Question

The initial pressures for I2 (g), H2(g), and HI(g) were Pi2 = 0.100 atm, Ph2 = 0.200 atm, and Phi = 0 atm, respectively. After the system came to equilibrium, the pressure of I2 (g) became very low, PI2 = 1.00 x 10^-5 atm. Calculate the equilibrium constant Kp for this reaction

Answer 1

for WHAT reaction?

Answer 2

The reaction is I2(g) + H2(g) -> 2HI(g)

Answer 3

...........H2 + I2 ==> 2HI
initial....0.2..0.1......0
change.....-x....-x.....+2x
equil...........1E-5........

I would see it this way. We reacted ALMOST 0.1 atm of I2 to leave such a small value at equilibrium. That means we used up 0.1 of the H2 to leave 0.1 at equil and we formed 0.2 atm HF. Substitute and solve for Kp. You want to substitute 1E-5 for I2.

Answer 4

So instead of using 0.1 for I2, I want to use 1E-5 for I2 when solving for KP?

Answer 5

Or do I solve for x, using the given pressure of I2 at equilibrium,
so should it be 1E-5 = 0.1 - x? (solve for x)

Answer 6

If you wish to solve for x then x =0.099999 but since that is so close to 0.1, you will call it 0.1 as the p of I2 used.

Answer 7

Kp = pHI^2/pH2*pI2
pI2 in this set up is 1E-5.

Answer 8

Hmm I'm still confused at what to plug in for pHI^2 and pH2 for solving Kp

Answer 9

Nevermind I got it, with x = 0.099999 then I just use what I have in the ICE chart and plug it in to solve for Kp,

Thanks for all the help!