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 \(\mathrm{B}\) ?

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

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.

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.

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.