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.