What is the effect of using vanadium (V) oxide as catalyst in the reaction represented by the following equation? 2SO2 + O2 ⇌ 2SO3; ∆H = - xKJmol-1
Decreases the value of ∆H
Increases the collision rate of reactant particles
Shifts equilibrium position to the right
Reduces the time for attainment of equilibrium
Explanation
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Discussions (7)
The enthalpy change for any reaction,whether exothermic or endothermic is not affected by the presence of a catalyst in the reaction.
The answer is D not A.
The use of a catalyst (i.e vanadium(v) oxide) increases the reaction rate which is directly proportional to collision rate, making it to increase as well.
A catalyst can NEVER tamper with the enthalpy change of a reaction. Please take this correction into consideration.
Thank you.
The answer is Option D please , A catalyst will have no effect on the enthalpy change of a reaction rather It will reduce the time needed to attain equillibrium
option D is accurate
catalyst affect the time taken for a reaction to reach equilibrium 
Catalyst does not have anything to do with the enthalpy change and since the reaction is exothermic it tends to increase the collision of the reactants. Is it not true?
The correct option is D
Catalyst alters the activation energy and thereby reduces the time for attainment of equilibrium
