Post a New Question

Chemistry Equilibrium Constant

posted by .

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

  • Chemistry Equilibrium Constant -

    for WHAT reaction?

  • Chemistry Equilibrium Constant -

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

  • Chemistry Equilibrium Constant -

    ...........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.

  • Chemistry Equilibrium Constant -

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

  • Chemistry Equilibrium Constant -

    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)

  • Chemistry Equilibrium Constant -

    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.

  • Chemistry Equilibrium Constant -

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

  • Chemistry Equilibrium Constant -

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

  • Chemistry Equilibrium Constant -

    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!

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question