Question

# 1) The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K: PCl5(g) <---...

1) The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K:
PCl5(g) <--- ----->PCl3(g) + Cl2(g) (reversible)
Calculate the equilibrium partial pressures of all species when PCl5(g) is introduced into an evacuated flask at a pressure of 1.52 atm at 500 K.

 PPCl5 = _____atm PPCl3 = ______atm PCl2 = ______atm

2) The equilibrium constant, Kp, for the following reaction is 55.6 at 698 K:

H2(g) + I2(g) <----- ----> 2HI(g) (reversible)

Calculate the equilibrium partial pressures of all species when H2 and I2, each at an intitial partial pressure of 2.18 atm, are introduced into an evacuated vessel at 698 K.

 PH2 = _____atm PI2 = _____atm PHI = ______atm

Kp = 0.497

Kp = PCl3*CL2/PCl5

initially

PCl5 = 1.52

PCl3 = 0

Cl2 = 0

in equilbirium

PCl5 = 1.52 - x

PCl3 = 0 + x

Cl2 = 0 + x

substitute

Kp = PCl3*CL2/PCl5

0.497= x*x/(1.52 - x)

x^2 + (0.497)x - (1.52*0.497) = 0

x = 0.6554

PCl5 = 1.52 - 0.6554 = 0.8646 atm

PCl3 = 0 + 0.6554 atm

Cl2 = 0 + 0.6554 atm

Q2

Kp = 55.6

Pinitial:

H2 = 2.18

I2 = 2.18

HI = 0

in equilbirium

H2 = 2.18 - x

I2 = 2.18 - x

HI = 0 +2x

Kp = HI^2 / (H2)(I2)

55.6 = (2x)^2 / (2.18 - x)(2.18 - x)

sqrt(55.6) = 2x/(2.18 - x)

2/7.456x = 2.18-x

(2/7.456 +1)x = 2.18

x = 2.18/((2/7.456 +1)) = 1.7189

H2 = 2.18 - 1.7189 = 0.4611 atm

I2 = 2.18 - 1.7189 = 0.4611 atm

HI = 0 +2*1.7189 = 3.4378 atm

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