Phosgene (COCl₂) is a toxic gas that was used as a chemical weapon in World War 1 - Leaving Cert Chemistry - Question 9 - 2018

The equilibrium constant (Kc) expression for the reaction is given by:

$K_c = \frac{[COCl_2]}{[CO][Cl_2]^{1/2}}$

where [COCl₂] is the concentration of phosgene, [CO] is the concentration of carbon monoxide, and [Cl₂] is the concentration of chlorine.

Increasing the pressure by reducing the container size shifts the equilibrium to the side with fewer moles of gas. In this case, since there are 2 moles of reactants (1 mole CO and 1/2 mole Cl₂) and 1 mole of product (1 mole COCl₂), the equilibrium will shift to the right, resulting in a darker green appearance as more phosgene is formed.

Using a higher temperature generally favors the endothermic reaction according to Le Châtelier's principle. Since the formation of phosgene from CO and Cl₂ is an exothermic process (as indicated by the negative ΔH), increasing the temperature will shift the equilibrium to the left, resulting in a smaller yield of phosgene.

The value of the equilibrium constant (Kc) is a function of temperature and is not affected by the presence of a catalyst. A catalyst speeds up the rate at which equilibrium is reached but does not alter the position of the equilibrium or the Kc value. Therefore, there is no effect on Kc when using a charcoal catalyst.

Initially, let:

• [CO] = 0.200 moles
• [Cl₂] = 0.200 moles

At equilibrium, 85.0% of the chlorine has reacted:

• Moles of Cl₂ reacted = 0.200 * 0.85 = 0.170 moles
• Moles of Cl₂ remaining = 0.200 - 0.170 = 0.030 moles
• Moles of COCl₂ formed = 0.170 moles (since 1:1 molar ratio with Cl₂)

Using the equilibrium concentrations in a 12.0L container:

• Concentration of CO: [ \frac{0.200 - 0.170}{12.0} = 0.0025 , ext{mol L}^{-1}
  ]
• Concentration of Cl₂: [ \frac{0.030}{12.0} = 0.0025 , ext{mol L}^{-1}
  ]
• Concentration of COCl₂: [ \frac{0.170}{12.0} = 0.01417 , ext{mol L}^{-1}
  ]

Substituting these values into the Kc expression:

$K_c = \frac{[COCl_2]}{[CO][Cl_2]^{1/2}} = \frac{0.01417}{0.0025 \times \sqrt{0.0025}} = 2266.67 \, ext{mol}^{-1}$

Using low temperatures and high pressures for chemical processes can be costly. Low temperatures may lead to slower reaction rates, requiring prolonged processing times, and high pressures can necessitate specialized equipment that is expensive to build and maintain. Moreover, at high pressures, safety risks such as vessel failure or leaks increase, making these conditions less desirable for large-scale industrial processes.