Question

Lexan is a plastic used to make compact discs, eyeglass lenses, and bulletproof glass. One of the compounds used to make Lexan is phosgene \(\left(\mathrm{COCl}_{2}\right),\) an extremely poisonous gas. Phosgene decomposes by the reaction,$$\operatorname{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)$$,for which \(K_{\mathrm{p}}=6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C}\). If pure phosgene at an initial pressure of 1.0 atm decomposes, calculate the equilibrium pressures of all species.

Short answer

The equilibrium pressures of the species involved in the decomposition of phosgene gas (\(\mathrm{COCl}_{2}\)) are approximately 0.99992 atm for COCl2, 8.2 × 10^{-5} atm for CO, and 8.2 × 10^{-5} atm for Cl2.

Step-by-step solution

Set up ICE table

An ICE table keeps track of the initial pressures of the species involved, the change in pressure, and the equilibrium pressures. The equilibrium constant expression for the given reaction is: \(K_\mathrm{p} = \frac{[\mathrm{CO}][\mathrm{Cl}_{2}]}{[\mathrm{COCl}_{2}]}\) The table is filled with the following information: - The initial pressure of phosgene is 1.0 atm, and the initial pressures of CO and Cl2 are 0 atm, as they are not initially present. - We will set the change in pressure for CO and Cl2 as x and phosgene as -x, since they are being formed and phosgene is decomposing. - The equilibrium pressures of the species are: COCl2 = 1.0 - x, CO = x, and Cl2 = x.

Write the expression for \(K_{\mathrm{p}}\)

Substituting the equilibrium pressures from the ICE table into the equilibrium constant expression, we have: \(K_{\mathrm{p}} = \frac{x^{2}}{(1.0 - x)}\) We are given that \(K_{\mathrm{p}} = 6.8 \times 10^{-9}\). So, we can solve for x: \(6.8 \times 10^{-9} = \frac{x^{2}}{(1.0 - x)}\)

Solve for x

Since \(K_{\mathrm{p}}\) is very small, we can use the approximation that x is much smaller than 1. Therefore, the denominator 1.0 - x can be approximated as 1.0. This simplifies the equation to: \(6.8 \times 10^{-9} = \frac{x^{2}}{1.0}\) Rearranging and taking the square root of both sides, we get: \(x = \sqrt{6.8 \times 10^{-9}}\) Now, calculate the value of x: \(x \approx 8.2 \times 10^{-5}\)

Find the equilibrium pressures

Using the value of x, we can now calculate the equilibrium pressures of each species from the equilibrium values in the ICE table: For COCl2: 1.0 - x ≈ 1.0 - 8.2 × 10^{-5} = 0.99992 atm For CO: x = 8.2 × 10^{-5} atm For Cl2: x = 8.2 × 10^{-5} atm The equilibrium pressures of the species are approximately 0.99992 atm for COCl2, 8.2 × 10^{-5} atm for CO, and 8.2 × 10^{-5} atm for Cl2.

Key concepts

ICE Table

The ICE Table is a useful tool in chemical equilibrium problems. It stands for Initial, Change, and Equilibrium. This table helps keep track of the concentrations, or in some cases pressures, of all species in a reaction throughout the equilibrium process.

• **Initial:** Start by writing the initial pressures or concentrations before any reaction occurs.
• **Change:** Determine how the pressures or concentrations change as the reaction proceeds toward equilibrium. We use variables like x to represent these changes.
• **Equilibrium:** Calculate the final pressures or concentrations when the system reaches equilibrium by adding the initial values and the changes.

In the original problem, pure phosgene (\(COCl_2\)) starts with an initial pressure of 1.0 atm. The pressures of carbon monoxide (CO) and chlorine (\(Cl_2\)) start at 0 atm, as they are not initially present. As phosgene decomposes, x amount of CO and \(Cl_2\) are created, while phosgene decreases by x atm. Thus, the equilibrium pressure expressions are \(COCl_2 = 1.0 - x\), \(CO = x\), and \(Cl_2 = x\).

Equilibrium Constant

An Equilibrium Constant (\(K_p\) or \(K_c\)) helps understand the ratio of the concentrations or partial pressures of products to reactants at equilibrium. For reactions involving gases, \(K_p\) is often used because it considers the partial pressures.
For the decomposition of phosgene: \[COCl_2(g) \rightleftharpoons CO(g) + Cl_2(g)\]the equilibrium constant expression based on partial pressures is:\[K_{p} = \frac{[CO][Cl_2]}{[COCl_2]}\]This expression tells us the ratio of the products' pressures to the reactants' pressure at equilibrium. The very small value of \(K_p\), like \(6.8 \times 10^{-9}\)in the exercise, indicates that the reaction favors the formation of reactants, meaning very little decomposition of phosgene occurs under the given conditions.

Pressure Calculations

To find the equilibrium pressures in gas-phase reactions, we can use the initial pressures, changes during the reaction, and the equilibrium constant expression. In equilibrium problems involving gases, pressure is often used instead of concentration.
After setting up the ICE table, we substitute the equilibrium expressions into the \(K_p\) equation:\[K_{p} = \frac{x^2}{1.0 - x}\]In cases where \(K_p\) is very small, we can make an approximation: since \(x\)is much smaller than initial pressures, \(1.0 - x\) can approximately equal 1.0. This simplifies the calculation and allows us to solve for \(x\). By substituting the calculated \(x\) back into the expressions from the ICE table, we determine the equilibrium pressures for all involved gases.

Decomposition Reaction

Decomposition reactions involve the breakdown of a compound into simpler products. These reactions can occur under various conditions such as heating, exposing to light, or in the presence of catalysts. In this exercise, phosgene decomposes into carbon monoxide and chlorine gas.
The reaction is reversible and reaches a state of equilibrium, where the rate of decomposition equals the rate of recombination of products to reform the reactant. Because the equilibrium constant \(K_p\) is extremely small, it suggests that at equilibrium, only a tiny fraction of phosgene decomposes. Thus, most of the mixture remains as \(COCl_2\), with only trace amounts of \(CO\)and\(Cl_2\) present at equilibrium.