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Is the Ostwald process endothermic?

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1) If something causes an endothermic reaction does this always cause a higher yield in products if the temperature is raised?

2) If something causes an exothermic reaction does this always cause a lower yield in products if the temperature is raised?

 

This is not homework, I'm just researching the Ostwald process and had some general questions.

The sum of the reaction enthalpies for the three steps leads to an overall enthalpy of about -1181kJ/mol. So the reaction is exothermic overall. All three steps are independently exothermic as well. Increasing T for a reaction can increase yield, particularly if te transformnation is an equilibrium process. All steps of the Ostwald process are practically irreversible, except for the last step (if carried out in air):

 

 

 

[ce]4NO_2(g) + O_2(g) + 2H_2 O(l) -> 4HNO_3(aq)[/ce]

 

This is the fuming nitric acid equilibrium

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Increasing the temperature in step 1 increases the yield of products. But in step 2 of the Ostwald process, why does decreasing the temperature increase the yield of products?

Not sure about all the specifics of this particular reaction [step 2], but I think I know what principle is at play here.

 

[ce] 2NO + O_2 -> 2NO_2[/ce]

 

molecular oxygen (in the [math]T_1 [/math] ground state) is a paramagnetic diradical. NO is a radical as well. This is probably a radical process where the desired product is kinetically favored over an alternative product that is more thermodynamically favored.

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