Reduction of a metal oxide in liquid state is | vedclass

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(A) The reduction of a metal oxide is represented by the reaction: $MO(s/l) + C(s) \rightarrow M(s/l) + CO(g)$.According to the Gibbs free energy equa

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The value of entropy change of the reduction process is more

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(A) The reduction of a metal oxide is represented by the reaction: $MO(s/l) + C(s) \rightarrow M(s/l) + CO(g)$.According to the Gibbs free energy equation,$\Delta G = \Delta H - T\Delta S$.For the reaction to be spontaneous,$\Delta G$ must be negative.When the metal oxide is in the liquid state,the entropy of the system is higher compared to the solid state.Consequently,the entropy change $(\Delta S)$ for the reduction process becomes more positive (or less negative) when the reactant is in the liquid state,making the term $-T\Delta S$ more negative,which facilitates the reduction process.

Reduction of a metal oxide in liquid state is easier than in the solid state because

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