Consider the following gaseous equilibrium in a closed container of volume 'V' at T(K). P2(g) + Q2(g) 2PQ(g) 2 moles each of P2(g), Q2(g) and PQ(g) are present at equilibrium. Now one mole each of 'P2' and 'Q2' are added to the equilibrium keeping the temperature at T(K). The number of moles of P2, Q2 and PQ at the new equilibrium, respectively, are
- 2.56, 1.62, 2.24
- 2.67, 2.67, 2.67
- 1.21, 2.24, 1.56
- 1.66, 1.66, 1.66
Answer
(2) 2.67, 2.67, 2.67
Solution
Kc = [PQ]2[P2][Q2] = (2/V)2(2/V)(2/V) = 1
After adding 1 mol each: P2 = 3, Q2 = 3, PQ = 2 mol
Qc = 223 3 = 49 < Kc reaction moves forward.
Let x mol of P2 react: P2 = 3 - x, Q2 = 3 - x, PQ = 2 + 2x
(2 + 2x)2(3 - x)2 = 1 ( ng = 0, V cancels)
2 + 2x = 3 - x
x = 13
P2 = Q2 = 3 - 13 = 2.67 mol; PQ = 2 + 23 = 2.67 mol
Answer: (2)