For the reaction: Br₂(l) + Cl₂(g) ⇌ 2BrCl(g); K_p = 1 atm. In a closed container of volume 164 L, initially 10 moles of Cl₂(g) are present at 27°C. What minimum mass (in g) of Br₂(l) must be introduce

Question

For the reaction: Br₂(l) + Cl₂(g) ⇌ 2BrCl(g); K_p = 1 atm. In a closed container of volume 164 L, initially 10 moles of Cl₂(g) are present at 27°C. What minimum mass (in g) of Br₂(l) must be introduced into this container so that above equilibrium is maintained at a total pressure of 2.25 atm. Vapour pressure of Br₂(l) at 27°C is 0.25 atm. Assume that volume occupied by the liquid is negligible. (R = 0.082 L-atm/K-mol, Atomic mass of Br = 80)

Prashant Kotian · Accepted Answer

Correct answer: 0800. K_p = P_BrCl² / P_Cl₂ = 1. Total P = P_Cl₂ + P_BrCl + P_Br₂(vap) = 2.25 atm. Since P_Br₂(vap) = 0.25 atm, P_Cl₂ + P_BrCl = 2.0 atm. Let P_BrCl = y, then P_Cl₂ = 2.0 - y. y² / (2.0 - y) = 1 => y² + y - 2.0 = 0 => (y+2)(y-1) = 0 => y = 1 atm. So P_BrCl = 1 atm, P_Cl₂ = 1 atm. Moles of Cl₂ at eq = P_Cl₂ V / RT = (1 × 164) / (0.082 × 300) = 164 / 24.6 = 20/3 mol. Moles of BrCl at eq = (1 × 164) / 24.6 = 20/3 mol. Bromine required for BrCl formation = (20/3) / 2 = 10/3 mol. Moles of Br₂ in vapor phase = P_Br₂ V / RT = (0.25 × 164) / 24.6 = 5/3 mol. Minimum total Br₂ consumed = 10/3 + 5/3 = 15/3 = 5.0 mol. Mass of Br₂ = 5.0 mol × 160 g/mol = 800 g. Therefore, correct answer is 0800.

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