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Lena772

  • one year ago

A 2.0 L vessel contains HI and H2. The following equilibrium is established: H2 (g) + I2 (g) ⇌ 2 HI (g) At equilibrium [H2] = 2.0 M [I2] = 1.5 M [HI] = 2.5 M The initial concentration of I2 was not mentioned so you must assume [I2]o = 0. Calculate the initial moles of: HI (Tol: ± 0.1) H2 (g) (Tol: ± 0.1)

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  1. Lena772
    • one year ago
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    @Abhisar @Photon336

  2. Photon336
    • one year ago
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    You could work backwards

  3. Photon336
    • one year ago
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    wait those +/- thats the change in concentration right? @Lena772

  4. Lena772
    • one year ago
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    |dw:1442112283855:dw|

  5. Lena772
    • one year ago
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    Thats just showing the Tolerance for the answer will only be 0.1 above or below @Photon336

  6. Lena772
    • one year ago
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    I don't know how to solve for x working backwards @Photon336

  7. Lena772
    • one year ago
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    @abhisar

  8. Photon336
    • one year ago
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    So this was all the information they gave you?

  9. Lena772
    • one year ago
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    yep

  10. Lena772
    • one year ago
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    I know M= moles solute/L of soln but i dont know if i could manipulate that equation with what I'm given here.

  11. Photon336
    • one year ago
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    @Woodward thoughts? i'm not sure

  12. anonymous
    • one year ago
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    Yeah I am just working through this myself, I haven't done these in a while but I think I can help you. For instance, your ICE table doesn't seem quite right, you're starting with 0 moles of \(I_2\) and then below you are losing x amount to get 1.5 moles! |dw:1442114000906:dw| This doesn't make sense, the equilibrium will shift the other way!

  13. anonymous
    • one year ago
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    Whoops I said moles but this is concentration, let me straighten myself out before I say anymore lol

  14. anonymous
    • one year ago
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    Ok, so here's how I solved it, write out the ICE table but we don't need to use the equilibrium constant expression! We just simply label our unknowns in Molarity: |dw:1442114537041:dw| So we get three equations: \[[H_2]_0 + x = 2\] \[0+x = 1.5 \] \[[HI]_0 -2x = 2.5\] That middle one tells us that x=1.5, so the rest will just be algebra! Once you find these concentrations, don't forget to multiply by 2.0L to get the number of moles. I think that does it!

  15. Photon336
    • one year ago
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    @Woodward interesting; well if the change in x must arguably be the same for both hydrogen gas and Iodine, does this mean that we can easily find out what the initial concentration of HI is. For HI it's just 5.5 M for Iodine initial concentration is just 0 And for H2 it's 0.5 M

  16. anonymous
    • one year ago
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    Yeah that's exactly what I got. :D

  17. Photon336
    • one year ago
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    @Woodward why didn't we need to use Kc in this case?

  18. anonymous
    • one year ago
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    We don't need to use Kc because the information we have is essentially Kc. Calculate Kc for fun to see if you can!

  19. Photon336
    • one year ago
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    Oh I see, because we're given the equilibrium concentrations already that's basically Kc, so all we need to do is to just find the initial concentrations. thanks for clarifying that

  20. Lena772
    • one year ago
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    I got 5.5 and 2.5 and those were wrong

  21. Lena772
    • one year ago
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    @Woodward @Photon336

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