For the reaction SO_{2(g)} + 1/2 O_{2(g)} rig | vedclass

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Question

(B) The given reaction is $SO_{2(g)} + 1/2 O_{2(g)} \rightleftharpoons SO_{3(g)}$.In this reaction,the number of moles of gaseous products is $1$ and

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$P_1 < P_2 < P_3$

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(B) The given reaction is $SO_{2(g)} + 1/2 O_{2(g)} \rightleftharpoons SO_{3(g)}$.In this reaction,the number of moles of gaseous products is $1$ and the number of moles of gaseous reactants is $1 + 0.5 = 1.5$.According to Le Chatelier's principle,an increase in pressure shifts the equilibrium towards the side with fewer moles of gas.Since the product side has fewer moles $(1 From the graph,for a given temperature,the yield follows the order $P_3 > P_2 > P_1$.Therefore,the correct relationship is $P_3 > P_2 > P_1$,which is equivalent to $P_1 < P_2 < P_3$.

For the reaction $SO_{2(g)} + 1/2 O_{2(g)} \rightleftharpoons SO_{3(g)}$,the percentage yield of product at different pressures is shown in the figure. Then,which among the following is true?

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