Ellingham Diagram

My textbook says the following about the [Ellingham diagram](https://en.wikipedia.org/wiki/Ellingham_diagram):



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> Each plot is a straight line except when some change in phase ($\ce{s -> liq}$ or $\ce{liq -> g}$) takes place. The temperature at which such change occurs is indicated by an increase in the slope on the positive side (e.g., in the $\ce{Zn}$, $\ce{ZnO}$ plot, the melting is indicated by an abrupt change in the curve).
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However, I do not understand this. If a solid metal oxide changes into a liquid or a liquid into a gas, shouldn't the entropy increase? And as a result, the change in entropy should become less negative with regards to the equation:



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> Ellingham diagram normally consists of plots of $\Delta\_\mathrm fG^\circ$ *vs* $T$ for formation of oxides of elements i.e., for the reaction,
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> $$\ce{2x M (s) + O2 (g) -> 2 M\_xO (s)}$$
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> In this reaction, the gaseous amount (hence molecular randomness) is decreasing from left to right due to consumption of gases leading to a negative value of $\Delta S$ which changes the sign of the second term in equation (6.14). Subsequently $\Delta G$ shifts towards higher side despite rising $T$ (normally, $\Delta G$ decreases i.e. goes to lower side with increasing temperature). The result is positive slope in the curve for most of the reactions shown above for formation of $\ce{M\_xO (s)}$.
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If this is the case, the slope should become less positive according to the equation:


$$\Delta G = \Delta H - T\,\Delta S$$


where $-\Delta S$ is the slope. Am I missing something?




Qualitative explanation


We can all agree that
$$S\_{solid}<S\_{liquid}<S\_{gas}$$


Metal oxides are generally ionic and have high melting points. Let us say that they do not change phase. So, let us assume $S\_{oxide}$ doesn't change as much.


Metals, however, may change their phase.


$$(S\_{oxide}-S\_{solid})>(S\_{oxide}-S\_{liquid})>(S\_{oxide}-S\_{gas})$$


Note that $(S\_{oxide}-S\_{metal})$ is your $\Delta S$. As $T$ increases, at the phase-change temperatures, $\Delta S$ decreases, $-\Delta S$ increases $\Rightarrow$ the slope increases.


 3

We can explain this with the example of $\ce{CO + O2 <=> CO2}$.
Here we have to focus that there are two reactants on one side and there is only one reactant on the other.


The sum of the entropy of the reactants is subtracted from the entropy of the products, which would result in a negative entropy change in the present example of ideal gases, and hence a positive slope.


 0

In the reaction of solid metal with oxygen the product is also solid. So CHANGE in entropy in the reaction is comparatively lower. Because solid has lower entropy and the product as well as one of the reactants are solid.
But when a molten metal and a gas (oxygen) react the initial entropy is much higher (liquids have more entropy than solids) and they​ react to form a solid, so change in entropy is much more negative (higher entropy to lower entropy).
So in the equation since $-\Delta S$ is the slope if $\Delta S$ is more negative slope becomes more positive


 -3