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Homework Help: Partial Pressure/ICE Chart

  1. Apr 9, 2008 #1
    1. The problem statement, all variables and given/known data
    A 1L flask is filled with 1 mole of H2 and 2 moles of I2 at 448degrees Celcius. The value of the Keq for the reaction
    H2 + I2 -><- 2HI at 448degrees Celcius is 50.5. What are the partial pressures of H2, I2, and HI in the flask at Equilibrium?

    All reactants and products are gases. Answer is to be in atm.

    2. Relevant equations

    Pressure of:
    H2 I2 HI

    3. The attempt at a solution
    First, the initial equilibrium of HI is 0 because it doesnt exist yet.
    I used PV=nRT to find pressure of H2 and I2
    H2) P= 1mole x .0821 x 721K
    ------------------------- = 59.19atm

    I2) P= 2mole x .0821 x 721K
    ---------------------- = 118.39atm

    I need help trying to find what exactly happens after the system reaches equilibrium and how to find the final pressures. Thought: could the pressure of H2 be the same before and after equilibrium because the temperature remains the same? Same for I2?
  2. jcsd
  3. Apr 9, 2008 #2
    I'm way past my bedtime so I'll check much much later but leaving the message here.

    H2 + I2 -><- 2HI is an important equation.

    You're thinking along the right track, there has to be some changes happening to the system to reach equilibrium.

    Thought: could the pressure of H2 be the same before and after equilibrium because the temperature remains the same? Same for I2? No definitely not. I'll explain why down there. Read on.

    You see, initially flask has 1 mole of H2, 2 mole of I2 and there is no HI at the start, just because the gases are assumed to be in there first by theory (i.e. not reacted yet). Of course the forward reaction of the equilibrium will happen; Moles of H2 and I2 will decrease and HI will be formed.

    By Keq, I kind of found out that it means equilibrium constant, we call it Kp over here.
    You should have some kind of equation on the Keq, try look it up, this is my simplified version:

    Kp = [Partial Pressure of Reactant(s)]^power of reactant coefficient / [Partial Pressure of Product]^power of product coefficient

    Try it out!
  4. Apr 9, 2008 #3
    You've found the initial pressures. Find what the change is for each gas when it goes toward equilibrium, and then find what the equilibrium pressures are. It will consist of a variable.
  5. Apr 9, 2008 #4
    Ok, so I went with Snazzy on this one. for the C in my ICE chart i have the change in Hydrogen as 0 because the moles of hydrogen did not change. I have the change in Iodine, I2, as -x because the number of I2 moles decreased from 2 to 1. Finally I have the change in HI as +2x because it started as 0 moles and increased to 2 moles.

    I took this and plugged it in

    50.5= Pressure HI
    Pressure H2 x Pressure of I2

    50.5= 2x
    59.19 x (118.39-x)

    50.5= 2x
    7007.5 - 59.19x

    353878.75 - 292989.095x = 2x

    353878.75 = 2991.5x

    x= 118.29

    P H2 I2 HI

    I 59.19 118.39 0

    C 0 -x +2x

    E 59.19 0.10 236.58
    Last edited: Apr 9, 2008
  6. Apr 9, 2008 #5
    Hold on, mate. You don't know how many moles exactly will be at equilibrium for each gas. You only know that the change for I2 AND H2 is -x, and the change for HI is +2x.

    Using that you can figure out what X is.
  7. Apr 9, 2008 #6
    just so i can understand it, how come H2 is -x AND I2 is -x?
  8. Apr 9, 2008 #7
    Both H2 and I2 start with one mole each. They both have 1 has their coefficient (1H2 + 1I2 -> 2HI).

    Both of them will react with each other to produce HI. Thus, an amount of H2 mixes with an equal amount of I2 to produce HI. You don't know how much H2 and I2 react with each other, that's why you leave it as "-X".
  9. Apr 9, 2008 #8

    (59.19-x)(118.39-x) ?

    I end up with

    x^2 + 177.58x - 7007.5

    What am I doing wrong?
    Last edited: Apr 9, 2008
  10. Apr 9, 2008 #9
    You're not doing anything wrong. Solve the quadratic that ensues. You're actually missing a squared term for the pressure of HI since the equation gives 2 moles HI for every mole of H2 and I2.
  11. Apr 10, 2008 #10
    The question said the flask started out with 1 mole of H2 and 2 moles of I2. So the final should be (1-x) mole of H2 and 2 moles of (2-x) mole of I2 if I'm not wrong. From your steps I think you're doing it fine.

    Only that you need to have that [2H]² somewhere as Snazzy says. It's a matter of the power of coefficient, for example if you have α HI instead you will be using [HI]^α....since the equation gives α moles HT for every mole of H2 and I2.
  12. Apr 10, 2008 #11
    oh so its x^2 and not 2x, thanks
  13. Apr 10, 2008 #12
    Is it x^2? Or is it (2x)^2?
  14. Apr 10, 2008 #13
    HI=x^2 H2=(1-x) and I2=(2-x)....I just have to plug in the numbers now and see what i get.
  15. Apr 10, 2008 #14
    Why is it x^2 if the change is 2x for HI?
  16. Apr 10, 2008 #15
    this is the ICE chart i have so far

    P H2 I2 HI

    I 59.19atm 118.39atm 0atm
    C (1-x) (2-x) x^2
    E 59.19(1-x) 118.39(2-x) x^2

    59.19(1-x) times 118.39(2-x)

    when i do that i get x=square root of 707757.5 - 1061636.25x + 353878.75x^2
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