Based on really quick look at thermochemistry using tabulated heats of formation and standard entropies for all substances in the condensed phase except CO, this is an endothermic reaction (est. ΔH ~ 650 kJ/mol) and entropically favorable due to gas production (ΔS ~ 0.7 kJ/mol K). At room temperature it's quite non spontaneous (ΔG ~ +460 kJ/mol), meaning equilibrium favors reactants. You can estimate the temperature at which it becomes favorable using ΔG=ΔH-TΔS, assuming H and S are temperature independent. Setting ΔG = 0, T ~ 915 K. And yes, using fine C particulate will probably speed the reaction up.
Of course, just because a reaction is unfavorable doesn't mean you can't form products at lower temperatures - you just need to find ways to shift the equilibrium in the desired direction (removing product CO for instance). And of course the thermodynamics don't necessarily have anything to do with the rate of the reaction. A reaction may yield product "on paper" but that doesn't mean it's efficient or that other products aren't formed under a set of reaction conditions.
Based on really quick look at thermochemistry using tabulated heats of formation and standard entropies for all substances in the condensed phase except CO, this is an endothermic reaction (est. ΔH ~ 650 kJ/mol) and entropically favorable due to gas production (ΔS ~ 0.7 kJ/mol K). At room temperature it's quite non spontaneous (ΔG ~ +460 kJ/mol), meaning equilibrium favors reactants. You can estimate the temperature at which it becomes favorable using ΔG=ΔH-TΔS, assuming H and S are temperature independent. Setting ΔG = 0, T ~ 915 K. And yes, using fine C particulate will probably speed the reaction up.
Of course, just because a reaction is unfavorable doesn't mean you can't form products at lower temperatures - you just need to find ways to shift the equilibrium in the desired direction (removing product CO for instance). And of course the thermodynamics don't necessarily have anything to do with the rate of the reaction. A reaction may yield product "on paper" but that doesn't mean it's efficient or that other products aren't formed under a set of reaction conditions.
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Of course, just because a reaction is unfavorable doesn't mean you can't form products at lower temperatures - you just need to find ways to shift the equilibrium in the desired direction (removing product CO for instance). And of course the thermodynamics don't necessarily have anything to do with the rate of the reaction. A reaction may yield product "on paper" but that doesn't mean it's efficient or that other products aren't formed under a set of reaction conditions.
Of course, just because a reaction is unfavorable doesn't mean you can't form products at lower temperatures - you just need to find ways to shift the equilibrium in the desired direction (removing product CO for instance). And of course the thermodynamics don't necessarily have anything to do with the rate of the reaction. A reaction may yield product "on paper" but that doesn't mean it's efficient or that other products aren't formed under a set of reaction conditions.
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