Physical Chemistry (2)

2007-08-25 7:18 am

2.08g of PCl5 are vaporized at 500K in a 0.001m^3 flask. Some dissociation occurs according to the equation PCl5<---->PCl3+Cl2. The partial pressure of Cl2 in the resulting mixture is 20.265kPa.Determine the partial pressures of the other constituents and the total pressure of the mixture. (Relative atomic masses:P:31;Cl:35.5;Universal gas constant=8.314/mol K)

回答 (1)

Uncle Michael

2007-08-25 10:04 am

✔ 最佳答案

PCl5 = PCl3 + Cl2
Mole ratio PCl3 : Cl2 = 1 : 1
Partial pressure of PCl3 = PPCl3 = PCl2 = 20.265 kPa

Consider Cl2 in the resulting mixture :
PCl2 = 20.265 kPa = 20.265 x 103 N m-2
V = 0.001 m3
T = 500 K

Number of moles of Cl2, nCl2
= PCl2V/RT
= (20.265 x 103) x 0.001 / (8.314 x 500)
= 0.004875 mol

PCl5 = PCl3 + Cl2
Mole ratio PCl5 : Cl2 = 1 : 1
No. of moles of PCl5 reacted = 0.004875 mol
Molar mass of PCl5 = 31 + 35.5x5 = 208.5 g mol-1
No. of moles of PCl5 added = 2.08 / 208.5 = 0.009976 mol
No. of moles of PCl5 left in the mixture,
nPCl5 = 0.009976 - 0.004875 = 0.005101 mol
Partial pressure of PCl5 in the mixture, PPCl5
= nPCl5RT/V
= 0.005101 x 8.314 x 500 / 0.001
= 21205 N m-2
= 21.205 kPa

Total pressure
= PPCl5 + PPCl3 + PCl2
= 21.205 + 20.265 + 20.265
= 61.735 kPa

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