At a certain temperature in a 5L vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction, CO + Cl$$_2$$ $$\rightleftharpoons$$ COCl$$_2$$. At equilibrium, if one mole of CO is present then equilibrium constant (K$$_c$$) for the reaction is:

The balanced chemical equation is $$\mathrm{CO + Cl_2 \rightleftharpoons COCl_2}$$. For every mole of carbon monoxide that reacts, exactly one mole of chlorine reacts and exactly one mole of phosgene ($$\mathrm{COCl_2}$$) is produced.

Initially the 5 L vessel contains 2 moles of CO and 3 moles of Cl$$_2$$, while no COCl$$_2$$ is present. Let $$x$$ be the number of moles of CO that actually react before equilibrium is established. Then the equilibrium mole numbers are obtained by simple subtraction or addition:

Initial moles: $$\mathrm{CO}=2,\; \mathrm{Cl_2}=3,\; \mathrm{COCl_2}=0$$

Change in moles: $$\mathrm{CO}=-x,\; \mathrm{Cl_2}=-x,\; \mathrm{COCl_2}=+x$$

Equilibrium moles: $$\mathrm{CO}=2-x,\; \mathrm{Cl_2}=3-x,\; \mathrm{COCl_2}=x$$

We are told that at equilibrium only one mole of CO is present. Therefore

$$2 - x = 1 \quad\Longrightarrow\quad x = 1.$$

Substituting $$x = 1$$ back into the equilibrium expressions gives

Equilibrium moles: $$\mathrm{CO}=1,\; \mathrm{Cl_2}=2,\; \mathrm{COCl_2}=1.$$

Because the volume of the vessel is 5 L, we convert these moles into molar concentrations using the definition $$\text{concentration} = \dfrac{\text{moles}}{\text{volume}}.$$

$$[\mathrm{CO}] = \dfrac{1}{5}\ \text{M} = 0.2\ \text{M}$$ $$[\mathrm{Cl_2}] = \dfrac{2}{5}\ \text{M} = 0.4\ \text{M}$$ $$[\mathrm{COCl_2}] = \dfrac{1}{5}\ \text{M} = 0.2\ \text{M}$$

The law of mass action states that the equilibrium constant $$K_c$$ for the reaction $$\mathrm{CO + Cl_2 \rightleftharpoons COCl_2}$$ is

$$K_c = \dfrac{[\mathrm{COCl_2}]}{[\mathrm{CO}]\,[\mathrm{Cl_2}]}.$$

Substituting the equilibrium concentrations calculated above, we have

$$K_c = \dfrac{0.2}{(0.2)(0.4)} = \dfrac{0.2}{0.08} = 2.5.$$

Hence, the correct answer is Option A.