NaCl dissociation not spontaneous? It's a well-known fact that the dissociation of $\text{NaCl}$ can be represented by the series of equations $$ \begin{eqnarray*} \text{Na} &amp;\rightarrow&amp; \text{Na}^+ + \text{e}^- \qquad \Delta H = 495 \, \text{kJ} \\ \text{e}^- + \text{Cl} &amp;\rightarrow&amp; \text{Cl}^- \qquad \qquad \Delta H = -348 \, \text{kJ} \end {eqnarray*} $$

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?
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).