Chemistry - Chemical Equilibrium Question with Solution | TestHub
Three ideal gases A, B and D were taken in a vessel of constant volume in molar ratio of 1:2:4, respectively, causing the reaction:
A(g) + 2B(g) ⇌ C(g)
After a long time, when the equilibrium is established the total pressure was 6/7th of the initial total pressure. At this point, the volume was reduced to one-third and a catalyst was added starting the reaction:
2C(g) + D(g) ⇌ 2F(g)
If at the second equilibrium, moles of A and C are equal and the ratio of equilibrium total pressure at second equilibrium to the equilibrium total pressure at first equilibrium is 'x:100', then the value of 'x' is
Answer:
Solution:
Take initial moles A = 1, B = 2, D = 4 in volume V (total 7 units) Total moles at first equilibrium = (6/7)(7) = 6, and each unit of extent lowers moles by 2, so extent = 0.5
First equilibrium: A = 0.5, B = 1, C = 0.5, D = 4
Kc = nC V2/(nA nB2) = (0.5)V2/[(0.5)(1)2] = V2
Let the total extent of the first reaction be u and that of the second be y
nA = 1 - u, nB = 2 - 2u, nC = u - 2y, nD = 4 - y, nF = 2y
Total moles = 7 - 2u - y
At volume V/3: Kc = nC (V/3)2/(nA nB2) = V2, so nC/(nA nB2) = 9
With nA = nC: nB2 = 1/9, so 2 - 2u = 1/3 and u = 5/6
1 - u = u - 2y gives 1/6 = 5/6 - 2y, so y = 1/3
Total moles at second equilibrium = 7 - 5/3 - 1/3 = 5
Pressure ratio = (5/(V/3))/(6/V) = 15/6 = 2.5
2.5 = x/100, so x = 250, written as 0250
