Answer: Option D
:
D
$2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right):\Delta H=-197kJ.mo{l}^{-1}$
In the contact process for the manufacture of Sulphuric acid, $V_2{O}_5$ is the preferred catalyst for the above reaction. Earlier, platinated asbestos was used. As can be seen from the above equation, the enthalpy change is negative which implies that the reaction should be favoured in the forward direction for lower temperatures. However, in real scenarios a lower temperature could bring down the rate of the oxidation to unfeasible levels. So we don’t mind conducting the reaction at ${450}^{\circ }C$ and 1-2 atm pressure. Do note that the catalyst only accelerates the rate of the reaction without altering the position of the chemical equilibrium.
First the $V^{5+}$ ions oxidize $S{O}_2$ (g) into $S{O}_3$ (g) and themselves get reduced to $V^{4+}$ ions. Naturally the doubt arises as to whether this involvement of Vanadium ions affects the equilibrium. Through catalyst regeneration, the $V^{4+}$ ions are oxidized back to $V^{5+}$ ions such that Vanadium ions are not part of the overall reaction!