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| 1. |
Account for the following fact. The reduction of a metal oxide is easier if the metal is formed in the liquid state at the temperature of reduction. |
| Answer» Solution :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_rG^(@) (DeltaG = DeltaH-TDeltaS)` becomes more-ve and HENCE the reduction becomes easier. | |