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JEE Main 2021 · Chemistry · Chemical Equilibrium · medium

Gaseous PCl5 dissociates as PCl5rightleftharpoonsPCl3+Cl2 with Kp=0.50 atm at a certain temperature. When the equilibrium total pressure is 2.0 atm…

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Gaseous PCl5\mathrm{PCl_5} dissociates as PCl5⇌PCl3+Cl2\mathrm{PCl_5}\rightleftharpoons\mathrm{PCl_3}+\mathrm{Cl_2} with Kp=0.50 atmK_p=0.50\ \mathrm{atm} at a certain temperature. When the equilibrium total pressure is 2.0 atm2.0\ \mathrm{atm}, the partial pressure of Cl2\mathrm{Cl_2} is closest to
  1. A. 0.45 atm0.45\ \mathrm{atm}
  2. B. 0.62 atm0.62\ \mathrm{atm}
  3. C. 0.50 atm0.50\ \mathrm{atm}
  4. D. 0.31 atm0.31\ \mathrm{atm}
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Answer: (B) 0.62 atm0.62\ \mathrm{atm}

Start with 11 mol PCl5\mathrm{PCl_5} and degree of dissociation α\alpha: total moles =1+α=1+\alpha, and pPCl3=pCl2=α1+αPp_{\mathrm{PCl_3}}=p_{\mathrm{Cl_2}}=\dfrac{\alpha}{1+\alpha}P, pPCl5=1−α1+αPp_{\mathrm{PCl_5}}=\dfrac{1-\alpha}{1+\alpha}P. Then Kp=α2P1−α2K_p=\dfrac{\alpha^2P}{1-\alpha^2}. Putting Kp=0.5K_p=0.5, P=2P=2: 2α21−α2=0.5\dfrac{2\alpha^2}{1-\alpha^2}=0.5, so α2=0.2\alpha^2=0.2, α=0.447\alpha=0.447. Therefore pCl2=0.4471.447×2=0.618≈0.62 atmp_{\mathrm{Cl_2}}=\dfrac{0.447}{1.447}\times2=0.618\approx0.62\ \mathrm{atm}. The value 0.450.45 is α\alpha itself, wrongly used as a pressure.
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