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

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1. Consider the gas-phase reaction of PCl5, for which theequilibrium constant has the value of 11.5 at 300 degreesCelsius:
PCl5 (g) <---> PCl3 (g) +Cl2 (g)
a. Suppose enough PCl5 is placed in a flask of a certain volumeat 300 degrees Celsius such that the initial pressure of PCl5 is1.500 atm. Calculate the equilibrium pressure for each of the threegases, and also calculate the fraction of PCl5 that has dissociatedto 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_ Cl2f=
b. Now Suppose that 1.500 atm PCl5 is placed in a flask thatalso contains an initial pressure of 1.000 atm Cl2, with theinitial pressure of PCl3=0. Calculate the equilibrium pressures,and the fraction related.
Ans: Equil pressure PCl5 PCl3 Cl2_f=______-

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

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Chemistry

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