Consider the following reaction:

H2+I2<->2HI

A reaction mixture at equilibrium at 175 K contains P(H2)=0.958 atm, P(I2)=0.877 atm, and P(HI)=0.020 atm. A second reaction mixture, also at 175 K, contains P(H2)=P(I2)=0.616 atm, and P(HI)=0.106 atm. What will the partial pressure of HI be when the reaction reaches equilibrium at 175 K?

1.Use the first set of conditions to calculate K for the reaction.

2. I would calculate reaction Q next with the second set of conditions to see which way the reaction will proceed to reach equilibrium.
3. Set up an ICE chart, after step 2, and substitute the values into K.

To find the partial pressure of HI when the reaction reaches equilibrium at 175 K, we need to use the equilibrium constant expression (Kp) for the given reaction and solve for the unknown.

The equilibrium constant expression (Kp) for the reaction is given by:

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

Given the first reaction mixture at equilibrium, we can substitute the known values into the equilibrium constant expression:

0.020^2 / (0.958 * 0.877) = Kp

Given the second reaction mixture, we can substitute the known values into the equilibrium constant expression:

(P(HI))^2 / (0.616 * 0.616) = Kp

Now, equate the two expressions for Kp:

0.020^2 / (0.958 * 0.877) = (P(HI))^2 / (0.616 * 0.616)

Now, solve for the unknown, which is the partial pressure of HI:

(P(HI))^2 = (0.020^2 / (0.958 * 0.877)) * (0.616 * 0.616)

(P(HI))^2 = 0.0005427

Now, take the square root of both sides to get the partial pressure of HI:

P(HI) = √0.0005427

P(HI) ≈ 0.023 atm

Therefore, the partial pressure of HI when the reaction reaches equilibrium at 175 K is approximately 0.023 atm.