Which of the following is incorrect on the basis of the above Ellingham diagram for carbon?

Up to $710^{\circ} C$ , the reaction of formation of $C O_{2}$ is energetically more favorable, but above $710^{\circ} C$ , the formation of $C O$ is preferred.

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In principle, carbon can be used to reduce any metal oxide at a sufficiently high temperature.

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$Δ S (C_{(s)} + \frac{1}{2} O_{2 (g)} \to C O_{(g)}) < Δ S (C_{(S)} + O_{2 (g)} \to C O_{2 (g)})$

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Carbon reduces many oxides at elevated temperature because $Δ G^{\circ}$ vs temperature line has a negative slope.

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Solution

The correct option is C
$Δ S (C_{(s)} + \frac{1}{2} O_{2 (g)} \to C O_{(g)}) < Δ S (C_{(S)} + O_{2 (g)} \to C O_{2 (g)})$

The line corresponding to the formation of $C O_{2}$ is nearly horizontal.
Till $710^{\circ} C$ , $Δ G_{f}^{\circ}$ for $C O_{2}$ is more negative.
But after $710^{\circ} C$ , $Δ G_{f}^{\circ}$ for $C O$ is more negative
and thus the formation of $C O$ 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 $O_{2}$ yields 1 mol of $C O$ .
Contrast this with the formation of $C O_{2}$ :
One mole of $O_{2}$ yields one mol of $C O_{2}$ . Hence the answer.

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Similar questions

Based on the above Ellingham diagram, the formation of carbon monoxide from carbon can reduce the metal oxide to metal at a temperature _______ the point of intersection of the two lines.
1. Equal to
2. Higher than
3. Lower than
4. None

Q. According to Ellingham diagram, the oxidation reaction of carbon to carbon monoxide may be used to reduce which one of the following oxides at the lowest temperature?

These equilibria can be discussed in terms of thermodynamic functions.

For the reaction :

(i)

M_{(s)} + \frac{1}{2} O_{2 (g)} \to M O_{(} s)

(ii)

\frac{1}{2} C_{(s)} + \frac{1}{2} O_{2 (g)} \to \frac{1}{2} C O_{2 (g)}

(iii)

C_{(s)} + \frac{1}{2} O_{2 (g)} \to C O_{(g)}

(iii)

C O_{(g)} + \frac{1}{2} O_{2 (g)} \to C O_{2 (g)}

The temperature dependence of

Δ ˚ G

of reaction (i) to (iv) is shown in the given diagram. This is known as the Ellingham diagram. With the help of the Ellingham diagram, one can easily predict the most suitable reducing agent for the reduction of metal oxides.

A I_{2} O_{3}

can be reduced by carbon at a temperature of:

Q. Carbon can reduce ferric oxide to iron at a temperature above 983 K because:

Q. A metal oxide

M O_{(s)}

can be reduced by carbon or carbon monoxide successfully when the line in Ellingham diagram for reaction (i):

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