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Equilibrum Concentration Problem

  1. Mar 9, 2005 #1
    H2(g) + I2(g) = 2HI at a temperature T


    If 1.40 mol H2 and 3.50 mol I2 are placed in a
    1.00 L vessel, what is the equilibrium concentration of I2 in the gaseous mixture?
    The equilibrium constant is K = 49.7 x 10 10

    HINT You can use the approximation for Large Equilibrium Constants


    this problem is giving me such a big headache, i mean i set up the table and it should be

    49.7X10^10=[2x]^2/[1.4-x][3.5-x]

    and then just use a quadratic fomula to find x, and subtract the number from 3.5, and i keep getting for x= 2.29x10^13, cause it cant be negative

    please help
     
  2. jcsd
  3. Mar 9, 2005 #2

    GCT

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    Homework Helper

    this seems correct, there must be something with your math.
     
  4. Mar 10, 2005 #3

    ShawnD

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    I entered that into my calculator (equation solver on a TI calculator) and it says the answer is x = 352491.1346 when the denominator in your equation is removed. x is so huge that subtracting it from something like 1.4 rounds to 0 and the calculator gets divide by 0 errors.

    If you have a TI calculator, set time aside to learn everything about the equation solver. It will save you an incredible amount of time in the future.
     
  5. Mar 15, 2005 #4
    do you go to SUNY stony brook? Because I do and that was one of the capa questions for this week, heh.

    You have to set up the table:
    H2 I2 2HI
    1.4 3.5 0
    -1.4 -1.4 +2(1.4)
    0 2.1 2.8

    The first row is the initial, second is the change to completion, third is the moles at completion...since H2 limits this reaction, you have to do your calculations based on the complete consumption of H2 and then you go from completion to equilibrium like so...

    H2 I2 2HI
    0 2.1 2.8
    +x +x -2x
    x 2.1+x 2.8-2x

    so thos're are your equilibrium equations and then you set up Keq like this:

    49.7x10^10 = (2.8-2x)^2/(x)(2.1-x)

    since Keq is so large, you can ignore the value of x for now...so then u get this:

    49.7x10^10 = (2.8)^2/(x)(2.1)

    and then you simply solve for x...no quadratic equation necessary. Then you go back to your equilibrium equation for I2 which is 2.1 - x and you plug in your x value. And guess what, since x is so darn small, chances are if you put in 2.1 for your answer, you'd be right.

    Hope that helps.
     
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