Problem 65 - Chemistry

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.

Step-by-Step Solution

Answer

The equilibrium pressures for COCl\(_2\), CO, and Cl\(_2\) in the decomposition of phosgene are approximately \(0.999974 \, \mathrm{atm}\), \(2.61 \times 10^{-5} \, \mathrm{atm}\), and \(2.61 \times 10^{-5} \, \mathrm{atm}\), respectively.

Step 1: Define the change in pressure

Let the change in pressure during the reaction be represented by \(x\). This means that the phosgene will decrease by \(x\) atm, and both CO and Cl\(_2\) will increase by \(x\) atm. Then, the equilibrium pressures of COCl\(_2\), CO, and Cl\(_2\) will be \((1.0-x)\) atm, \(x\) atm, and \(x\) atm respectively.

Step 2: Write the expression for \(K_{p}\) and solve for \(x\)

Now we can write the expression for \(K_{p}\): \[K_{p} = \frac{[\mathrm{CO}][\mathrm{Cl}_{2}]}{[\mathrm{COCl}_2]}\] Since we are working with pressures, the brackets represent the partial pressures of the species: \[K_{p} = \frac{(x)(x)}{(1 - x)}\] Given the value of \(K_{p} = 6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C}\), we can now substitute: \[\begin{equation} 6.8 \times 10^{-9} = \frac{x^{2}}{(1 - x)} \end{equation}\] This is a quadratic equation in \(x\). We can solve it by rearranging and simplifying: \[\begin{equation} x^2 + (6.8 \times 10^{-9})(x) - 6.8 \times 10^{-9} = 0 \end{equation}\] Now we can solve for \(x\) using the Quadratic Formula: \[\begin{equation} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{equation}\] where \(a = 1\), \(b = 6.8 \times 10^{-9}\), and \(c = -6.8 \times 10^{-9}\). Plugging these values, we only consider the positive root since the pressure cannot be negative: \[x \approx 2.61 \times 10^{-5}\]

Step 3: Determine the equilibrium pressures

Using the value of \(x\), we can now determine the equilibrium pressures for each species in the reaction: \[\mathrm{COCl}_{2}: \: 1.0 - 2.61 \times 10^{-5} \approx 0.999974 \, \mathrm{atm}\] \[\mathrm{CO}: \: 2.61 \times 10^{-5} \, \mathrm{atm}\] \[\mathrm{Cl}_{2}: \: 2.61 \times 10^{-5} \, \mathrm{atm}\] Hence, the equilibrium pressures for COCl\(_2\), CO, and Cl\(_2\) are \(0.999974 \, \mathrm{atm}\), \(2.61 \times 10^{-5} \, \mathrm{atm}\), and \(2.61 \times 10^{-5} \, \mathrm{atm}\), respectively.

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to a state where the concentrations or pressures of the reactants and products remain constant over time. This occurs when the forward and reverse reactions balance each other out. In our example, phosgene decomposes into CO and Cl\(_2\). Initially, only phosgene is present at 1.0 atm, and as it decomposes, both CO and Cl\(_2\) start forming until equilibrium is reached.

Key points about chemical equilibrium include:

• It is dynamic, meaning reactions continue to occur at the molecular level.
• No net change in pressures or concentrations is observed at equilibrium.
• The equilibrium constant, like our given \(K_{p} = 6.8 \times 10^{-9}\), quantifies the ratio of product pressures to reactant pressures at equilibrium.

Understanding chemical equilibrium helps predict how reactions behave under different conditions and is crucial for controlling industrial processes, ensuring safety, and optimizing product yields.

Quadratic Equation

The quadratic equation is a vital tool in solving equilibrium problems, especially when dealing with unknown pressures or concentrations. In the decomposition of phosgene, we derived a quadratic equation to find the equilibrium pressures. The initial setup was based on the expression for \(K_{p}\):

\[6.8 \times 10^{-9} = \frac{x^{2}}{(1 - x)}\]

This equation represents a typical quadratic form \(ax^2 + bx + c = 0\). Using the solutions of quadratic equations, particularly the quadratic formula:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

We solved for \(x\) considering our specific values of \(a\), \(b\), and \(c\), discarding the negative root since pressure cannot be negative. Solving quadratic equations in this way helps us determine how the system behaves and reach the correct equilibrium pressures. Thus, mastering this math skill is essential when dealing with equilibrium scenarios.

Le Chatelier's Principle

Le Chatelier's Principle is a principle used to predict the effect of a change in conditions on chemical equilibria. It states that if a dynamic equilibrium is disturbed by changing the conditions, the system will respond to counteract the change and re-establish equilibrium.

In the phosgene decomposition example, if we were to increase the pressure or add more CO or Cl\(_2\), the equilibrium would shift to counteract this change. For instance, adding more Cl\(_2\) would drive the reaction towards forming more phosgene, reducing the disturbance.

Key applications of Le Chatelier's Principle include:

• Predicting how changes in pressure, temperature, or concentration will affect the equilibrium position.
• Understanding the reversibility of reactions and how to manipulate product yields.
• It is especially helpful in designing chemical processes and optimizing industrial reactions.

Le Chatelier's Principle is an indispensable concept for chemists and engineers, allowing them to control and optimize various chemical systems.