$Δ G^{o}$ versus $T$ plot in the Ellingham's diagram slopes downward for which of the reaction?

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

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

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2 A g + \frac{1}{2} O_{2} \to A g_{2} O

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M g + \frac{1}{2} O_{2} \to M g O

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Solution

The correct option is A $C + \frac{1}{2} O_{2} \to C O$

$C (s) + O_{2} (g) \to C O_{2} (g)$ : Entropy of solids is negligible. So here one molecule of gas is resulting in one molecule of gas. Hence there is almost no net entropy. So there will be no slope, it is completely horizontal.

$C (s) + \frac{1}{2} O 2 (g) \to 2 C O (g)$ : Here one mole of gas is giving you two moles of gas as products. So here the entropy will be positive. And as a result, this curve will go downwards.

Hence, the correct option is $A$

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Similar questions

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 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:

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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