For the formation of which compound in the El | vedclass

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(A) In the Ellingham diagram,the slope of the line for the formation of $CO$ is negative because the reaction $2C(s) + O_2(g) \rightarrow 2CO(g)$ invo

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$CO$

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(A) In the Ellingham diagram,the slope of the line for the formation of $CO$ is negative because the reaction $2C(s) + O_2(g) \rightarrow 2CO(g)$ involves an increase in the number of moles of gas $(\Delta n_g = 2 - 1 = 1)$.According to the equation $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$,since $\Delta S^{\circ}$ is positive,the term $-T\Delta S^{\circ}$ becomes more negative as temperature $(T)$ increases.Therefore,the formation of $CO$ becomes more thermodynamically stable and favorable at higher temperatures compared to metal oxides like $FeO$,$ZnO$,or $Cu_2O$,which show positive slopes.

For the formation of which compound in the Ellingham diagram does $\Delta G^{\circ}$ become more and more negative with an increase in temperature?

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