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Pure Hellspawn

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About Pure Hellspawn

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  1. Not sure where to put this, but I'm 100% stuck on two homework problems for chemistry and have spent at least 3 hours on it. The problems are as follows:

    #1:

    Consider the following reaction:
    2 O3 ---> 3 O2
    The rate law for this reaction is as follows:
    Rate = k [O3]^2 / [O2]
    Suppose that a 1.0 L reaction vessel initially contains 1.0 mol of O3 and 1.0 mol of O2.

    What fraction of the O3 will have reacted when the rate falls to one-half of its initial value?

    For this one I created an ICE chart, it looks like this:
    O3 O2
    I 1 1
    C -2x +3x
    E 1-2x 1+3x

    I'm stuck after that. Oh boy.

    #2:
    The data below were collected for the following reaction at a certain temperature:

    X2Y --> 2X + Y

    Time Molarity
    0.0 0.100
    1.0 0.0856
    2.0 0.0748
    3.0 0.0664
    4.0 0.0598
    5.0 0.0543

    I know this is a second order reaction but I need help finding the rate constant. I tried using the integrated rate law to no success.

    Anyways, this is how the last three hours of my life have been going. Please help me. I need it badly.

    1. Show previous comments  7 more
    2. fraggle

      fraggle

      Pure Hellspawn said:

      Not sure where to put this, but I'm 100% stuck on two homework problems for chemistry and have spent at least 3 hours on it. The problems are as follows:

      #1:

      Consider the following reaction:
      2 O3 ---> 3 O2
      The rate law for this reaction is as follows:
      Rate = k [O3]^2 / [O2]
      Suppose that a 1.0 L reaction vessel initially contains 1.0 mol of O3 and 1.0 mol of O2.

      What fraction of the O3 will have reacted when the rate falls to one-half of its initial value?


      Surely this is easy. They're initially in a 1:1 ratio, so the rate is k. You need to calculate the ratio when the rate is k/2, so

      (O3)^2 = (O2)/2. [1]

      The number of atoms never changes regardless of the reaction, so

      O2 + O3 = 1 + 1
      O2 = 2 - O3. [2]

      Substitute [2] into [1]:

      (O3)^2 = (2 - (O3)) / 2
      (O3)^2 + (O3) / 2 - 1 = 0

      Solve by the quadratic formula:

      O3 = (-0.5 + sqrt(0.25 + 4)) / 2 = 0.78078

      So you have 0.78078 mol of O3, 1.21922 mol of O2. This means that 0.21922 mol has been converted.

    3. Creaphis

      Creaphis

      Csonicgo said:

      if you remebmer the little formula for half life you should be able to figure the rest out.


      Okay, let's see... scripted fights, linear levels, one-sided dialogue, more scripted fights... how does this help again?

    4. Csonicgo

      Csonicgo

      fraggle said:

      Genius


      dude, I may need your help on some misc. aequilibria soon.


      dammmn.

      Creaphis said:

      Okay, let's see... scripted fights, linear levels, one-sided dialogue, more scripted fights... how does this help again?

      hahaha, I dunno man. I've been out of it the past few weeks, I've actually not been eating much(nothing but fruits and maybe a muffin) and in protest my body somehow gains weight. 185 to 215 eating fruit? Am I sleep eating?

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