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(A) The reaction is $CO + Cl_2 \rightleftharpoons COCl_2$. Initially,$CO = 2\,mol$,$Cl_2 = 3\,mol$,$COCl_2 = 0\,mol$. At equilibrium,$CO = 1\,mol$. Ch

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$2.5$

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(A) The reaction is $CO + Cl_2 ightleftharpoons COCl_2$. Initially,$CO = 2\,mol$,$Cl_2 = 3\,mol$,$COCl_2 = 0\,mol$. At equilibrium,$CO = 1\,mol$. Change in $CO = 2 - 1 = 1\,mol$ reacted. According to stoichiometry,$1\,mol$ of $CO$ reacts with $1\,mol$ of $Cl_2$ to produce $1\,mol$ of $COCl_2$. At equilibrium: $[CO] = \frac{1\,mol}{5\,L} = 0.2\,M$,$[Cl_2] = \frac{3-1\,mol}{5\,L} = \frac{2\,mol}{5\,L} = 0.4\,M$,$[COCl_2] = \frac{1\,mol}{5\,L} = 0.2\,M$. $K_c = \frac{[COCl_2]}{[CO][Cl_2]} = \frac{0.2}{0.2 	imes 0.4} = \frac{1}{0.4} = 2.5$.

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