Chemistry

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The initial concentration for the compounds involved in the reaction shown were determine to be [CO]=1.80, [CL2]=9.70; [Cl2CO]=0. Calculate the value of the equilibrium constant (Kc) at 527C, if the equilibrium concentration of CO was 1.05E-3.

CO(g) + CL2(g) <--> Cl2CO(g)




so what i am thinking is i first set it as:
Kc= [Cl2CO]/[CO][CL2]

but after that i am lost and don't know what to do...

  • Chemistry -

    CO(g) + CL2(g) <--> Cl2CO(g)

    Set up the ICE chart and substitute the equilibrium values into the Kdq you have writte. Calculate Keq from that. Can you set up the ICE chart?

  • Chemistry -

    What is an ICE chart?

  • Chemistry -

    Is the ICE chart is the same as a list of initial, change and equilibrium values?

  • Chemistry -

    By the way, you need to specify if these concns are partial pressures, molarity, or mols in a particular volume.

  • Chemistry -

    it is molarity

  • Chemistry -

    CO(g) + Cl2(g) <--> Cl2CO(g)

    K= what you wrote.

    I = initial concns (before any reaction):
    (CO) = 1.80 M from the problem.
    (Cl2) = 9.70 M from the problem.
    (Cl2CO) = 0

    C = change in concn:
    (Cl2CO) = +x
    (Cl2) = -x
    (CO) = -x

    E = equilibrium concns:
    (Cl2CO) = 0+x = x
    (Cl2) = 9.70-x
    (CO) = 0.00105 (from the problem)

    If we started with 1.80 M for (CO) and we ended up with 0.0105, then how much reacted? That MUST be 1.80-0.0105 (that makes x = 0.0105 doesn't it) = 1.7995 which you might keep as 1.799 or round to 1.80--not much difference.
    Now go through and assign equilibrium values to CL2 (that is 9.70-x) and to Cl2CO (that is +x)
    Plug those values into Keq that you have set up and solve for K. I don't know how picky your teacher is about significant figures, but if so you need to watch them at the end. You may have 3 s.f.
    Any questions?

  • Chemistry -

    i set up the equation as:
    Keq= x/(17.36-1.7895x)

    where would i go from here?

  • Chemistry -

    I made two typos in my response. I typed If we started with 1.80 M for (CO) and we ended up with 0.0105, then how much reacted? That MUST be 1.80-0.0105 (that makes x = 0.0105 doesn't it) = 1.7995 which you might keep as 1.799 or round to 1.80--not much difference. but it should have read, If we started with 1.80 M for (CO) and we ended up with 0.00105, then how much reacted? That MUST be 1.80 - 0.00105 (that makes x = 1.80 - 0.00105 = 1.80 for all practical purposes.
    Keq = (Cl2CO)/(Cl2)(CO)
    You KNOW (Cl2CO). It's x. Didn't we put a number on x? Put the number in, not x. You know what it is.
    Now redo your E part of the ICE table and redo the Keq. I don't have the foggiest where you found (17.36-1.7895x).
    (Cl2O) = 1.80 M in round numbers but you may want to make it closer than that.
    (CO) = 0.00105 M
    (Cl2) = 9.70-1.80=7.90
    Check my work. It's late and I'm bleary eyed.

  • Chemistry -

    thanks for your help

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