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Homework Help: [CHEM] Equlibrium Help !

  1. Nov 16, 2005 #1
    [CHEM] Equlibrium Help plz!

    Hi, I am having trouble with a few questions on my chapter review and I have a test tomorrow. I wasn't at school today so I didn't get the chance to ask questions.

    1. The oxidation of sulfur dioxide to sulfur trioxide is an important reaction. At 1000K, the value of Kc is 3.6x10^-3.
    2 SO2 + O2 <-> 2 SO3
    A closed flask originally contains 1.7 mol/L SO2 and 1.7 mol/L O2. What is [SO3] at equilibrium when the reaction vessel is maintained at 1000K?
    Ok So I set up the ice table and for equilbrium i get 1.7-2x [SO2], 1.7-2x [O2], and 2x for [SO3]. Then...

    Kc = -____[SO3]^2____
    3.6x10^-3 = ____(2x)^2____
    ------------(1.7-2x)^2 (1.7-x)
    (3.6x10^-3)(2.89-6.8x-4x^2)(1.7-x) = 4x^2
    (3.6x10^-3)(4.913-14.45x+13.6x^2-4x^3)= 4x^2

    I didn't go any further because I don't know what to do now that there is an x cubed. Unless I'm doing it wrong, could you please correct me.

    2. Write the chemical equation for the reversible reaction that has the following equilibrium expression.
    Kc = blah blah blah
    Now, the eqn would be: 4 NH3 + 5 O2 <-> 4 NO + 6 H2O
    Assume that, at a certain temperature, [NO] and [NH3] are equal. Also assume that [H2O] = 2.0 mol/L and [O2] = 3.0 mol/L. What is the value of Kc at this temperature.

    What do I do for this question?
    Last edited: Nov 16, 2005
  2. jcsd
  3. Nov 16, 2005 #2

    Physics Monkey

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    Homework Helper

    The equation is cubic and would generally be solved numerically (though a nasty formula, analogous to quadratic formula, does exist for cubic polynomials). There are ways to find a good approximate answer which is still correct to several decimal places.

    What would you naively expect the value of x to be? In other words, will x be large (close to 1.7 say), will x be small (near zero), or something more in the middle? To rephrase it one more way, do you expect a lot or a little S03 to be produced?
    Last edited: Nov 16, 2005
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