Question

# Which of the following is incorrect on the basis of the above Ellingham diagram for carbon?In principle, carbon can be used to reduce any metal oxide at a sufficiently high temperature.Up to 710∘C, the reaction of formation of CO2 is energetically more favorable, but above 710∘C, the formation of CO is preferred.ΔS(C(s)+12O2(g)→CO(g))<ΔS(C(S)+O2(g)→CO2(g))Carbon reduces many oxides at elevated temperature because ΔG∘ vs temperature line has a negative slope.

Solution

## The correct option is C ΔS(C(s)+12O2(g)→CO(g))<ΔS(C(S)+O2(g)→CO2(g))The line corresponding to the formation of CO2 is nearly horizontal. Till 710∘C, ΔG∘f for CO2 is more negative.  But after 710∘C, ΔG∘f for CO is more negative and thus the formation of CO is thermodynamically favoured.  Don't forget - the reactions are all for the formation of one mole of the oxide.  In the formation of the monoxide, 0.5 mol of a O2 yields 1 mol of CO. Contrast this with the formation of CO2: One mole of O2 yields one mol of CO2. Hence the answer.

Suggest corrections

Similar questions
View More