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09/10/2021 10:20 am
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The value of Kp for the equilibrium reaction
N2O4(g) ⇌ 2NO2(g) is 2.
The percentage dissociation of N2O4(g) at a pressure of 0.5 atm is
(a) 25
(b) 88
(c) 50
(d) 71
1 Answer
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09/10/2021 10:35 am
Correct answer: (d) 71
Explanation:
N2O4(g) ⇌ 2NO2(g)
Initial moles 1 0
Moles at equil. (1 - α) 2α
(α = degree of dissociation)
Total number of moles at equil.
= (1 - α) + 2α
= (1 + α)
PN2O4 = \(\frac{1 - α}{1+ α}\) x P
PNO2 = \(\frac{2α}{1 + α}\) x P
Kp = \(\frac{(P_{NO_2})^2}{P_{N_2O_4}}\)
= \(\frac{(\frac{2α}{1+α}\times P)^2}{(\frac{1 -α}{1 + α}) \times P}\)
= \(\frac{4α^2P}{1-α^2}\)
Given, Kp = 2, P = 0.5 atm
∴ Kp = \(\frac{4α^2P}{1-α^2}\)
= \(\frac{4α^2 \times 0.5}{1-α^2}\)
α = 0.707 = 0.71
∴ Percentage dissociation
= 0.71 × 100 = 71