Question

Why \[{V_2}{O_5}\] is used In Contact Process?

Answer 1

Hint: Contact process is an industrial method for production of sulphuric acid (\[{H_2}S{O_4}\]) in which sulphur is burnt in air to form sulphur dioxide (\[S{O_2}\]) which further reacts with oxygen to form sulphur trioxide (\[S{O_3}\]). The oxidation of \[S{O_2}\] is a reversible and exothermic reaction. Then finally sulphur trioxide is converted into sulphuric acid.

Complete step-by-step answer: \[{V_2}{O_5}\] or vanadium pentaoxide acts as a catalyst in the contact process in form of granular or small particles under very high temperature condition of \[450\] degree Celsius and pressure of \[1 - 2\] atm.
\[2S{O_2}(g) + {O_2}(g)\xrightarrow[\begin{subarray}{l}
  450^\circ C \\
  1 - 2atm
\end{subarray} ]{{{V_2}{O_5}}}2S{O_3}(g)\]
Vanadium is a d block element and the d block elements are widely used as a catalyst because of their variable oxidation state.
Oxidation state of vanadium in \[{V_2}{O_5}\] is \[ + 5\]. During the redox reaction, \[{V_2}{O_5}\] reduces to \[{V_2}{O_4}\](\[ + 4\] oxidation state) whereas \[S{O_2}\] is oxidized to\[S{O_3}\].
\[S{O_2} + {V_2}{O_5} \to S{O_3} + {V_2}{O_4}\]
But \[{V_2}{O_5}\] is a catalyst and remains unchanged at the end of the process, hence it is finally oxidized in air, back to form \[{V_2}{O_5}\].
\[{V_2}{O_4} + 1/2{O_2} \to {V_2}{O_5}\]

The role of \[{V_2}{O_5}\] is not to produce a greater amount of end product (that is sulphuric acid) but increases the rate of the reaction by lowering the activation energy required for the reaction. Thus, \[{V_2}{O_5}\] does not alter the position of equilibrium rather complete the reaction in sensible time.

Earlier platinum was used as the catalyst but it appeared to react with arsenic impurities present with sulphur to produce poisonous gases. But no such reactions are seen with \[{V_2}{O_5}\]. Moreover, \[{V_2}{O_5}\] is much cheaper.

Note: \[S{O_3}\] reacts with water to form \[{H_2}S{O_4}\] as the following reaction.
\[S{O_3}(g) + {H_2O}(g) \to {\text{ }}{H_2}S{O_4}(aq)\]
But the above reaction is highly uncontrollable with a large amount of heat release. Hence first \[S{O_3}\] is dissolved in concentrated sulphuric acid to form oleum having formula \[{H_2}{S_2}{O_7}\]by the reaction
\[{H_2}S{O_4}(aq) + S{O_3}(g) \to {H_2}{S_2}{O_7}(aq)\].
Then further it is oleum that undergoes reaction with water to form\[{H_2}S{O_4}\]. \[{H_2}{S_2}{O_7}(aq) + {H_2}O(l) \to 2{H_2}S{O_4}(aq)\]
Oleum is also called as fuming sulphuric acid and it produces double amount of \[{H_2}S{O_4}\] as much used to make oleum hence it is widely used in industries as the yield is very high.