Where are my Chemistry people at?

by RavenPearl 8 Replies latest jw friends

  • RavenPearl
    RavenPearl

    Dude.......i'm like zoning on this. I can't figure it out. I got 1.79 and .11 and .11 ...but Its not working.

    Consider the following equilibrium

    COBr2(g) ↔ CO(g) + Br2(g) Kc = 0.190 at 73 °C

    Consider the following equilibrium:

    COBr2(g) ↔ CO(g) + Br2(g) Kc = 0.190 at 73 °C

    A 3.8 mol sample of COBr2 is transferred to a 2.0 L flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species.

    [COBr2] = M

    [CO] = M

    [Br2] = M

    And the second part of this is instead of 2L ...it's 1L and I have to find it. Does anyone know this? I'm totally stuck. I did google and ice boxed it and everything and i'm stuck. Any help?

  • bohm
    bohm

    I know jack shit about chemistry but here goes:

    Go to wikipedia:

    https://en.wikipedia.org/wiki/Equilibrium_constant

    Lets write it as A <-> B + C

    The initial concentration of A is

    [A] = 3.8/2 M/L = z

    Then you know that whatever happens

    [A] + ([B] + [C])/2 = z and [B] = [C]

    so:

    [A] + [B] = z

    Then looking at wikipedia I guess:

    Kc = [B] [C] / [A] = [B]^2 / [A]

    Plug in [A] we get

    Kc = [B]^2 / [A] = [B]^2 / (z-[B])

    If we let x = [B] we get:

    Kc = x^2 / (z-x)

    This gives you the 2d poly:

    x^2 + Kc * x - z Kc = 0

    solving for x gives:

    x = [B] = 0.1741

    and so

    [A] = 1.726

    Keep in mind this is just something I am jotting down from wikipedia...

  • waton
    waton

    wow bohm.you are booming! make no bones about it!

  • RavenPearl
    RavenPearl

    First off, wow, dude thank you. You went and actually 'studied' up on it. That's freaking selflessness right there.right on I gotta praise you on that

    unfortunately those answers don't work. There are no coefficents for the elements so it can't be squared. ...again awesome job though!

  • bohm
    bohm

    ah yes that makes sense. So it should probably be:

    [A] <-> 1/2 [B] + 1/2 [C]

    to keep Kc dimensionless and

    Kc = [B]^.5 * [C]^.5 / [A]

    and same procedure as above can be applied to solve for [B], [A]. Does that make sense?

  • bohm
    bohm

    The coefficients on wikipedia (sigma, tau, etc. on wikipedia) keeps track of the number of elements. So if you start out with A=1000 of A's in a bottle and they all get made into B's and C's you end up with 1000 of B and 1000 of C. So in general it holds that

    A = 1/2 B + 1/2 C

    That you divide by volume doesn't change that and so

    [A] = 1/2 [B] + 1/2 [C]

  • LevelThePlayingField
    LevelThePlayingField

    In the JW world, remember, 2+2=5. That's my 2 cents. Oh, and don't mix ammonia and bleach :)

  • RavenPearl
    RavenPearl

    Thank you all! I appreciate your help. I was able to figure it out. I had to use the quadratic formula in order to get it right. And unfortunately, a slight miscalculation in the 2a portion was changing everything. So Instead of dividing the whole thing by 2am, i was only doing it under the square root!...messed up everything.

    Thank you so much guys...i got it =D you guys are awesome!

    p.s. ......yes mustard gas sucks

  • sir82
    sir82

    In the JW world, remember, 2+2=5.

    Actually, that's 1 + 1 = 1 (overlapping generation).

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