1. Tardigrade
  2. Question
  3. Chemistry
  4. For a reaction 2SO2(s)+O2(g)leftharpoons 2SO3(g); Δ H=-188.3 kJ. The number of moles of SO3 formed is increased if

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Q. For a reaction $ 2S{{O}_{2}}(s)+{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g); $ $ \Delta H=-188.3\,kJ. $ The number of moles of $ S{{O}_{3}} $ formed is increased if

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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.