Question

For a reaction $2S{O}_2(s)+O_2(g)\rightleftharpoons 2S{O}_3(g);$ $\Delta H=-188.3kJ.$ The number of moles of $S{O}_3$ formed is increased if

• A. temperature is increased at constant volume
• B. Inert gas is added to the mixture
• C. ${\text{O}}_{\text{2}}$ is removed from the mixture
• D. volume of the reaction flask is decreased

Answer 1

Answer: (D)

$2S{O}_2(g)+O_2(g)\rightleftharpoons 2S{O}_3(g);$ $\Delta H=-188.3kJ$ Since, the reaction is exothermic, according to Le-Chatelier principle, increase in temperature shifts the equilibrium in backward direction, ie, less moles of $\text{S}{\text{O}}_{\text{3}}$ 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 $\left(\because p\propto \frac{1}{V}\right)$ shifts the equilibrium in forward direction, ie, more moles of $\text{S}{\text{O}}_{\text{3}}$ are obtained.