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09/10/2021 10:08 am
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Gaseous N2O4 dissociates into gaseous NO2 according to the reaction
[N2O4(g) ⇌ 2NO2(g)]
At 300 K and 1 atm pressure, the degree of dissociation of N2O4 is 0.2. If one mole of N2O4 gas is contained in a vessel, then the density of the equilibrium mixture is :
(a) 1.56 g/L
(b) 6.22 g/L
(c) 3.11 g/L
(d) 4.56 g/L
1 Answer
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09/10/2021 10:17 am
Correct answer: (c) 3.11 g/L
Explanation:
N2O4(g) ⇌ 2NO2(g)
t = 0 1 0
t = eq. 1-α 2α
Where α = Degree of dissociation.
Mol. wt. of mixture
= \(\frac{(1 - α) \times M_{N_2O_4} + 2α \times M_{NO_2}}{(1 - α)}\)
= \(\frac{(1-0.2)92 + 2 \times 0.2 \times 46}{1 + 0.2}\)
= 76.66
Now, As per ideal gas equation
PV = nRT
PMmixture = dRT
∴ d = \(\frac{PM_{mix}}{RT}\)
= \(\frac{1 \times 76.66}{0.0821 \times 300}\)
= 3.11 g/L