Question

The $\frac{{\Delta }_f{G}^{\circ }}{T}$ line for $C+O_2$ $\to$ $C{O}_2$ is nearly parallel to the ${\Delta }_f{G}^{\circ }$ = 0 – True or False?

• A. True
• B. False

Answer 1

The correct option is A

True

The $\frac{{\Delta }_f{G}^{\circ }}{T}$ line for $C+O_2$ $\to$ $C{O}_2$ is nearly parallel to the ${\Delta }_f{G}^{\circ }$ = 0

On the left hand side of the above reaction, there is one mole of gas. How many moles of gas are there on the products side? What can we infer about the change in entropy (assuming we are talking about a particular temperature)? For the same temperature, there are equal number of moles of gas on either side. In other words, volumes of gases on either side are equal. Without any loss in generality, the entropy change is negligible as given by the formula ∆G = ∆H - T.∆S

So in the ∆G vs T graph (which is what Ellingham diagram is - isn't it?) the line C + O₂ ⟶ CO₂ is nearly parallel to the ∆_f Gº = 0