Select ALL correct statements about the oxidation of SO2 in the contact process using V2O5.

SO2 is oxidised from S(+4) to S(+6)

V2O5 is reduced to V2O4 by SO2, with vanadium going from +5 to +4

V2O4 is re-oxidised to V2O5 by O2

SO2 is oxidised from S(+4) to S(+6)

The catalyst shifts the equilibrium position towards SO3 by making the reaction more exothermic

V2O5 is reduced to V2O4 by SO2, with vanadium going from +5 to +4

V2O4 is re-oxidised to V2O5 by O2

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3df052

Select ALL correct statements about the oxidation of $SO X_{2}$ in the contact process using $V X_{2} O X_{5}$ .

A $SO X_{2}$ is oxidised from S(+4) to S(+6)
BThe catalyst shifts the equilibrium position towards $SO X_{3}$ by making the reaction more exothermic
C $V X_{2} O X_{5}$ is reduced to $V X_{2} O X_{4}$ by $SO X_{2}$ , with vanadium going from +5 to +4
D $V X_{2} O X_{4}$ is re-oxidised to $V X_{2} O X_{5}$ by $O X_{2}$

The cycle

V X_{2} O X_{5} + SO X_{2} V X_{2} O X_{4} + SO X_{3}

, then

2 V X_{2} O X_{4} + O X_{2} 2 V X_{2} O X_{5}

shows V(+5)/V(+4) and S(+4) to S(+6). A catalyst changes the rate but not

Δ H

or the equilibrium position, so the last statement is false.