Problem 77 - Chemistry: The Central Science in SI Units, Global Edition

Question

Consider the hypothetical reaction $$\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)+2 \mathrm{C}(g)$$ A flask is charged with \(100 \mathrm{kPa}\) of pure \(\mathrm{A}\), after which it is allowed to reach equilibrium at \(25^{\circ} \mathrm{C}\). At equilibrium, the partial pressure of \(\mathrm{B}\) is \(25 \mathrm{kPa}\). (a) What is the total pressure in the flask at equilibrium? (b) What is the value of \(K_{p} ?(\mathbf{c})\) What could we do to maximize the yield of \(\mathrm{B}\) ?

Step-by-Step Solution

Answer

Total pressure at equilibrium is 175 kPa. The value of Kp for this reaction is 833.33. To maximize the yield of B, we can reduce the pressure and adjust the concentration of reactants and products.

Step 1: Understand given information and calculate partial pressures of other components

We are given that initial pressure of A is 100 kPa, and at equilibrium, the partial pressure of B is 25 kPa. We are also given the hypothetical reaction: $$\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) + 2 \mathrm{C}(g)$$ Let's find the change in the pressure of A, B, and C. Since we know the partial pressure of B, we can calculate the decrease in pressure for A and the increase in pressure for B and C in terms of the unit (ΔP). From the reaction, for each mole of A consumed, one mole of B is produced and 2 moles of C are produced. So, the change in pressures will be: ΔP_A = -ΔP ΔP_B = ΔP ΔP_C = 2ΔP

Step 2: Calculate total pressure at equilibrium

Now, let's calculate the total pressure at equilibrium by adding the partial pressures of components A, B, and C: P_total = P_A + P_B + P_C We know that the initial pressure of A is 100 kPa and the pressure of B at equilibrium is 25 kPa. We can use the change in pressure equations mentioned above to find the pressure of C and the total pressure: P_A = 100 kPa - ΔP P_B = ΔP + 25 kPa P_C = 2ΔP Substitute these values in the total pressure equation: P_total = (100 - ΔP) + (ΔP + 25) + 2ΔP Since ΔP is equal to the equilibrium pressure of B and we know that the equilibrium pressure of B is 25 kPa, we can substitute ΔP = 25 kPa: P_total = (100 - 25) + (25 + 25) + 2 * 25 P_total = 75 + 50 + 50 P_total = 175 kPa Total pressure at equilibrium is 175 kPa.

Step 3: Calculate Kp for the reaction

We can now use the total pressure and partial pressures of components to calculate Kp for the reaction. Kp can be found using the following equation: Kp = \(\frac{P_B * P_C^2}{P_A}\) Substitute the values of partial pressures: Kp = \(\frac{(25)*(50^2)}{75}\) Kp = 833.33 The value of Kp for this reaction is 833.33.

Step 4: Maximize the yield of B

To maximize the yield of B, we can analyze the effects of changing pressure, temperature, and concentration of reactants/products according to Le Chatelier's principle. 1. Pressure: Since there are more gas moles on the right side of the reaction (B and C), increasing the pressure will shift the equilibrium to the left side, which will produce more A and decrease the yield of B. So, reducing the pressure will maximize the yield of B. 2. Temperature: Since the information about whether the reaction is endothermic or exothermic is not given, we cannot decide the effect of temperature change on the reaction to maximize the yield of B. 3. Concentration: Increasing the concentration of A or decreasing the concentration of B or C will shift the equilibrium towards the right side, increasing the yield of B. In conclusion, to maximize the yield of B, we can reduce the pressure and adjust the concentration of reactants and products.

Key Concepts

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in chemistry that describes how a chemical system at equilibrium responds to a disturbance. When a system is at equilibrium and an external change, such as pressure, concentration, or temperature, is applied, the system will adjust to minimize the effect of the disturbance.

For this exercise, consider the hypothetical gas reaction: \( \mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) + 2 \mathrm{C}(g) \). According to Le Chatelier's Principle:

• If pressure is increased, the equilibrium will shift towards the side with fewer gas moles. For this reaction, this would be the left side because it has one mole (\(\mathrm{A}\)) compared to three moles on the right (\(\mathrm{B} + 2\mathrm{C}\)). Therefore, reducing pressure favors more moles being produced, shifting it to the right and increasing \(\mathrm{B}\)'s yield.


• If a reactant or product's concentration changes, the equilibrium will shift to oppose that change. To maximize \(\mathrm{B}\), increase the concentration of \(\mathrm{A}\) or decrease the concentration of \(\mathrm{B}\) or \(\mathrm{C}\).


• Given no temperature change details, predictions about thermal shifts can't currently apply. But typically, if the reaction releases heat (exothermic), decreasing temperature could shift equilibrium to the right.

Understanding Le Chatelier’s Principle allows for manipulation of reaction conditions to optimize desired products.

Partial Pressure

Partial pressure is a crucial concept in understanding gaseous equilibria. It refers to the pressure that a single gas component in a mixture of gases would exert if it occupied the entire volume alone. In a reaction system, the total pressure is the sum of the partial pressures of all the components present.

In this exercise, the system starts with \(\mathrm{A}\) at \(100 \mathrm{kPa}\). At equilibrium, \(\mathrm{B}\) levels at \(25 \mathrm{kPa}\). Using stoichiometry of the balanced reaction \( \mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) + 2 \mathrm{C}(g) \), you can determine changes in individual pressures:

• The partially consumed \(\mathrm{A}\) decreases by the amount that becomes \(\mathrm{B}\), multiplying the change for \(\mathrm{C}\) by its stoichiometric factor of 2.
• Total pressure sums: \(\mathrm{P}_{\text{total}} = (100 - \Delta \mathrm{P}) + 25 + 2\Delta \mathrm{P} \), simplifying with \(\Delta \mathrm{P} = 25 \mathrm{kPa}\) yields \(175 \mathrm{kPa}\).

Recognizing atomic interactions and changes in partial pressures explains how gases behave under conditions like equilibrium.

Equilibrium Constant (Kp)

The equilibrium constant, \(K_p\), is a key indicator of where a reaction balance lies under a given set of conditions. For reactions involving gases, \(K_p\) is defined using the partial pressures of the involved components, reflecting the ratios of products to reactants at equilibrium.

In our hypothetical reaction \( \mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) + 2 \mathrm{C}(g) \), the value of \(K_p\) can be determined from the equation:

• \[K_p = \frac{(P_{\mathrm{B}})(P_{\mathrm{C}})^2}{P_{\mathrm{A}}} \]


• Plugging the known values: \((25)(50^2) / 75\), results in \(K_p = 833.33\).

This provides insights into the favorability and extent of the reaction under equilibrium conditions. A higher \(K_p\) suggests products are vastly predominant while a lower one hints at a reactant-favored scenario. Understanding \(K_p\) enables predictions about how changing conditions might shift equilibrium.