Physical Chemistry

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

shalini

12 Years agoGrade

4 Answers

Aarti Gupta

11 Years ago

The entropy of a metal is higher in its liquid state than in its solid state. Therefore,
entropy change, ∆S of the reduction process is more on the positive side when the metal formed in liquid state and metal oxide being reduced is in the solid state. As a result, the value of ∆G becomes more on negative side and therefore, reduction becomes easier.

Kratika sharma

9 Years ago

The entropy of metal is more in its liquid state as compared to solid state . As a result the delta G in more negative , ans the reaction is more feasible

Moyukh Chakrabarti

8 Years ago

The entropy of a metal is higher in its liquid state than in its solid state. Therefore,entropy change, ∆S of the reduction process is more on the positive side when the metal formed in liquid state and metal oxide being reduced is in the solid state. As a result, the value of ∆G becomes more on negative side and therefore, reduction becomes easier.

The entropy of metal is more in its liquid state as compared to solid state . As a result the delta G in more negative , ans the reaction is more feasible

Kushagra Madhukar

5 Years ago

Dear student,

Please find the answer to your question.

The entropy of a metal is higher in its liquid state than in its solid state. Therefore, entropy change, ∆S of the reduction process is more on the positive side when the metal formed in liquid state and metal oxide being reduced is in the solid state. As a result, the value of ∆G becomes more on negative side and therefore, reduction becomes easier.

Thanks and regards,

Kushagra