Question

Consider the reaction:
H₂(g)+I₂(g)⇌2HI(g)
A reaction mixture at equilibrium at 175 K contains P_H2=0.958atm, P_I2=0.877atm, and P_HI=0.020atm. A second reaction mixture, also at 175 K, contains P_H2=P_I2=0.629 atm , and P_HI= 0.101 atm .

Is the second reaction at equillibrium? I found that the kp=4.76X10^-4 and Qp = .0257 So no, not at equillibrium.

If not, what is the partial pressure of HI when the reaction reaches equilibrium at 175 K? I need help figuring out the ICE chart and how to set up the equation to solve for the partial pressure of HI for the second reaction.

Answer 1

We see that Qp > Kp for the reaction, hence the reaction will move in Backward direction

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

0.629....0.629.........0.101

+X.........+X.............-2X

0.629+X..0.629+X...0.101-2X

Kp = 4.76 x 10^-4 = (0.101-2X)^2 / (0.629+X)^2

Taking square roots on both sides

=> (0.101-2X) / (0.629+X) = 0.0218

=> 0.101 - 2X = 0.0137 + 0.0218 X

=> X = 0.0432 atm

partial pressure of HI when the reaction reaches equilibrium at 175 K = 0.101 - 2X = 0.0147 atm