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JEE Main 2022 · Chemistry · Chemical Equilibrium · medium

The vapour density of N2O4 (taken before any dissociation) is 46. At 1 atm total pressure the equilibrium mixture N2O4(g)rightleftharpoons 2NO2(g)…

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The vapour density of N2O4\mathrm{N_2O_4} (taken before any dissociation) is 4646. At 11 atm total pressure the equilibrium mixture N2O4(g)⇌2NO2(g)\mathrm{N_2O_4(g)\rightleftharpoons 2NO_2(g)} shows an observed vapour density of 34.534.5. If the same mixture is allowed to re-equilibrate at 22 atm (same temperature), the degree of dissociation will be about
  1. A. 0.330.33
  2. B. 0.170.17
  3. C. 0.240.24
  4. D. 0.500.50
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Answer: (C) 0.240.24

α=(D−d)/d=(46−34.5)/34.5=1/3\alpha=(D-d)/d=(46-34.5)/34.5=1/3. For N2O4⇌2NO2\mathrm{N_2O_4\rightleftharpoons2NO_2}, Kp=4α2P1−α2=4(1/9)(1)8/9=0.5K_p=\dfrac{4\alpha^2P}{1-\alpha^2}=\dfrac{4(1/9)(1)}{8/9}=0.5 atm. At P=2P=2 atm: 4α2(2)1−α2=0.5⇒8α2=0.5−0.5α2⇒α2=0.5/8.5=0.0588\dfrac{4\alpha^2(2)}{1-\alpha^2}=0.5\Rightarrow 8\alpha^2=0.5-0.5\alpha^2\Rightarrow\alpha^2=0.5/8.5=0.0588, so α=0.243≈0.24\alpha=0.243\approx0.24. Higher pressure favours fewer moles of gas, so α\alpha must fall below 0.330.33.
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