\( V_2O_5 \) catalyst is used in the manufacturing of
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Sulphuric acid
Transition metals and their compounds are good catalysts because they can show several oxidation states and form complexes. Vanadium(V) oxide is the catalyst of the Contact process for sulphuric acid, where it speeds up the oxidation of sulphur dioxide to sulphur trioxide:
\( \displaystyle \ce{2SO2 + O2} \xrightarrow{\ce{V2O5}} \ce{2SO3} \)
Sulphur trioxide is absorbed in concentrated sulphuric acid to give oleum, which is diluted with water:
\( \displaystyle \ce{SO3 + H2SO4 -> H2S2O7} \)
\( \displaystyle \ce{H2S2O7 + H2O -> 2H2SO4} \)
During the reaction, V(V) is reduced by \( \ce{SO2} \) and oxidised back by \( \ce{O2} \); this change of oxidation state is what makes \( \ce{V2O5} \) a catalyst.
Why the other options are incorrect
Option (1) Ammonia — Incorrect. It is made by the Haber process, with finely divided iron as the catalyst and molybdenum as a promoter: \( \ce{N2 + 3H2 <=> 2NH3} \).
Option (2) Urea — Incorrect. It is made from ammonia and carbon dioxide under pressure, without \( \ce{V2O5} \): \( \ce{2NH3 + CO2 -> NH2CONH2 + H2O} \).
Option (4) Nitric acid — Incorrect. It is made by the Ostwald process, in which ammonia is oxidised over a platinum-rhodium gauze: \( \ce{4NH3 + 5O2 -> 4NO + 6H2O} \).
NCERT (2026-27 edition):
Catalytic Properties
