2SO2(g)+O2(g)⇌2SO3(g) ∆H=−188.3kJ Since, the reaction is exothermic, according to Le-Chatelier principle, increase in temperature shifts the equilibrium in backward direction, i.e. less moles of SO3 are formed. Similarly, if any of the reactant is removed, the equilibrium shifts in backward direction. For the above reaction, number of gaseous products =2 number of gaseous reactants =3 Hence, increase in pressure or decrease in volume (∵p∝
1
V
) shifts the equilibrium in forward direction, i.e. more moles of SO3 are obtained.