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(D) The decomposition of a metal oxide is represented as: $2MO(s) \rightarrow 2M(s/l) + O_2(g)$.$1$. Since a gaseous product $(O_2)$ is formed from a

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(D) The decomposition of a metal oxide is represented as: $2MO(s) \rightarrow 2M(s/l) + O_2(g)$.$1$. Since a gaseous product $(O_2)$ is formed from a solid reactant,the entropy increases $(\Delta S^o > 0)$.$2$. Breaking chemical bonds requires energy,making the process endothermic $(\Delta H^o > 0)$.$3$. According to the Gibbs free energy equation,$\Delta G^o = \Delta H^o - T \Delta S^o$. For the reaction to be spontaneous,$\Delta G^o$ must be negative,which occurs when $T \Delta S^o > \Delta H^o$ at high temperatures.These principles are the basis of the Ellingham diagram.

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