In the Ellingham diagram for the formation of metal oxides, the plot for the reaction exhibits a uni — Metallurgy and Isolation of Elements Chemistry Question
Question
In the Ellingham diagram for the formation of metal oxides, the plot for the reaction $2C_{(s)} + O_{2(g)} \rightarrow 2CO_{(g)}$ exhibits a unique characteristic compared to most metal oxidation curves. What explains this specific slope?
💡 Solution & Explanation
In the reaction $2C_{(s)} + O_{2(g)} \rightarrow 2CO_{(g)}$, 1 mole of oxygen gas is consumed to produce 2 moles of carbon monoxide gas. Since the number of gaseous moles increases, the entropy change ($\Delta S$) is positive. According to $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, a positive $\Delta S^\circ$ multiplied by a negative sign yields a negative slope for the $\Delta G^\circ$ vs $T$ plot.