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What's wrong with my calculations?

  1. Jun 21, 2009 #1
    1.The equilibrium constant is Kc = 85.0 at 460°C for the reaction
    SO2(g) + NO2(g)[tex]\Updownarrow[/tex]NO(g) + SO3(g)
    If 2.24L SO2, 4.48 L NO2, 2.24 L NO and 3.36 L SO3 (ALL AT STP) are transferred to a 5.00 L reaction vessel and heated to 460°C, what will be the equilibrium molar concentrations?

    2. because it is at STP it's safe to assume that 1 mole will equal 22.4 L, therefore I could divide all the initial volumes by 22.4L to get mol then divide by 5.00L to get mol L-1.
    3. I made an initial, change, equilibrium chart.
    SO2 NO2 NO SO3
    I: .02, .04, .02, .03
    C: -x, -x, +x, +X
    E: .02-x, .04-X, .02+x, .03+x

    85= (.02+x) (.03+x)/(.02-x)(.04-X) -> 85= .0006+.05X+x2/.0008-.06x-x2 -> 85(.0008-.06x-x2)=.0006+.05X+x2-> .068-5.1x-85x2=.0006+.05X+x2-> 84x2-5.15x+.00674=0
    -(-5.15) +-[tex]\sqrt{(-5.15)^2-4(84*.0674)}[/tex] divided by 2(84)
    x=.0424, x=.0189

    now when equated back into the calculations for equilib. concentrations
    E: .0011, .0211,.0389,.0489
    those numbers can't be correct because in order to be a valid assumption doesn't the number i got for x/by the initial given*100 have to be less than 5%...mine are not therefore there must be something mistaken in my calculations. can someone please explain this to me.
  2. jcsd
  3. Jun 22, 2009 #2


    User Avatar

    Staff: Mentor

    Check your math.

    Where have you used assumptions that work if 5% rule is obeyed?
  4. Jun 22, 2009 #3
    this is the other eq. that i got


    -(-5.15) +- [tex]\sqrt{(-5.15)^2*4(-86*.0674)}[/tex] divided by 2(-86)

    is there a rule of mathematics that i might not be following...i am a little rusty on algebra etc.
    the 5% rule was out of my textbook but it doesn't seem to work on any of my other problems so i think there is something i'm not factoring into it.
  5. Jun 22, 2009 #4


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    Staff: Mentor

    85= (.02+x) (.03+x)/(.02-x)(.04-X) is OK, you make some error later.

    State 5% rule.
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