Q.How is nitric acid manufactured by Ostwald's process? How does it react with the following?

Ostwald's process makes HNO3 by the catalytic oxidation of ammonia in three stages; concentrated/dilute HNO3 then oxidizes Cu, Zn, S8, and P4, itself being reduced to NO2/NO.

Ostwald's process for manufacture of HNO3

Step 1 — Catalytic oxidation of ammonia: a mixture of ammonia and excess air is passed over a Pt-Rh gauze catalyst at about 500 K and 9 bar pressure:

$4NH_3(g) + 5O_2(g) \xrightarrow[9\ bar]{Pt/Rh,\ 500\ K} 4NO(g) + 6H_2O(g)$

Step 2 — Oxidation of NO: the nitric oxide formed is further oxidized by excess air to nitrogen dioxide:

$2NO(g) + O_2(g) \rightarrow 2NO_2(g)$

Step 3 — Absorption in water: NO2 is dissolved in water to give nitric acid; the NO produced is recycled back to step 2:

$3NO_2(g) + H_2O(l) \rightarrow 2HNO_3(aq) + NO(g)$

The dilute nitric acid (~68%) obtained is concentrated by distillation.

Reactions of nitric acid

a) With Copper

Dilute HNO3 oxidizes copper, being itself reduced to NO:

$3Cu + 8HNO_3(dilute) \rightarrow 3Cu(NO_3)_2 + 2NO\uparrow + 4H_2O$

Concentrated HNO3 oxidizes copper, being reduced to NO2:

$Cu + 4HNO_3(conc.) \rightarrow Cu(NO_3)_2 + 2NO_2\uparrow + 2H_2O$

b) With Zn

With concentrated HNO3:

$Zn + 4HNO_3(conc.) \rightarrow Zn(NO_3)_2 + 2NO_2\uparrow + 2H_2O$

With very dilute HNO3 (as zinc is more reactive, reduction can go further to ammonium salt):

$4Zn + 10HNO_3(very\ dilute) \rightarrow 4Zn(NO_3)_2 + NH_4NO_3 + 3H_2O$

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