In the context of the Ellingham diagram, evaluate the theoretical change in the number of gaseous mo — Metallurgy and Isolation of Elements Chemistry Question

💡 Solution & Explanation

The reaction is $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$. The number of gaseous moles of products is 1 ($CO_2$), and the number of gaseous moles of reactants is 1 ($O_2$). Therefore, $\Delta n_g = 1 - 1 = 0$. This corresponds to $\Delta S \approx 0$, explaining why the line for this reaction on the Ellingham diagram is nearly horizontal.