Question

1) The equilibrium constant, K_p, 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, K_p, 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

Answer 1

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