The equilibrium 2SO_{2(g)} + O_{2(g)} rightle | vedclass

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(B) According to Le Chatelier's principle,if the concentration of a product is decreased,the equilibrium shifts in the forward direction to counteract

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$B$. An adsorbent is used to remove $SO_3$ as soon as it is formed

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(B) According to Le Chatelier's principle,if the concentration of a product is decreased,the equilibrium shifts in the forward direction to counteract the change.$A$. $A$ catalyst only increases the rate of both forward and backward reactions equally,so it does not shift the equilibrium.$B$. Removing $SO_3$ as soon as it is formed decreases the concentration of the product,which shifts the equilibrium forward to produce more $SO_3$.$C$. The reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ involves $3$ moles of gaseous reactants and $2$ moles of gaseous products. According to Le Chatelier's principle,low pressure shifts the equilibrium towards the side with more moles of gas,which is the backward direction.$D$. Removing reactants shifts the equilibrium in the backward direction.

The equilibrium $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ shifts forward,if

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