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+1 vote
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in Chemistry by (24.2k points)

Ellingham diagrams are plots of ΔGº vs temperature which have applications in metallurgy

2H2 + O2 = 2H2O ΔG(J) = –247500 +55.85 T ........ (I)

2CO + O2 = 2CO2 ΔG(J) = –282400 +86.81 T.......... (II)

The Ellingham diagrams for the oxidation of H2(I) and CO (II) are given below.

The two lines intersect (TE) at 1125 K.

Which of the following is correct ?

I. Δ Gº for reaction (i) is more negative at T < 1125K

II. Δ G° for the reduction of CO is more negative at T < 1125 K

III. H2 is a better reducing agent at T > 1125 K

IV. H2 is a better reducing agent at T < 1125 K

(A) I and II 

(B) I and III 

(C) III only 

(D) I and IV

1 Answer

+2 votes
by (23.5k points)
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Best answer

Correct answer is (C) III only

According to the given data, Plot I is for CO as it has more negative ΔG value Plot II is for H2. And as we know in the Ellingham diagram, compound for which, the Plot lies below acts as a better reducing agent. So, at T > 1125 K, H2 is a better reducing agent. 

Note: The given graph in the question is not according to given data in the question.

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