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Why is the reduction of a metal oxide ea...

Why is the reduction of a metal oxide easier if the metal formed is in the liquid state at the temperature of reduction ?

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Entropy is higher when a metal is in the liquid state than when it is in the solid state. Therefore, the value of entropy change (`Delta`S) of the reduction process is more on the +ve side when the metal formed is in the liquid state and the metal oxide being reduced is in the solid state. Since the value of T`Delta ` S increases and that of `Delta ` H remains constant, therefore, the value of `Delta_r G^(@) (DeltaG = DeltaH-TDeltaS)` becomes more-ve and hence the reduction becomes easier.
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Why is the reduction of a metal oxide easier if the metal formed is in liquid state at the temperature of reduction?

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Why is the reductio of a meal oxide easier if the metal foremd is in liquid state in the temoerature of reduction?

Statement-1: The reduction of a metal oxide is easier if the metal formed is in liquid state at he temperature of reduction. Statement-2: The value of entropy change DeltaS of the reduction process is more on +ve side when the metal formed is in liquid state and the metal oxide being reduced in a solid state. Thus, the value of the DeltaG becomes more or negative side.