In Ellingham diagram, the following graph (X) corresponds to
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a
2C(s)+O2(g)→ 2CO(g)
b
2Mg(s)+O2(g)→ 2MgO(s)
c
2CO(g)+O2(g)→ 2CO2(g)
d
C(s)+O2(g)→ CO2(g)
answer is A.
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Detailed Solution
2C(s)+O2(g)→ 2CO(g) ∆S=+ve ∆G=∆H-T∆S ∆G=∆H-T(+) As 'T' increases, ∆G decreases, ∆H remains unaffected by change in 'T' ; A straight line with negative slope is obtained;