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I have been learning thermodynamics for quite some time and I fail to relate Gibbs free energy and stability.

For example, how do I compare stability of metal oxides using the following data? Also, do stable compounds decompose easily or hard?

$$ \begin{array}{clrr} \hline \text{#} & \text{Compound} & \Delta_\mathrm{f}G^\circ/\pu{kJ mol^-1} & \Delta_\mathrm{f}H^\circ/\pu{kJ mol^-1} \\ \hline 1 & \ce{ZnO(s)} & -318.4 & -348.3 \\ 2 & \ce{Cu2O(s)} & -146.0 & -168.8 \\ 3 & \ce{HgO(s)} & -58.5 & -90.8 \\ 4 & \ce{PbO(s)} & -187.9 & -217.3 \\ \hline \end{array} $$

andselisk
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Meet Lalwani
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  • Generally the compound with more negative Gibbs free energy of formation is the most stable one. – S R Maiti Feb 22 '21 at 10:59
  • can you elaborate??? – Meet Lalwani Feb 22 '21 at 12:22
  • Gibbs free energy is basically an indicator of how far and in which direction the equilibrium is pushed. So if the Gibbs free energy is lower (i.e. more negative) that means the equilibrium is pushed towards the product side. Here the free energy of formation means the formation of the metal oxide from its elements in standard state. So the lower the Gibbs free energy, the metal oxide is more stable w.r.t the elements. – S R Maiti Feb 22 '21 at 12:38

1 Answers1

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The thermodynamic data refers to formation of the given compounds from the elements in their standard states. For instance for $ \ce{ ZnO(s)}$ it refers to the reaction

$ \ce{ Zn(s) +1/2O2(g) ->ZnO(s)} \tag{1}$

with each element or compound present at a pressure of 1 bar and (unless noted otherwise) a temperature of 298.15 K.

A more negative value of $\Delta G^\circ$ means greater product stability. This follows from the relationship

$$\Delta G^\circ = -RT\log K \tag{2}$$

where K is the equilibrium constant for the reaction at that temperature. An increased value of K implies an increased amount of products relative to reagents. By the relation in equation (2) a larger value of K also means a more negative value of $\Delta G^\circ$.

By definition a more stable compound decomposes less readily. Decomposition here refers to the reverse of reaction (1).

When comparing the tabulated oxides, a more stable oxide has a more negative free energy of formation. Therefore $ \ce{ZnO(s)}$ is the most stable of the oxides relative to oxygen and the reduced metal. Of the metals the mercury(II) oxide would require the least energy to reduce.

Note that for all of the reactions here $$K = p^\circ/p_\ce{O2}$$ where $p^\circ=\pu{1 bar}$ (assuming the pressure is low). When starting from metal oxides, allowed to decompose until at equilibrium, a less stable oxide is at equilibrium with a higher partial pressure of oxygen (smaller K) due to more of the oxide decomposing. Therefore imagine you have an evacuated box containing the oxide of one metal and an amount of another metal in pure (reduced) form. If the equilibrium partial pressure $p_\ce{O2}(\text{eq})$ for the oxide is higher than $p_\ce{O2}(\text{eq})$ for the oxide of the second metal, then (if you wait long enough) all of the oxide will be reduced and the second metal oxidized until either the original oxide or pure metal is exhausted.

The enthalpies of formation refer to how much heat evolves when 1 mole of the oxide is formed from the reagents at constant T and p, with all compounds in their standard states. The more negative the value the more heat is released (the reaction is more exothermic). The oxidation of zinc is the most exothermic reaction (per mole of product formed) of those shown in the table.

Buck Thorn
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