Question

Lexan is a plastic used to make compact discs, eyeglass lenses, and bullet- proof 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 $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ for which \(K_{\mathrm{p}}=6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C}\). If pure phosgene at an jnitial pressure of \(1.0\) atm decomposes, calculate the equilibrium pressures of all species.

Short answer

The equilibrium pressures of all species are \(\mathrm{COCl}_{2}: 1.0\; \mathrm{atm}\), \(\mathrm{CO}: 2.6\times10^{-5}\; \mathrm{atm}\), and \(\mathrm{Cl}_{2}: 2.6\times10^{-5}\; \mathrm{atm}\).

Step-by-step solution

Write the equilibrium expression

The equilibrium expression for the given reaction is: $$ K_{\mathrm{p}} = \frac{[\mathrm{CO}] [\mathrm{Cl}_{2}]}{[\mathrm{COCl}_{2}]} $$ where \(K_{\mathrm{p}} = 6.8 \times 10^{-9}\), \([\mathrm{CO}]\) is the partial pressure of \(\mathrm{CO}\), \([\mathrm{Cl}_{2}]\) is the partial pressure of \(\mathrm{Cl}_{2}\), and \([\mathrm{COCl}_{2}]\) is the partial pressure of \(\mathrm{COCl}_{2}\).

Set up an ICE table

An ICE (Initial, Change, and Equilibrium) table is a way to organize information about concentrations or pressures for a reaction at equilibrium. In this case, we will organize the partial pressures of the three species. It looks like this: ``` Initial Change Equilibrium COCl2 1.0 atm -x 1.0-x CO 0 atm +x x Cl2 0 atm +x x ``` Here, "Initial" refers to the initial pressures, "Change" refers to the changes in pressures, and "Equilibrium" refers to the equilibrium pressures.

Use the equilibrium expression with the ICE table values

Now, we can plug in the equilibrium pressures from the ICE table into our equilibrium expression: $$ 6.8 \times 10^{-9} = \frac{x \cdot x}{1.0-x} $$

Solve for x

To simplify the expression, we can rewrite it as: $$ x^2 = 6.8 \times 10^{-9}(1.0-x) $$ Since the value of \(K_{\mathrm{p}}\) is quite small, we can make an approximation that x will be much smaller than 1.0. Therefore, we can write the equation as: $$ x^2 = 6.8 \times 10^{-9} $$ Taking the square root of both sides, we get: $$ x = \sqrt{6.8 \times 10^{-9}} = 2.6\times10^{-5} $$

Calculate the equilibrium pressures of all species

Based on our ICE table, the equilibrium pressures of the three species are as follows: $$ [\mathrm{COCl}_{2}]_{\text{eq}} = 1.0 - x = 1.0 - 2.6\times10^{-5} \approx 1.0\; \mathrm{atm} $$ $$ [\mathrm{CO}]_{\text{eq}} = x = 2.6\times10^{-5}\; \mathrm{atm} $$ $$ [\mathrm{Cl}_{2}]_{\text{eq}} = x = 2.6\times10^{-5}\; \mathrm{atm} $$ So, the equilibrium pressures of all species are \(\mathrm{COCl}_{2}: 1.0\; \mathrm{atm}\), \(\mathrm{CO}: 2.6\times10^{-5}\; \mathrm{atm}\), and \(\mathrm{Cl}_{2}: 2.6\times10^{-5}\; \mathrm{atm}\).

Key concepts

ICE Table

An ICE table is a helpful tool used in chemistry to organize the initial, change, and equilibrium values of reactants and products. This method simplifies the process of solving equilibrium problems by clearly laying out all the necessary data.

In the context of the decomposition of phosgene, we use an ICE table to track the partial pressures:

• Initial: We start with 1.0 atm of COCl₂, while CO and Cl₂ are at 0 atm.
• Change: As phosgene decomposes, it loses pressure 'x' while CO and Cl₂ gain pressure 'x'.
• Equilibrium: At equilibrium, we have (1.0 - x) atm for COCl₂ and 'x' atm each for CO and Cl₂.

By setting up this table, we clearly define the pathway from the start of the reaction to its equilibrium mixture, making it easier to substitute values into equilibrium expressions.

Equilibrium Expression

The equilibrium expression is a mathematical representation used to relate the concentrations or pressures of reactants and products at equilibrium. It helps to quantify the state of the reaction when it has reached stability, meaning the rates of the forward and reverse reactions are equal.

For the decomposition of phosgene into carbon monoxide and chlorine, the expression is written as:\[K_{\text{p}} = \frac{[\text{CO}] [\text{Cl}_{2}]}{[\text{COCl}_{2}]}\]
Here, \(K_{\text{p}}\), the equilibrium constant for gas-phase reactions, is given as \(6.8 \times 10^{-9}\). This formula lets us know how far the reaction proceeds before reaching equilibrium:

• A small \(K_{\text{p}}\) value, like in this reaction, means that at equilibrium, most of the reactant remains, and very little product forms.
• To find the equilibrium pressures, we substitute the pressures from the ICE table into this expression.

Understanding this expression is crucial because it predicts which way the reaction favours, reactants or products, under given conditions.

Partial Pressures

Partial pressure is the pressure exerted independently by a particular gas in a mixture. Understanding partial pressures is essential for solving equilibrium problems because these values help indicate how gases behave and interact in a closed system.

In our reaction, the partial pressures are denoted by concentrations in brackets. At the beginning, the partial pressure of COCl₂ is 1.0 atm, while CO and Cl₂ both start at 0 atm.

• As the reaction proceeds to equilibrium, the pressure of COCl₂ decreases by an amount 'x', resulting in partial pressures of \(1.0 - x\).
• In contrast, the pressures of both CO and Cl₂ rise by 'x', finishing at equilibrium with pressures of 'x' each.

These partial pressures are critical for applying them in the equilibrium expression to determine if the reaction favours reactants heavily, indicated by the small change or high remaining pressure of the reactant, COCl₂.

Kp Constant

The Kp constant is specific to reactions involving gases. It connects the equilibrium pressures of gaseous reactants and products in the equilibrium expression. This constant is calculated or provided at specific temperatures for a specific reaction.

For the decomposition of phosgene:

• The given \(K_{\text{p}}\) is \(6.8 \times 10^{-9}\) at 100°C. This small value tells us the reaction heavily favours the reactant side (COCl₂ state) at equilibrium, meaning very little decomposition occurs.
• Using \(K_{\text{p}}\), we set up the equilibrium expression with the calculated partial pressures.
• A low \(K_{\text{p}}\) shows the tendency of the system to remain mostly as the reactants, as shown with negligible product formation.

This constant allows chemists to predict the behaviour of gaseous reactions under varying conditions and understand the position of equilibrium in practical contexts.