Two systems PCl_{5(g)} rightleftharpoons PCl_ | vedclass

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(C) $1$. Consider the two equilibrium reactions occurring in the same container: $(i)$ $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ (ii) $CO

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$PCl_5$ decreases.

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(C) $1$. Consider the two equilibrium reactions occurring in the same container: $(i)$ $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ (ii) $COCl_{2(g)} \rightleftharpoons CO_{(g)} + Cl_{2(g)}$ $2$. When $CO_{(g)}$ is added to the container,according to Le Chatelier's principle,the equilibrium of reaction (ii) will shift in the backward direction to consume the added $CO_{(g)}$. $3$. As reaction (ii) shifts backward,the concentration of $Cl_{2(g)}$ decreases. $4$. Now,consider reaction $(i)$. Since the concentration of $Cl_{2(g)}$ has decreased,the equilibrium of reaction $(i)$ will shift in the forward direction to compensate for the loss of $Cl_{2(g)}$. $5$. As reaction $(i)$ shifts forward,$PCl_{5(g)}$ dissociates to form more $PCl_{3(g)}$ and $Cl_{2(g)}$. $6$. Therefore,the amount of $PCl_{5(g)}$ decreases.

Two systems $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ and $COCl_{2(g)} \rightleftharpoons CO_{(g)} + Cl_{2(g)}$ are in equilibrium simultaneously in a container of constant volume. If some $CO_{(g)}$ is added to the container at constant volume,then at the new equilibrium:

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