Chemistry - Chemical Equilibrium Question with Solution | TestHub
For the reaction: Br2(l) + Cl2(g) ⇌ 2BrCl(g); KP = 4 atm. In a closed container of volume 123 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 3.25 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)
Answer:
Solution:
Gases present: Cl2, BrCl and Br2 vapour (0.25 atm) PCl2 + PBrCl = 3.25 - 0.25 = 3.00 atm
KP = PBrCl2/PCl2 = 4
Let PBrCl = y: y2 = 4(3 - y), so y2 + 4y - 12 = 0 and y = 2 atm, PCl2 = 1 atm
Moles per atm = V/RT = 123/[(0.082)(300)] = 5.0 mol
BrCl = 10 mol, so Cl2 used = 5 mol and Br2(l) used = 5 mol
Br2 vapour = (0.25)(5.0) = 1.25 mol
Minimum Br2 = 5 + 1.25 = 6.25 mol
Mass = 6.25 × 160 = 1000 g
