Chemistry - Chemical Equilibrium Question with Solution | TestHub
An amount of 0.3 mole of each A2(g) and B2(g) is introduced in a sealed flask and heated to 1500 K where following equilibrium is established:
A2(g) + B2(g) ⇌ 2AB(g)
At equilibrium, moles of AB is 0.3. At this stage, 0.15 mole of C2(g) is added and a new equilibrium is also established as:
A2(g) + C2(g) ⇌ 2AC(g)
At the new equilibrium, the moles of AB becomes 0.24. What is the equilibrium constant for the second reaction?
Answer:
Solution:
First equilibrium: AB = 0.3, A2 = B2 = 0.3 - 0.15 = 0.15 mol K1 = (0.3)2/[(0.15)(0.15)] = 4 (Δn = 0, so moles can be used)
AB falls from 0.3 to 0.24, so 0.06 mol AB reacts back, giving 0.03 mol each of A2 and B2
B2 = 0.15 + 0.03 = 0.18 mol (B₂ takes no part in the second reaction)
K1 = (0.24)2/[(A2)(0.18)] = 4, so A2 = 0.0576/0.72 = 0.08 mol
A2 used by C2 = 0.15 + 0.03 - 0.08 = 0.10 mol, so AC = 0.20 mol
C2 = 0.15 - 0.10 = 0.05 mol
K2 = (0.20)2/[(0.08)(0.05)] = 0.04/0.004 = 10
Answer is 0010
