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JEE Advanced 2016 Paper 2 · Chemistry · Chemical Equilibrium · hard

One mole of PCl5(g) is introduced into a vessel fitted with a frictionless movable piston at 500 K and allowed to reach equilibrium…

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One mole of PCl5(g)\mathrm{PCl_5(g)} is introduced into a vessel fitted with a frictionless movable piston at 500 K500\ \mathrm{K} and allowed to reach equilibrium PCl5(g)⇌PCl3(g)+Cl2(g)\mathrm{PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)}. At equilibrium the total pressure is 1.001.00 atm and PCl5\mathrm{PCl_5} is 50%50\% dissociated. Take R=0.0821 L atm K−1mol−1R=0.0821\ \mathrm{L\,atm\,K^{-1}mol^{-1}}. Select the correct statement(s).
  1. A. Kp=0.50K_p=0.50 atm at 500 K500\ \mathrm{K}.
  2. B. If the gas mixture is compressed isothermally to an equilibrium total pressure of 2.002.00 atm, the degree of dissociation becomes about 0.380.38.
  3. C. If argon is added at constant temperature while the total pressure is held at 1.001.00 atm, the degree of dissociation of PCl5\mathrm{PCl_5} decreases.
  4. D. KcK_c at 500 K500\ \mathrm{K} is 13.7 mol L−113.7\ \mathrm{mol\,L^{-1}}.
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Answer: B

With α=0.5\alpha=0.5 and P=1P=1 atm, the mole fractions are PCl5=0.5/1.5=1/3\mathrm{PCl_5}=0.5/1.5=1/3 and each product 1/31/3, so Kp=(1/3)(1/3)1/3×1=α2P1−α2=0.250.75=0.333K_p=\dfrac{(1/3)(1/3)}{1/3}\times1=\dfrac{\alpha^2P}{1-\alpha^2}=\dfrac{0.25}{0.75}=0.333 atm. (A) says 0.500.50 atm: incorrect. (B) At P=2P=2 atm, α2=Kp/(P+Kp)=0.3333/2.3333=1/7\alpha^2=K_p/(P+K_p)=0.3333/2.3333=1/7, so α=0.378≈0.38\alpha=0.378\approx0.38: correct (compression shifts the equilibrium towards fewer moles). (C) Adding inert gas at constant total pressure increases the volume, which lowers the partial pressures of the reacting gases and favours the side with more moles, so α\alpha increases: incorrect. (D) Kc=Kp/(RT)ΔnK_c=K_p/(RT)^{\Delta n} with Δn=1\Delta n=1 gives 0.3333/(0.0821×500)=8.1×10−3 mol L−10.3333/(0.0821\times500)=8.1\times10^{-3}\ \mathrm{mol\,L^{-1}}; 13.713.7 comes from multiplying by RTRT: incorrect.
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