# Why is zinc oxide reduced at a temperature of 1400 degrees Celsius?

I am attaching a graph of the Ellingham diagram on which my question is based. If we interpret the graph, $\ce{ZnO}$ is reduced by $\ce{C}$ at a temperature of $\pu{800 ^\circ C}$ to $\pu{1000 ^\circ C}$. However, the actual temperature required is about $\pu{1400 ^\circ C}$. Why is this the case?

• Kinetics? What is the source of the 1400 C value? – airhuff Feb 21 '17 at 7:14
• Which Carbon are you talking about, The carbon which is oxidizing into CO or CO2..? – Mitchell Feb 21 '17 at 10:36

If the answer is $1400\ \mathrm{^\circ C}$ it must imply that the user does not want $\ce{CO}$ in their product. Based on the diagram, the intersection of $\ce{ZnO + C -> Zn + CO}$ occurs around $1000\ \mathrm{^\circ C}$. Full conversion of $\ce{CO}$ to $\ce{CO2}$ requires a temperature of around $1400\ \mathrm{^\circ C}$ as shown in the diagram.
As to why the reaction does not follow the $\ce{C -> CO2}$ curve (no $\ce{CO}$ involved), this is a kinetic reason. Subtract the $\Delta G$ values of each line at a specific temperature (where the lines intersect, the value is 0, meaning the reductions of the two species are in equilibrium). You will find that at the temperature of intersection between $\ce{2ZnO -> 2Zn + O2}$ and $\ce{C + O2 -> CO2}$ $(\Delta G = 0)$, there is sizeable energy driving force for $\ce{ZnO + C -> Zn + CO}$ $(G = -300\ \mathrm{kJ/mol}\ \ce{ZnO})$. The difference of the Gibbs energies is related to the ratios of the $\ce{CO}$ and $\ce{CO2}$ products by
$$\Delta G_\mathrm r(\ce{ZnO + C -> Zn + CO})-\Delta G_\mathrm r(\ce{2ZnO + C -> 2Zn + CO2}) =-RT\ln\frac{[\ce{CO}]}{[\ce{CO2}]}$$
Through plugging in the number it is a very high proportion that goes through the $\ce{CO}$ pathway. Again, the only way any of this is important is if it mentions that $\ce{CO}$ must be eliminated, else the reaction proceeds at low temperature as you say.