reduction of metal oxide is easier if the metal in liquid state?
reduction of metal oxide is easier if the metal in liquid state?
Being in liquid state means that the temperature of the substance is higher. According to the ellingham diagram, the value of ΔG for the formation of metal oxides increases as temperature increases. At these higher temperatures, many metal graphs can be observed below this one (except for very high reactive metals of course) and therefore the reduction of the metal oxide is easier.
For reduction of metal oxide the gibbs energy should be negative, this is a mandatory condition. In solid state, the entropy or randomness will be less due high intermolecular force of attraction while in liquid state, the entropy will be more comparatively, therefore gibbs energy will be more negative.
Being in liquid state means that the temperature of the substance is higher. According to the ellingham diagram, the value of ΔG for the formation of metal oxides increases as temperature increases. At these higher temperatures, many metal graphs can be observed below this one (except for very high reactive metals of course) and therefore the reduction of the metal oxide is easier.
For reduction of metal oxide the gibbs energy should be negative, this is a mandatory condition. In solid state, the entropy or randomness will be less due high intermolecular force of attraction while in liquid state, the entropy will be more comparatively, therefore gibbs energy will be more negative.
