Which of the following reactions has positive ΔS° value?

The correct answer is Option (2) → \(2C (s) + O_2(g) \rightarrow 2CO (g)\)

Entropy measures disorder, and gases have far more entropy than solids. So for a reaction involving gases, the sign of ΔS° is decided by the change in the number of moles of gas: ΔS° is positive when the moles of gas increase and negative when they decrease.

    Option (1): 1 mol gas → 0 mol gas, ΔS° negative
    Option (2): 1 mol gas → 2 mol gas, ΔS° positive
    Option (3): 3 mol gas → 0 mol gas, ΔS° negative
    Option (4): 3 mol gas → 2 mol gas, ΔS° negative

So the reaction with positive ΔS° is 2C(s) + O₂(g) → 2CO(g). This is why its line in the Ellingham diagram slopes downwards (ΔG° = ΔH° − TΔS° becomes more negative as T rises), which makes carbon a good reducing agent at high temperature.

Why the other options are incorrect

• Option (1) \(6FeO(s) + O_2(g) \rightarrow 2Fe_3O_4 (s)\) — Incorrect. A gas is used up and only a solid forms, so the disorder decreases.

• Option (3) \(4Al (s) + 3O_2 (g) \rightarrow 2Al_2O_3 (s)\) — Incorrect. Three moles of gas are consumed; ΔS° is strongly negative, as for most metal oxides in the Ellingham diagram (lines with positive slope).

• Option (4) \(2CO (g) + O_2(g) \rightarrow 2CO_2 (g)\) — Incorrect. Three moles of gas become two, so ΔS° is negative.

NCERT (Pre-rationalised edition, 2022-23):
Ellingham Diagram – “… (a) Ellingham diagram normally consists of plots of Δ_fG° vs T for formation of oxides of elements i.e., for the reaction, \(\ce{2xM(s) + O2(g) -> 2M_xO(s)}\) In this reaction, the gaseous amount (hence molecular randomness) is decreasing from left to right due to the consumption of gases leading to a –ve value of ΔS which changes the sign of the second term in equation (6.14).”