$Δ G$ vs $T$ plot in the Ellingham diagram slopes downward for the reaction.

M g + \frac{1}{2} O_{2} ⟶ M g O

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2 A g + \frac{1}{2} O_{2} ⟶ A g_{2} O

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C + \frac{1}{2} O_{2} ⟶ C O

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C O + \frac{1}{2} O_{2} ⟶ C O_{2}

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Solution

The correct option is C $C + \frac{1}{2} O_{2} ⟶ C O$
In Ellingham diagram,
higher is the slope higher is the entropy change so
$Δ S (C_{(s)} + 1 / 2 O_{2 (g)} ⟶ C O_{(g)}) > Δ S (C_{(S)} + O_{2 (g)} ⟶ C O_{2 (g)}$
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.

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Similar questions

From the thermo chemical reactions,

$C_{(g r a p h i t e)} + \frac{1}{2} O_{2} \to C O; Δ H = - 110.5 K J$

$C O + \frac{1}{2} O_{2} \to C O_{2}; Δ H = - 283.2 K J .$

$Δ H$ for the reaction, $C_{(g r a p h i t e)} + O_{2} \to C O_{2}$ is:

Which one of the following chemical equations represent combustion reactions?

C + O_{2} \to C O_{2}

B: $C + \frac{1}{2} O_{2} \to C O$

C: $C O + \frac{1}{2} O_{2} \to C O_{2}$

From the thermochemical reactions,

C + 1/2O₂ → CO; ∆H = -110.5 kJ

CO + 1/2O₂→ CO₂; ∆H = -283.2 kJ

The heat of reaction C + O₂ → CO₂ is:

C

+ \frac{1}{2} O_{2} \to C O; Δ H = - 110.5

C O + \frac{1}{2} O_{2} \to C O_{2}; Δ H = - 283.2

C (graphite) + O_{2} \to C O_{2}

C + O_{2} \to C O_{2}

C + \frac{1}{2} O_{2} \to C O

C O + \frac{1}{2} O_{2} \to C O_{2}

k J / m o l

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