Chemical EquilibriumHard
Question
If for 2A2B(g) ⇋ 2A2(g) + B2(g), Kp = TOTAL PRESSURE (at equilibrium) and starting the dissociation from 4 mol of A2B then :
Options
A.degree of dissociation of A2B will be (2/3).
B.total no. of moles at equilibrium will be (14/3).
C.at equilibrium the no. of moles of A2B are not equal to the no. of moles of B2.
D.at equilibrium the no. of moles of A2B are equal to the no. of moles of A2.
Solution
2A2B(g) ⇋ 2A2(g) + B2(g) KP = P
4(1 - α) 4α 2α ∑ n = 4 + 2α
KP =
= P ⇒ 2α3 = (1 - α)2 (4 + 2α)
2α3 = (1 + α2 - 2α) (4 + 2α)
2α3 = 4 + 2α + 4α2 + 2α3 - 8α - 4α2
α =
4(1 - α) 4α 2α ∑ n = 4 + 2α
KP =
= P ⇒ 2α3 = (1 - α)2 (4 + 2α)2α3 = (1 + α2 - 2α) (4 + 2α)
2α3 = 4 + 2α + 4α2 + 2α3 - 8α - 4α2
α =
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