# NaCl dissociation not spontaneous?

It's a well-known fact that the dissociation of $\text{NaCl}$ can be represented by the series of equations $$\begin{eqnarray*} \text{Na} &\rightarrow& \text{Na}^+ + \text{e}^- \qquad \Delta H = 495 \, \text{kJ} \\ \text{e}^- + \text{Cl} &\rightarrow& \text{Cl}^- \qquad \qquad \Delta H = -348 \, \text{kJ} \end {eqnarray*}$$

Adding gives $\text{Na} + \text {Cl} \rightarrow \text{Na}^+ + \text{Cl}^-$ with $\Delta H = 147 \, \text{kJ}$, which is positive so it is not spontaneous. What explains this?

• The reactions of elements to give ions is irrelevant to the dissolution of NaCl. NaCl is already ionic and the dissolution is dependent on the interaction of those ions with water molecules. The initial formation of NaCl from its elements is irrelevant. – matt_black Aug 14 '16 at 13:06
• That's not a well-known fact because it's not a fact... – orthocresol Aug 14 '16 at 13:26
• Why atomic chlorine? – Greg Aug 14 '16 at 17:12

You might be confusing $\ce{\Delta G}$ and $\ce{\Delta H}$.

$$\ce{\Delta G = \Delta H - T \Delta S}$$

$\ce{\Delta G}$, free energy, measures the "usefulness" of a system. If $\ce{\Delta G \gt 0}$, then the reaction is not thermodynamically favored - it's NOT SPONTANEOUS.

$\ce{\Delta H}$ being positive means that the reaction is endothermic, or makes the surroundings cooler. While this would CONTRIBUTE to a positive value of $\ce{\Delta G}$, it DOES NOT GUARANTEE it.

In the case of your example above, I would assume the $\ce{T \Delta S}$ factor would make $\ce{\Delta G \lt 0}$ if that reaction really is SPONTANEOUS (I don't have an entropy table with me).

• What is this?? The equation $\ce{Na + Cl -> Na+ + Cl- }$ does not represent the dissolution of NaCl. – orthocresol Aug 14 '16 at 13:25

Sodium chloride does not spontaneously dissociate into ions; table salt pretty much remains as it is. ($\Delta G$ is positive even allowing for entropy $RTln(2)$). Perhaps you are thinking about aqueous solution? In aqueous solution the molecule does dissociate, and this is due to the stabilisation given to the ions by the highly polar water. The Coulomb energy between two ions of charge $z_1, z_2$ is $w=z_1z_2e^2/(4\pi\epsilon_0\epsilon r)$ and is reduced by $1\epsilon$ the dielectric constant. In water $\epsilon= 78$ compared to approx 2 for hexane and 1 for a vacuum. Thus in water, ions can exist separately, as their electric field is 'quenched' and their interaction is weak vs thermal energy (random collisional disruptive motion) and so dissolve, whereas in a non-polar solvent the ions can interact with one another even when well separated (since their electric field spreads out significantly) and so recombine more easily, and with greater stability, and do not dissolve