91Ó°ÊÓ

We can obtain the desired reaction by adding the given reactions in such a way that the reactants become C(g) + D(g) and the product becomes 2 B(g). Reaction 1: A(g) + B(g) ⇌ C(g) [K1 = 3.50 at 45°C] Reaction 2: 2 A(g) + D(g) ⇌ C(g) [K2 = 7.10 at 45°C] Reverse reaction 1: C(g) ⇌ A(g) + B(g) [K1' = 1/K1 = 1/3.50] Add reaction 2: C(g) + D(g) ⇌ 2 A(g) + B(g) [K3 = K1' * K2]

We need to calculate K3, the equilibrium constant of the combined reaction: K3 = K1' * K2 K3 = (1/3.50) * 7.10 K3 = 2.03 Now we need to get the desired reaction by removing 2 A(g) from both sides of the combined reaction: 2 A(g) + B(g) ⇌ C(g) + D(g) [K3 =2.03] Add -2 times reaction 1: C(g) + D(g) ⇌ 2 B(g) [K = K4]

Recall that the equilibrium constants multiply when adding reactions: K4 = K3 * (K1)^2 K4 = 2.03 * (3.50)^2 K4 = 24.9 So, the value of K for the reaction C(g) + D(g) ⇌ 2 B(g) at 45°C is 24.9.

Start with partial pressures of C and D both equal to 1.50 atm, and let x be the change in B's pressure at equilibrium. The reaction can be represented by: C(g) + D(g) ⇌ 2 B(g) 1.50 - x 1.50 - x 2x We can then use the equilibrium expression for K4: K4 = [B]^2 / ([C][D]) 24.9 = (2x)^2 / ((1.50 - x)(1.50 - x)) Solve this quadratic equation to find the value of x: x ≈ 0.796 atm Now, we can find the mole fractions of C, D, and B at equilibrium: PC = 1.50 - x ≈ 0.704 atm PD = 1.50 - x ≈ 0.704 atm PB = 2x ≈ 1.592 atm Partial pressures at equilibrium add up: Ptotal = PC + PD + PB ≈ 0.704 + 0.704 + 1.592 ≈ 3.0 atm Finally, calculate the mole fraction of B at equilibrium: Mole fraction of B = PB / Ptotal ≈ 1.592 / 3.0 ≈ 0.531 #Answer# The value of K for the reaction C(g) + D(g) ⇌ 2 B(g) at 45°C is 24.9. The mole fraction of B when equilibrium is reached is approximately 0.531.