The reaction PCl3(g)+Cl2(g)???PCl5(g)
has Kp=0.0870 at 300 ?C. A flask is charged with 0.50atmPCl3, 0.50atmCl2, and 0.20atmPCl5 at this temperature.

1.Calculate the equilibrium partial pressures of PCl3.Express your answer using two significant figures.

2.Calculate the equilibrium partial pressures of Cl2. .Express your answer using two significant figures.

3.Calculate the equilibrium partial pressures of PCl5.Express your answer using two significant figures.

1) p(PCl3) =  0.66 atm

2) p(Cl2) =  0.66 atm

3) p(PCl5) = 0.038 atm

Explanation:

Step 1: Data given

Kp = 0.0870

Temperature = 300 °C

Pressure of PCl3 = 0.50 atm

Pressure of Cl2 = 0.50 atm

Pressure of PCl5 = 0.20 atm

Step 2: The balanced equation

PCl3(g) + Cl2(g) ⇆ PCl5(g)

Step 3: The pressure at equilibrium

p = [PCl5] / [PCl3] [Cl2]

Q = [0.20] / [0.50] [0.50]  = 0.8

Q >> Kp

The reaction will shift to the left to decrease the numerator PCl5

p(PCl3) = (0.50 + x) atm

p(Cl2) = (0.50 + x) atm

p(PCl5) = (0.20 - x) atm

Kp = (pPCl5) / (pPCl3) (pCl2)

0.0870 = (0.20-x) / (0.50+x) (0.50+x)

0.0870 (0.50+x)(0.50+x)  = (0.20-x)

0.0870 (0.25 + 1.0x + x²)= (0.20-x)

0.02175  + 0.087x + 0.087x² = 0.2 -x

0.087x² + 1.087x -0.17825 = 0

x= 0.1619

Step 4: The equilibrium partial pressures

p(PCl3) = (0.50 + x) atm = 0.50 + 0.1619 =  0.6619 atm

p(Cl2) = (0.50 + x) atm = 0.50 + 0.1619 =  0.6619 atm

p(PCl5) = (0.20 - x) atm = 0.20 - 0.1619 = 0.0381 atm

201cm/s÷5: 18

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