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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 liquid state at the temperature of reduction?

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The reduction of a metal oxide with a reducing agent such as coke may be represented as :
`M_(x)O(s) +C (s) overset("heat") to xM(l)+CO(g)`
In case the metal formed as a result of reduction is a liquid, `DeltaS^(@)` will be higher than when it is in the solid state. Under the conditions, `DeltaG^(@)=DeltaH^(@)-T DeltaS^(@)` will become more negative and the reduction will be easier.
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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.