In a sealed container at 25°C, the gas-phase equilibrium 2BrCl(g) ⇌ Br2(g) + Cl2(g) has Kp = 0.130. If initially BrCl(g) = 0.400 atm, Br2(g) = 0.800 atm, and Cl2(g) = 0.800 atm, what is the BrCl(g) partial pressure at equilibrium?

In a sealed container at 25°C, the gas-phase equilibrium 2BrCl(g) ⇌ Br2(g) + Cl2(g) has Kp = 0.130. If initially BrCl(g) = 0.400 atm, Br2(g) = 0.800 atm, and Cl2(g) = 0.800 atm, what is the BrCl(g) partial pressure at equilibrium?

• A. 0.781 atm
• B. 0.419 atm
• C. 1.16 atm
• D. 1.21 atm

Answer: (C)

The key idea is using the expression for Kp and the stoichiometry to relate how the pressures change as the system shifts toward equilibrium. For 2 BrCl ⇌ Br2 + Cl2, Kp = (P_Br2)(P_Cl2) / (P_BrCl)^2. Since initially BrCl is 0.400 atm while Br2 and Cl2 are both 0.800 atm, the system will shift to form more BrCl (the reverse direction), so let x be the amount of Br2 and BrCl that react in the reverse direction. Then Br2 and Cl2 decrease by x, and BrCl increases by 2x: P_Br2 = 0.800 − x P_Cl2 = 0.800 − x P_BrCl = 0.400 + 2x Plug into Kp: 0.130 = (0.800 − x)(0.800 − x) / (0.400 + 2x)^2 Taking the square root gives: √0.130 ≈ 0.3606 = (0.800 − x) / (0.400 + 2x) Solve: 0.3606(0.400 + 2x) = 0.800 − x → 0.14424 + 0.7212x = 0.800 − x → 1.7212x = 0.65576 → x ≈ 0.381 Thus the equilibrium BrCl pressure is: P_BrCl = 0.400 + 2x ≈ 0.400 + 0.762 ≈ 1.16 atm So the BrCl partial pressure at equilibrium is about 1.16 atm.

The key idea is using the expression for Kp and the stoichiometry to relate how the pressures change as the system shifts toward equilibrium. For 2 BrCl ⇌ Br2 + Cl2, Kp = (P_Br2)(P_Cl2) / (P_BrCl)^2.

Since initially BrCl is 0.400 atm while Br2 and Cl2 are both 0.800 atm, the system will shift to form more BrCl (the reverse direction), so let x be the amount of Br2 and BrCl that react in the reverse direction. Then Br2 and Cl2 decrease by x, and BrCl increases by 2x:

P_Br2 = 0.800 − x

P_Cl2 = 0.800 − x

P_BrCl = 0.400 + 2x

Plug into Kp:

0.130 = (0.800 − x)(0.800 − x) / (0.400 + 2x)^2

Taking the square root gives:

√0.130 ≈ 0.3606 = (0.800 − x) / (0.400 + 2x)

Solve: 0.3606(0.400 + 2x) = 0.800 − x → 0.14424 + 0.7212x = 0.800 − x → 1.7212x = 0.65576 → x ≈ 0.381

Thus the equilibrium BrCl pressure is:

P_BrCl = 0.400 + 2x ≈ 0.400 + 0.762 ≈ 1.16 atm

So the BrCl partial pressure at equilibrium is about 1.16 atm.