Consider the hypothetical reaction \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{B}(g) . \mathrm{A}\) flask is charged with 0.75 atm of pure \(\mathrm{A},\) after which it is allowed to reach equilibrium at \(0^{\circ} \mathrm{C}\) . At equilibrium, the partial pressure of \(\mathrm{A}\) is 0.36 atm. (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 B?

The reaction is given by: \[A(g) \rightleftharpoons 2B(g)\] Set up the initial, change, and equilibrium pressures in relation to the reaction: For A: Initial Pressure: 0.75 atm Change: -x Equilibrium Pressure: 0.36 atm For B: Initial Pressure: 0 atm Change: 2x Equilibrium Pressure: 2x

As we know the equilibrium pressure of A is 0.36 atm, we can find the change in pressure (x): \[0.75-x=0.36 \Rightarrow x=0.39\]

Now that we have the value of x, we can find the equilibrium pressure of B: Equilibrium Pressure of B = 2x = 2(0.39) = 0.78 atm

Finally, the total pressure at equilibrium is the sum of the equilibrium pressures of A and B Total Pressure at equilibrium = 0.36 atm (A) + 0.78 atm (B) = 1.14 atm (b) Finding the value of \(K_p\)

Since the reaction is given by A(g) ⟷ 2B(g), the expression for the equilibrium constant will be: \[K_p=\frac{P(B)^2}{P(A)}\]

Substitute the equilibrium pressures of A and B in the expression: \[K_p=\frac{(0.78)^2}{0.36}\]

Now, find the value of \(K_p\): \[K_p=\frac{0.6084}{0.36} \approx 1.69\] (c) Maximizing the yield of B

Le Chatelier’s principle states that if an equilibrium system is subjected to a change in temperature, volume, or pressure, the system will adjust the equilibrium position to counteract the change. To maximize the yield of B, we can: 1. Increase the pressure of the system: Since there are fewer moles of gas on the right side of the reaction (2 moles of B) compared to the left side of the reaction (1 mole of A), increasing the pressure will shift the equilibrium in the direction of B, increasing its formation. 2. Remove B from the system as it is formed: Continuously removing B from the system will shift the equilibrium towards the right side, thus producing more B. 3. Use a catalyst: A catalyst can increase the rate of the reaction without affecting equilibrium; however, it can help in reaching equilibrium faster. Note, changes in temperature will not maximize the yield of B since no information is given about the heat associated with the reaction.