The equilibrium of the reaction 2A_{(s)} + 3B | vedclass

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(B) According to Le Chatelier's principle,if the pressure on a gaseous system at equilibrium is decreased,the equilibrium shifts in the direction wher

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The concentrations of $C$ and $D$ increase.

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(B) According to Le Chatelier's principle,if the pressure on a gaseous system at equilibrium is decreased,the equilibrium shifts in the direction where the number of moles of gaseous species increases.In the given reaction: $2A_{(s)} + 3B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)} + Q$.The number of moles of gaseous reactants is $3$ (from $3B_{(g)}$).The number of moles of gaseous products is $3 + 1 = 4$ (from $3C_{(g)} + D_{(g)}$).Since the number of moles of gaseous products $(4)$ is greater than the number of moles of gaseous reactants $(3)$,a decrease in pressure will shift the equilibrium towards the product side (right side).Therefore,the concentrations of products $C$ and $D$ will increase.

The equilibrium of the reaction $2A_{(s)} + 3B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)} + Q$ is represented as shown. If the pressure on the system is reduced to half of its original value,then:

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