On heating a mixture of SO_2Cl_2 and CO,two e | vedclass

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(D) According to Le Chatelier's principle,adding $SO_2$ (a product of the first equilibrium) will shift the first equilibrium $SO_2Cl_{2(g)} \rightlef

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Amount of $SO_2Cl_2$ and $CO$ will increase

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(D) According to Le Chatelier's principle,adding $SO_2$ (a product of the first equilibrium) will shift the first equilibrium $SO_2Cl_{2(g)} \rightleftharpoons SO_{2(g)} + Cl_{2(g)}$ to the left. This results in a decrease in the concentration of $Cl_2$. The second equilibrium is $CO_{(g)} + Cl_{2(g)} \rightleftharpoons COCl_{2(g)}$. Since the concentration of $Cl_2$ decreases,the second equilibrium will shift to the left to compensate for the loss of $Cl_2$. As a result,$COCl_2$ will decompose,leading to an increase in the amount of $CO$ and $Cl_2$. Therefore,the amount of $CO$ will increase.

List-$I$ (Equilibrium)	List-$II$ (Conditions for the process)
$P. A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$ (Endothermic)	$1. \text{High temperature}$
$Q. 2AB_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{3(g)}$ (Exothermic)	$2. \text{Low temperature}$
$R. 2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$ (Endothermic)	$3. \text{High temperature}$
	$4. \text{Low temperature}$
	$5. \text{Independent of pressure}$

On heating a mixture of $SO_2Cl_2$ and $CO$,two equilibria are simultaneously established: $SO_2Cl_{2(g)} \rightleftharpoons SO_{2(g)} + Cl_{2(g)}$ and $CO_{(g)} + Cl_{2(g)} \rightleftharpoons COCl_{2(g)}$. On adding more $SO_2$ at equilibrium,what will happen?

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