1. Consider the gas-phase reaction of PCl5, for which the equilibrium constant has the value of...

3.7 Ratings (335 Votes)

PCl5 (g) <-----------------> PCl3 (g) +Â Â Cl2 (g)

Â Â 1.50Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

Â Â 1.50 - xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x

Kp = P_PCl3 x P_Cl2 / P_PCl5

11.5 = x^2 / 1.50 - x

x = 1.343

Equilibrium pressures

PPCl5 = 0.157 atm

PPCl3 = 1.343 atm

PCl2Â Â = 1.343 atm

fraction pf PCl5 = 0.157 / 1.500 = 0.105

PCl5 (g) <-----------------> PCl3 (g) +Â Â Cl2 (g)

Â Â 1.50Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1

Â Â 1.50 - xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1+x

Kp = (x)(1+x) / (1.50 - x)

11.5 = x + x^2 / (1.50 - x)

x = 1.254

PPCl5 = 0.246 atm

PPCl3 = 1.254 atm

PCl2Â Â = 1.254 atm

fraction pf PCl5 = 0.246 / 1.500 = 0.164

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