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← JEE Main 2022 · 24 Jun · Shift 2
ChemistryInteger+4 / -1Question 83 of 90

JEE Main 2022 · 24 Jun · Shift 2, Chemistry question 83

PCl5\mathrm{PCl_5} dissociates as
PCl5(g)⇌PCl3(g)+Cl2(g)\mathrm{PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)}
5 moles of PCl5\mathrm{PCl_5} are placed in a 200 litre vessel which contains 2 moles of N2\mathrm{N_2} and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant KpK_p for the dissociation of PCl5\mathrm{PCl_5} is ______ ×10−3\times10^{-3}. (nearest integer)

(Given: R=0.082 L atm K−1mol−1R=0.082\ \mathrm{L\,atm\,K^{-1}mol^{-1}}; Assume ideal gas behaviour)

Answer and worked solution

Official answer: 1107

Total moles at equilibrium. n=PVRT=2.46×2000.082×600=10n=\dfrac{PV}{RT}=\dfrac{2.46\times200}{0.082\times600}=10; less 2 mol N2\mathrm{N_2} leaves 8.

Extent. 5−x+x+x=85-x+x+x=8, so x=3x=3: PCl5\mathrm{PCl_5} 2, PCl3\mathrm{PCl_3} 3, Cl2\mathrm{Cl_2} 3 mol.

KpK_p. 3×32×Pn=4.5×2.4610=1.107=1107×10−3\dfrac{3\times3}{2}\times\dfrac P{n}=4.5\times\dfrac{2.46}{10}=1.107=1107\times10^{-3}.

Answer: 1107