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Question: The catalyst used in making \[{H_2}S{O_4}\]​ in contact process is: A.\[{V_2}{O_5}\]​ B.\[F{e_2}...

The catalyst used in making H2SO4{H_2}S{O_4}​ in contact process is:
A.V2O5{V_2}{O_5}​
B.Fe2O3F{e_2}{O_3}
C.Cr2O3C{r_2}{O_3}
D.CrO3Cr{O_3}​

Explanation

Solution

Contact process is the most commonly used and commercially important method used for preparation of concentrated sulphuric acid in industries, it involves production of sulphuric acid using Sulphur trioxide as raw material in presence of a catalyst.

Complete step-by-step answer: During contact process we require sulphur trioxide, oxygen and a catalyst as raw materials for the reaction. A catalyst is a species that alters the rate of reaction without itself getting affected during the reaction, any species which fits this criteria is ready to go for use in reactions.
The use of catalyst in contact process is mainly to allow the reactions to take place at temperatures lower that what is normally used as according to Le-chatlier’s principle if we reduce the temperature of a exothermic reaction we enhance its product yield on the right hand side thus catalyst play a crucial role during contact process. However when we talk about industrial production we need to take care of the cost of managing the catalyst. Initially when the process was developed platinum metal was used as catalyst however due to its high susceptibility to arsenic which is present in sulphur trioxide raw material as impurity we replaced it with Vanadium pentoxide i.e.V2O5{V_2}{O_5} which is highly preferred and used today in industries.

Additional Information: Mechanism for the action of the catalyst comprises two steps:
Oxidation of SO2S{O_2}into SO3S{O_3}by V5+{V^{5 + }}:
2SO2  + 4V5+  + 2O2−  → 2SO3  + 4V4+2S{O_2}\; + {\text{ }}4{V^{5 + }}\; + {\text{ }}2{O^{2 - }}\; \to {\text{ }}2S{O_3}\; + {\text{ }}4{V^{4 + }}
Oxidation of V4+{V^{4 + }} back into V5+{V^{5 + }} by dioxygen (catalyst regeneration):
4V4+  + O2  → 4V5+  + 2O2−4{V^{4 + }}\; + {\text{ }}{O_2}\; \to {\text{ }}4{V^{5 + }}\; + {\text{ }}2{O^{2 - }}

Thus option (A) V2O5{V_2}{O_5} is the correct answer.

Note: To be prepared for these types of reactions in exams we need to practise more and more questions dealing with various preparation methods as this is a topic that needs to be memorized there is no way out of it.