Question

Given the following information:
2A(g) + B(g) ⇌ A2B(g) Kp1
2A(g) + C2(g) ⇌ 2AC(g) Kp2
3/2 A2 + B(g) + C(g) ⇌ AC(g) + A2B(g) Kp3
Which relationship represents the equilibrium constant for the reaction:
4A(g) + C2(g) + A2B(g) ⇌ 3A2(g) + B(g) + 2C(g)

Answer 1

To find the equilibrium constant expression for the given reaction, we need to use the equilibrium constants of the three reactions given and manipulate them using the stoichiometry of the overall reaction.

First, let's reverse the first two reactions and multiply them together to obtain the overall reaction:

A2B(g) + 2AC(g) ⇌ 2A(g) + B(g) + 2C2(g) (multiplied and reversed reactions 1 and 2)

Now, let's multiply reaction 3 by 2 and add it to the above reaction:

3A2 + 2AC + 2A2B ⇌ 2AC + 2A(g) + 3A2B(g) + B(g)

Simplifying, we get:

3A2 + 2A2B(g) + B(g) ⇌ 4A(g) + C2(g) + 2AC(g)

Now, we can use the equilibrium constant expressions for the three reactions:

Kp1 = (P_A2B) / (P_A^2 * P_B)

Kp2 = (P_AC)^2 / (P_A^2 * P_C2)

Kp3 = (P_AC * P_A2B) / (P_A^(3/2) * P_B * P_C)

Multiplying Kp1 and Kp2 and dividing by Kp3:

Kp1 * Kp2 / Kp3 = (P_A2B / P_AC)^2 * (P_A^2 * P_C2) / (P_AC * P_A2B * P_A^(3/2) * P_B * P_C)

Now, we can use the stoichiometry of the overall reaction to rewrite the expression in terms of the given reactants and products:

Kp4 = (P_A^(3/2) * P_B * P_C^2) / (P_A2B)^2

Therefore, the equilibrium constant for the given reaction is Kp4 = (P_A^(3/2) * P_B * P_C^2) / (P_A2B)^2.