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

Question

1. Consider the gas-phase reaction of PCl5, for which the equilibrium constant has the value of 11.5 at 300 degrees Celsius:

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

a. Suppose enough PCl5 is placed in a flask of a certain volume at 300 degrees Celsius such that the initial pressure of PCl5 is 1.500 atm. Calculate the equilibrium pressure for each of the three gases, and also calculate the fraction of PCl5 that has dissociated to reach equilibrium. (If "x" is the amount that reacted, and "c" is the initial amount, then "x/c" is the fraction reacted.)

Ans: Equil Pressure PCl5______ PCl3_______ Cl2______ f=________

b. Now Suppose that 1.500 atm PCl5 is placed in a flask that also contains an initial pressure of 1.000 atm Cl2, with the initial pressure of PCl3=0. Calculate the equilibrium pressures, and the fraction related.

Ans: Equil pressure PCl5 ______ PCl3 ______ Cl2_______ f=________-

Explanation / Answer

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

   1.50                                     0                0

   1.50 - x                                x                 x

Kp = PPCl3 x PCl2 / PPCl5

11.5 = x^2 / 1.50 - x

x = 1.343

Equilibrium pressures

PPCl5 = 0.157 atm

PPCl3 = 1.343 atm

PCl2   = 1.343 atm

fraction pf PCl5 = 0.157 / 1.500 = 0.105

b)

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

   1.50                                     0                1

   1.50 - x                                x                 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

PCl2   = 1.254 atm

fraction pf PCl5 = 0.246 / 1.500 = 0.164

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