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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Up to 710oC, the reaction of formation of CO2, is energetically more favourable, but above 710oC the formation of CO is preferred.

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

ΔSC(s)+1/2O2(g)⟶CO(g)<ΔSC(s)+O2(g)⟶CO2(g)

Carbon reduces many oxides at elevated temperature because ∆fG⊝ vs temperature line has a negative slope.

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

In Ellingham diagram, higher is the slope higher is the entropy change so ΔS(C(s)+1/2O2(g)⟶CO(g))>ΔS(C(S)+O2(g)⟶CO2(g)as CO have higher slope.Also, Carbon monoxide is a more effective reducing agent than carbon below 983 K but above this temperature the reverse is true.

Watch 3-min video & get full concept clarity