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Solutions
2SO2(g)+O2(g)⇌2SO3(g) In this reaction three moles (or volumes`) of reactants are converted into two moles (or volumes) of products i.e. there is a decrease in volume and so if the volume of the reaction vessel is halved the equilibrium will be shifted to the right i.e. more product will be formed and the rate of forward reaction will increase i.e.double that of reverse reaction.