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Question


Which of the following is incorrect on the basis of the above Ellingham diagram for carbon?
  1. In principle, carbon can be used to reduce any metal oxide at a sufficiently high temperature.
  2. Up to 710C, the reaction of formation of CO2 is energetically more favorable, but above 710C, the formation of CO is preferred.
  3. ΔS(C(s)+12O2(g)CO(g))<ΔS(C(S)+O2(g)CO2(g))
  4. 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 710C, ΔGf for CO2 is more negative. 
But after 710CΔGf 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.

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