A) temperature is increased at constant volume
B) inert gas is added to the mixture
C) \[{{O}_{2}}\] is removed from the mixture
D) volume of the reaction flask is decreased
Correct Answer: D
Solution :
\[2S{{O}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g);\]\[\Delta H=-188.3kJ\]Since, the reaction is exothermic, according to Le-Chatelier principle, increase in temperature shifts the equilibrium in backward direction, ie, less moles of \[S{{O}_{3}}\] 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 \[S{{O}_{3}}\] are obtained.
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