A) Reduction of a metal oxide is easier if the metal formed is in liquid state at reduction temperature
B) Reduction of metal oxide is normally less feasible at high temperature
C) The oxidation of CO into \[C{{O}_{2}}\] will have a negative slope on Ellingham diagram
D) The reactive metals have positive slope for oxidation on Ellingham diagram whereas non-reactive metals have a negative slope
Correct Answer: A
Solution :
[a] \[\Delta {{G}^{o}}=\Delta H-T\Delta S\] The criterion of feasibility is that at a given temperature change in Gibbs energy for the reaction must be -ve. For \[\Delta {{G}^{o}}\] to be -ve, \[T\Delta S\] should be +ve. \[\Delta S\], entropy change, depends much on the physical state of the compound. It will increase on changing the phase from solid to liquid or from liquid to gas. Reduction of metal oxide is less economical at high temperature. Oxidation of CO to \[C{{O}_{2}}\] proceeds with decrease in entropy \[\Delta S\].
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