1) The equilibrium constant, K p , for the following reaction is 0.497 at 500 K:
ID: 691967 • Letter: 1
Question
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.
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.
Explanation / Answer
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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