considerable rise in the ordering. The highly organized phase should be present when hydrated water surrounds these ions.

Step-by-step solution

Step 1: Introduction to the Concept

The third rule of thermodynamics states that when an object's temperature reaches absolute zero ($\mathbf{\text{0}}\mathbf{K}\mathbf{=}\mathbf{-}\mathbf{273}\mathbf{.}\mathbf{15}\mathbf{C}\mathbf{=}\mathbf{-}\mathbf{459}\mathbf{.}\mathbf{67}\mathbf{°}\mathbf{F}$), its atoms stop moving.

Step 2: Solution Explanation

Although the third law of thermodynamics applies to precisely crystalline substances, it does not apply to aqueous solutions.

It should be emphasised that these species are not found in the solid phase, but they are found in the aqueous phase, along with water molecules. There is a noticeable increase in the ordering when these ions are kept in the water. When hydrated water surrounds these ions, it should be in the highly ordered phase. As a result, the standard entropy values for these species are negative.