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
Correct Answer: D
Solution :
\[2S{{O}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons \,2S{{O}_{3}}(g);\] \[\Delta H=-188.3\,kJ\] 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.
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