Answer:
The reduction of a metal oxide with
a reducing agent such as coke may be represented as:
\[{{M}_{x}}O(g)+C(g)\xrightarrow{\,}xM(l)+CO(g)\]
In case the metal formed as a
result of reduction is a liquid, \[\Delta {{S}^{o}}\] will be higher than when
it is in the solid state. Under the conditions, \[\Delta {{G}^{o}}=\Delta
{{H}^{o}}-T\Delta {{S}^{o}}\] will become more negative and the reduction will
be easier.
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