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 initial pressure of \(1.0\) atm decomposes, calculate the equilibrium pressures of all species.

Short answer

At equilibrium, the pressures for all species involved in the decomposition of phosgene \(\left(\mathrm{COCl}_{2}\right)\) are: \(P_{COCl_2} \approx 1.00\) atm, \(P_{CO} \approx 2.61 \times 10^{-5}\) atm, and \(P_{Cl_2} \approx 2.61 \times 10^{-5}\) atm.

Step-by-step solution

Write the balanced chemical equation and the expression for Kp

\[ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{Cl}_{2}(g) \] The equilibrium constant expression is: \[ K_p = \frac{P_{CO} P_{Cl_2}}{P_{COCl_2}} \]

Set up the ICE table

Construct an ICE table using the initial pressure of phosgene (1.0 atm) and assume some amount of phosgene dissociates 'x'. Then, determine the change and equilibrium pressures for each species. | | COCl2 | CO | Cl2 | |-------|-------|----|-----| | I | 1.0 | 0 | 0 | | C | -x | +x | +x | | E | 1-x | x | x |

Substitute the equilibrium pressures into the Kp expression

Substitute the equilibrium pressures from the ICE table into the Kp expression: \[ K_p = \frac{x^2}{1-x} \]

Solve for x

Now, substitute the given value of Kp in the equation and solve for x: \[ 6.8 \times 10^{-9} = \frac{x^2}{1-x} \] To simplify the calculations, we can make the assumption that x in the denominator is negligible and simplify the equation as follows: \[ 6.8 \times 10^{-9} \approx \frac{x^2}{1} \] Now, solve for x: \[ x = \sqrt{6.8 \times 10^{-9}} \approx 2.61 \times 10^{-5} \]

Calculate the equilibrium pressures for each species

Using the value of x found in Step 4, determine the equilibrium pressures for each species: \[ P_{COCl_2} = 1-x = 1 - 2.61 \times 10^{-5} \approx 1.00 \text{ atm} \] \[ P_{CO} = P_{Cl_2} = x = 2.61 \times 10^{-5} \text{ atm} \] At equilibrium, the pressures for all species are: \(P_{COCl_2} \approx 1.00\) atm, \(P_{CO} \approx 2.61 \times 10^{-5}\) atm, and \(P_{Cl_2} \approx 2.61 \times 10^{-5}\) atm.

Key concepts

Understanding ICE Tables in Equilibrium Reactions

ICE tables are extremely helpful in solving problems related to equilibrium reactions. When a chemical reaction starts, the reactants and products may not be in equilibrium immediately. An ICE table helps track these changes through a series of stages: Initial, Change, and Equilibrium.

• Initial: This row represents the initial conditions or pressures of each species before any reaction has occurred. Often, the products start at zero because no reaction has yet occurred.
• Change: This is where 'x' comes into play! It represents the change in pressure as the system shifts towards equilibrium. The reactants decrease, while the products form, so their change is represented both as negative and positive values respectively.
• Equilibrium: Finally, this row shows the pressures when the system has reached a state of equilibrium. These are calculated based on the changes happening in the reaction as per the coefficients in the balanced chemical equation.

Inserting these values into the ICE table helps clarify how different species within the reaction behave as equilibrium is approached, giving a clear and logical framework to tackle further calculations.

The Role of the Equilibrium Constant

Let’s delve into the significance of the equilibrium constant, denoted as \( K_p \). It's essential for understanding to what extent a reaction proceeds at a certain temperature.
The equilibrium constant \( K_p \) relates the pressures of gases at equilibrium for a particular reaction. In our case:

• The formula used is \( K_p = \frac{P_{CO} P_{Cl_2}}{P_{COCl_2}} \).
• This expression is derived from the law of mass action, where the concentration of products raised to their stoichiometric powers is divided by the concentration of reactants raised to their powers.

The value of \( K_p \) gives us insight:

• A large \( K_p \) indicates that at equilibrium, a reaction favors the formation of products.
• A smaller \( K_p \), like in the case of phosgene decomposition with \( K_p = 6.8 \times 10^{-9} \), means the system largely favors the reactants.

In essence, \( K_p \) helps predict the position of equilibrium and assess the dominance of reactants or products once equilibrium is reached.

Exploring Reaction Decomposition

Decomposition reactions are a fundamental type of chemical reaction. These involve the breakdown of a single compound into two or more products.
In the given exercise, phosgene decomposes into carbon monoxide \((CO)\) and chlorine gas \((Cl_2)\):\[ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\]This type of reaction is key in various industrial processes and occurs under specific conditions of pressure and temperature.Understanding decomposition helps in predicting how much of each product forms from the decomposition of a given amount of reactant. This is precisely where equilibrium concepts come into play. They assist in finding out whether the reaction truly takes place significantly enough, which is controlled by parameters including equilibrium constants.Decomposition can also be influenced by factors like

• Temperature: which affects the \( K_p \) value and thus the extent of the reaction.
• Pressure: changing pressure can shift the equilibrium position in favor of more moles or fewer moles of gas depending on conditions outlined by Le Chatelier’s Principle.

In essence, diving into the decomposition aspect underscores the careful dance required between chemical species as they reach balance in given conditions almost akin to how phosgene decomposes.