Finding partial pressure?

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. 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?

I got the first part of the question using the ICE chart, but i don't know how to do the second part? Can anyone help me out?

Answer 1

[HI]²/[H2][I2] = 6

[x]²/[1-x][1-x] = 6

[x]² = 6[1-x][1-x]

[x]² = 6[x]^2-12x+6

5[x]²-12x+6=0

x = (12 ±√(144-120))/10

x = (1/5)√6+(6/5),(6/5)-(1/5)√6

x = {[x=0.7101],[x=1.6899]}

x=1.6899 doesn't make sense as this would give a negative concentration to [H2] and [I2].

Therefore [HI] = 0.7101 = 6/5-(1/5)√6 and

[H2] = [I2] = 0.2899 = -1/5 + (1/5)√6 = (1/5)(√6-1)

Second part:

1.000 atm => 1/24 mole/l

[HI] = 0.7101 is already greater than 1 atm.

There must be a mistake in the question.