Equlibirum question?

Question

At a 25oC, the equilibrium constant for the formation of HI (g) is 6.

H2 (g) + I2 (g) ó 2HI (g)

If one mole of H2 and 1 mole of I2 are introduced into a 1.00 L flask what is the composition at equilibrium? Consider the equilibrium mixture of H2, I2 and HI.

i found it to be:

[H2]= 0.45mol/L

[I2]= 0.45 mol/L

[HI]= 1.10 mol/L

If the equilibrium is disturbed by adding more HI so that the partial pressure of HI is suddenly increased to 1.000 atm, what will be the partial pressures of all the gases when the system returns to equilibrium?

How would i go about doing the second question?

Answer 1

I agree with your answer to part 1.

Calculate Kp from Kc.

Kp = Kc(RT)^delta n where delta n is the gas moles of product minus the gas moles of reactant. In our problem, the gas moles on both sides of the equation are the same (2), so delta n = 0 and Kp = Kc.

You can calculate the partial pressures using the ideal gas law.

PV = nRT

For H2, P = nRT / V = (0.45)(0.0821)(298) / (1) = 11.0 atm

For I2, same as H2 = 11.0 atm

For HI, P = nRT / V = (1.10)(0.0821)(298) / (1) = 26.9 atm

I don't get it! How can the partial pressure of HI be increased to 1.000 atm when it's already at 26.9 atm? Maybe increased to 100 atm. If that were the case:

Create an ICE table showing the partial pressures.

Pressure  . . . . . .H2   +    I2    ==>    2HI

initial . . . . . . . . .11.0  . .11.0 . . . . . . 100

change  . . . . . . .+x . . . .+x . . . . . . . -2x

final . . . . . . . .11.0+x    11.0+x  . . .100-2x

Kp = (P HI)^2 / (P H2)(P I2) = (100 - 2x)^2 / (11.0 + x)^2

6 = (100 - 2x)^2 / (11.0 + x)^2

2.4 = (100 - 2x) / (11.0 + x)

2.4 (11.0 + x) = 100 - 2x

26.4 + 2.4x = 100 - 2x

4.4x = 73.6

x = 16.7

P H2 = P I2 = 11.0 + x = 11.0 + 16.7 = 27.0 atm

P HI = 100 - 2x = 100 - 2(16.7) = 66.6 atm

Again, I'm not sure what gives with this problem.