Problem 89 - Chemistry The Science in Context

Question

Phosgene, \(\mathrm{COCl}_{2},\) gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) $$ The value of \(K_{\mathrm{c}}\) for this reaction is 5.0 at \(600 \mathrm{K} .\) What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which \(P_{\mathrm{CO}}=P_{\mathrm{Cl}_{2}}=0.265\) atm and there is no \(\mathrm{COCl}_{2} ?\)

Step-by-Step Solution

Answer

Answer: The equilibrium partial pressures are: CO: 0.119 atm Cl₂: 0.119 atm COCl₂: 0.146 atm

Step 1: Calculate the initial moles of each gas

The initial pressures of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) are equal at \(0.265\) atm. There is no \(\mathrm{COCl}_{2}\) present in the mixture initially.

Step 2: Set up the reaction quotient expression (Qc)

Since we don't know the change in the pressures occurring at equilibrium, we can represent the equilibrium partial pressures of \(\mathrm{CO}\), \(\mathrm{Cl}_{2}\), and \(\mathrm{COCl}_{2}\) as \(0.265 - x\), \(0.265 - x\), and \(x\) respectively, where \(x\) is the change in pressure due to the reaction. Now write the reaction quotient expression (Qc) using the partial pressures: $$Q_\mathrm{c} = \frac{P_{\mathrm{COCl}_2}}{P_{\mathrm{CO}} \times P_{\mathrm{Cl}_2}} = \frac{x}{(0.265 - x)^2}$$

Step 3: Set the reaction quotient expression (Qc) equal to the given Kc value

Since the given equilibrium constant value, \(K_\mathrm{c}\), is 5.0 at 600 K, we can set the reaction quotient expression equal to the \(K_\mathrm{c}\) value: $$5.0 = \frac{x}{(0.265 - x)^2}$$

Step 4: Solve for x

To solve for \(x\), we can first multiply both sides of the equation by \((0.265 - x)^2\) to get rid of the denominator: \(5.0(0.265 - x)^2 = x\) Next, expand the equation and rearrange it by moving all terms to one side: \(5.0(0.070225 - 0.53x + x^2) - x = 0\) \(0.351125 - 2.65x + 5x^2 - x = 0\) Now, rearrange the equation and simplify: \(5x^2 - 3.65x + 0.351125 = 0\) This is a quadratic equation, so you can solve for \(x\) by using the quadratic formula or by factoring. For this exercise, we will use the quadratic formula: $$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ where \(a = 5\), \(b = -3.65\), and \(c = 0.351125\). Plugging in these values: $$x = \frac{3.65 \pm \sqrt{(-3.65)^2 - 4(5)(0.351125)}}{2(5)}$$ Solving for \(x\), we get two possible values: \(x \approx 0.146\) and \(x \approx 0.481\). However, given the initial pressures of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\), the second value of \(x\) is not physically possible (since \(0.265 - 0.481\) would result in negative pressure). Therefore, we choose the first value: $$x \approx 0.146$$

Step 5: Determine the equilibrium partial pressures

Now that we have the value of \(x\), we can find the equilibrium partial pressures of the three gases: $$P_{\mathrm{CO}} = 0.265 - x = 0.265 - 0.146 \approx 0.119 \, \mathrm{atm}$$ $$P_{\mathrm{Cl}_{2}} = 0.265 - x = 0.265 - 0.146 \approx 0.119 \, \mathrm{atm}$$ $$P_{\mathrm{COCl}_{2}} = x = 0.146 \, \mathrm{atm}$$ So, the equilibrium partial pressures are: \(\mathrm{CO}: 0.119 \, \mathrm{atm}\) \(\mathrm{Cl}_{2}: 0.119 \, \mathrm{atm}\) \(\mathrm{COCl}_{2}: 0.146 \, \mathrm{atm}\)

Key Concepts

Equilibrium Constant

Understanding the equilibrium constant (\(K\textsubscript{c}\) or \(K\textsubscript{p}\)) is crucial in grasping the nuances of chemical equilibrium. This numerical value quantifies the ratio of concentration (or pressure, in the case of gases) of the products to the reactants at equilibrium, raised to the power of their stoichiometric coefficients. The beauty of the equilibrium constant lies in its constancy for a particular reaction at a given temperature.

For example, in the reaction of carbon monoxide with chlorine to form phosgene (\( \text{CO}(g) + \text{Cl}\textsubscript{2}(g) \rightleftharpoons \text{COCl}\textsubscript{2}(g) \)), the equilibrium constant at 600 K is 5.0. This tells us that at equilibrium, the ratio of the concentration (or partial pressures, \(K\textsubscript{p}\)) of phosgene to the product of the concentrations of carbon monoxide and chlorine is 5.0, irrespective of the initial amounts, provided the system is at 600 K.

Exercise Improvement Advice

To further understand this concept, students should practice with various equilibrium constants at different temperatures to see how temperature affects the value. Additionally, calculations involving different initial concentrations or partial pressures of reactants can solidify their grasp on how equilibria shift in response to changes in reaction conditions.

Reaction Quotient

The reaction quotient (\(Q\textsubscript{c}\) or \(Q\textsubscript{p}\)) often confuses students, but it's simply a real-time snapshot of a reaction that hasn't yet reached equilibrium. Calculated the same way as the equilibrium constant, the reaction quotient is the ratio of the product concentrations (or pressures) to the reactant concentrations (or pressures), each raised to the power of their stoichiometric coefficients, at any point during the reaction.

When we compare \(Q\textsubscript{c}\) to \(K\textsubscript{c}\) (or \(Q\textsubscript{p}\) to \(K\textsubscript{p}\)), we can predict the direction in which the reaction will shift to reach equilibrium. If \(Q\textsubscript{c} < K\textsubscript{c}\) (or \(Q\textsubscript{p} < K\textsubscript{p}\)), the forward reaction is favored; if \(Q\textsubscript{c} > K\textsubscript{c}\) (or \(Q\textsubscript{p} > K\textsubscript{p}\)), the reverse reaction is favored; and if they are equal, the system is at equilibrium.

Navigating the Reaction Quotient

To improve their grasp on this, students can practice by computing \(Q\textsubscript{c}\) or \(Q\textsubscript{p}\) at different stages of a reaction to observe how it changes over time and approaches \(K\textsubscript{c}\) or \(K\textsubscript{p}\) as the system reaches equilibrium.

Partial Pressure

In the realm of gases, partial pressure plays a leading role in chemical equilibrium. It’s the individual pressure exerted by a specific gas within a mixture of gases. Dalton's Law eloquently states that the total pressure in a mixture is equal to the sum of the partial pressures of all the gases present.

For a gas at equilibrium, knowing the partial pressures can be vital to calculate both \(Q\textsubscript{p}\) and \(K\textsubscript{p}\). In our exercise concerning phosgene production, the changes in partial pressures of reactants and products have been the key to solving for the equilibrium condition. The initial partial pressures set the stage, and then we use the concept of equilibrium to calculate the changes leading to the final partial pressures at equilibrium.

Hands-on Practice

A recommendation for students to gain more insight into this area would be to engage in exercises that involve changing the total pressure of the system or adding inert gases. Seeing how these manipulations affect the partial pressures of reactants and products can deepen the understanding of how gases interact within confined spaces in the context of chemical equilibrium.