A) \[\text{145}0\text{ kJ mo}{{\text{l}}^{\text{-1}}}\]
B) \[\text{86}%\text{M}{{\text{g}}^{+}}+\text{14}%\text{M}{{\text{g}}^{\text{2}}}^{+}\]
C) \[69%M{{g}^{+}}+31%M{{g}^{2+}}\]
D) \[\text{14}%\text{M}{{\text{g}}^{+}}+\text{86}%\text{M}{{\text{g}}^{\text{2+}}}\]
Correct Answer: A
Solution :
In the reaction, \[C+\frac{1}{2}{{O}_{2}}\to CO\], the \[\Delta S\] increases. Therefore as the temperature increases, \[T\Delta S\]increases and thus \[\Delta G(\Delta H-T\Delta S)\]decreases. In other words, the slope of the curve for the formation of CO decreases, whereas for all other oxides, it increases.
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