Write the balanced chemical equation and the equilibrium constant expression (K_c).

The reaction is:

$$\text{CO}(\text{g}) + \text{Cl}_2(\text{g}) \rightleftharpoons \text{COCl}_2(\text{g})$$

The equilibrium constant expression in terms of concentrations is:

$$K_c = \frac{[\text{COCl}_2]}{[\text{CO}][\text{Cl}_2]}$$

Given concentrations:
$[\text{CO}] = 0.623 \text{ M}$
$[\text{Cl}_2] = 0.068 \text{ M}$
$[\text{COCl}_2] = 0.25 \text{ M}$

$$K_c = \frac{0.25 \text{ M}}{(0.623 \text{ M})(0.068 \text{ M})}$$ $$K_c = \frac{0.25}{0.042364}$$ $$K_c \approx 5.899$$

The units for $K_c$ are $\frac{\text{M}}{\text{M} \cdot \text{M}} = \text{M}^{-1}$ or $\text{L/mol}$.

$$K_c = \boxed{5.90 \text{ L/mol}}$$

$\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$
From the balanced equation:
Moles of gaseous products = 1 (for $\text{COCl}_2$)
Moles of gaseous reactants = 1 (for $\text{CO}$) + 1 (for $\text{Cl}_2$) = 2

$$\Delta n = 1 - 2 = -1$$

Given temperature $T = 75^\circ\text{C}$.

$$T(\text{K}) = 75 + 273.15 = 348.15 \text{ K}$$

Given $R = 0.0821 \text{ atm K}^{-1} \text{ mol}^{-1}$.

$$K_p = K_c(RT)^{\Delta n}$$ $$K_p = (5.899 \text{ L/mol}) \left( (0.0821 \text{ atm K}^{-1} \text{ mol}^{-1})(348.15 \text{ K}) \right)^{-1}$$ $$K_p = (5.899 \text{ L/mol}) \left( 28.580315 \text{ atm L mol}^{-1} \right)^{-1}$$ $$K_p = (5.899 \text{ L/mol}) \left( \frac{1}{28.580315 \text{ atm L mol}^{-1}} \right)$$ $$K_p = (5.899 \text{ L/mol}) (0.034989 \text{ atm}^{-1} \text{ L}^{-1} \text{ mol})$$ $$K_p \approx 0.2064 \text{ atm}^{-1}$$ $$K_p = \boxed{0.206 \text{ atm}^{-1}}$$