Need major help

  1. I'm learning equilibirum in class, but I am stuck on this question. DOes anyone how how to do this problem? I use the quadratic formula but end end with answers A and B, but the correct answer is A. Can someone show me step by step how to do this problem? thanks

    9. At a certain temperature, T, K for the reaction below is 7.5 liters/mole.
    2NO2 <===> N2O4
    If 2.0 moles of NO2 are placed in a 2.0-liter container and permitted to react at this temperature, what will be the concentration of N2O4 at equilibrium?

    a) 0.39 moles/liter
    b) 0.65 moles/liter
    c) 0.82 moles/liter
    d) 7.5 moles/liter
    e) none of these
     
  2. jcsd
  3. You did the problem correctly. You end up having to solve the quadratic equation 30.x2 - 31x + 7.5 = 0. You get the answers of x = .39 and x = .65.

    Now, think about the problem. You are starting with 1.0 M NO2, correct? You should notice that the equilibrium concentration of NO2 will be 1.0 - 2x, and the equilibrium concentration of N2O4 will be x.

    If x was .65, then 1.0 - 2x would yield a negative concentration, which is physically impossible. Therefore, the only correct answer can be [N2O4] = .39 M because 1.0 - 2(.39) gives you a positive concentration.
     
    Last edited: Apr 15, 2006
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