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Lena772

  • one year ago

Hydrogen reacts with iodine to form hydrogen iodide as follows: H2 (g) + I2 (g) ⇌ 2 HI (g) At 445 °C, Kc = 50.2. If one starts with 0.100 M H2 (g), 0.100 M I2 (g) and 0.0500 M HI (g) what is the equilibrium concentration of HI?

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  1. Lena772
    • one year ago
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    @Photon336

  2. Photon336
    • one year ago
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    \[Kc = \frac{ [HI]^{2} }{ [H_{2}][I_{2}] }\]

  3. Photon336
    • one year ago
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    I would also calculate K to be sure of what direction our reaction is in. \[\frac{ [0.05]^{2} }{ [0.10][0.10] } = 0.25 Q \] Q < K so the reaction is going to go to the right

  4. Photon336
    • one year ago
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    I guess we could do an ice table for this

  5. Lena772
    • one year ago
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    0.25

  6. Photon336
    • one year ago
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    |dw:1442107231301:dw|

  7. Photon336
    • one year ago
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    \[\frac{ [0.05+2x] }{ [0.10-x][0.10-x] } = 50.2\]

  8. Photon336
    • one year ago
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    sorry the numerator should be squared

  9. Lena772
    • one year ago
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    0.01-0.10x-0.10x+x 0.01-0.20x+x (0.0025+2x^2)/(0.01-0.20x+x) = 50.2

  10. Photon336
    • one year ago
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    \[\frac{ [0.05+2x]^{2} }{ [0.1-x][0.1-x] } = 50.2\]

  11. Photon336
    • one year ago
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    I think you can disregard the x in the denominator that's how I was taught it.

  12. Photon336
    • one year ago
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    \[\frac{ (0.05+2x)^{2} }{ (0.01) } = 50.2\]

  13. Photon336
    • one year ago
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    \[(0.05+2x)^{2} = 0.502\]

  14. Abhisar
    • one year ago
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    That's a very neat explanation photon c:

  15. Photon336
    • one year ago
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    Thanks @abhisar, @Lena772 had another question that I couldn't answer before, I was hoping that she would reopen it, to see what you thought about it

  16. Abhisar
    • one year ago
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    Sure, give me the link c:

  17. Photon336
    • one year ago
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    This was from wolfram alpha app lol. \[(2x+0.05)^{2} = 4x ^{2}+0.2x+0.0025 = 0.502\] so wolfram alpha says the solution to this is x = 0.3296 \[\frac{ [(0.05+2(0.3296)]^{2} }{ 0.01 }\] = 50.2

  18. Photon336
    • one year ago
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    Thanks to the wolfram alpha app this has made our job(s) so much easier When I put this answer back into the equation it gives me the Kc of 50.2

  19. Photon336
    • one year ago
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    So the equilibrium concentration will be \[0.05+2(0.3296) = 0.7092\]

  20. Photon336
    • one year ago
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    \[\frac{ 0.502 }{ 0.01 } = 50.2\]

  21. Photon336
    • one year ago
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    So with ALL that, the equilibrium concentration should be 0.7092 for HI don't know if they want any sig figures. I guessed that I assumed that the concentration for the reactants changed by a small amount, My math skills are okay but not great If the answer happens to be wrong I guess you would have to find a way to solve this for x I disregarded x in the denominator. |dw:1442108700281:dw| that would be set to 50.2

  22. Lena772
    • one year ago
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    0.05 +( 2 * 0.3296)=0.7902 HI

  23. Lena772
    • one year ago
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    OH YOU ALREADY DID THAT LOL THANK YOU SO MUCH @Photon336 :')

  24. Lena772
    • one year ago
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    was incorrect :c

  25. Lena772
    • one year ago
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    @Photon336

  26. Photon336
    • one year ago
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    It must be, that in our answer we had to account for the change in the reactants which I disregarded

  27. Photon336
    • one year ago
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    I think I see the problem, we could not disregard the change in x for the reactants because the concentrations were large enough so that they couldn't be disregarded. not a problem I enter this in on wolfram alpha \[\frac{ [0.05+2x]^{2} }{ [(0.1-x)^{2}} = 50.2\] and I get two answers x = 0.0724827 x = 0.149162 Only one of these answers makes sense we have 0.10 moles of each reactant, so clearly 0.149162 cant be the answer otherwise we would have negative reactants. it must be x = 0.073

  28. Photon336
    • one year ago
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    @Lena772 my first assumption was to disregard the change -x in reactants but then I realized that I couldn't do this

  29. Photon336
    • one year ago
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    Before we go any further let's plug x = 0.073 into what I had set up above

  30. Lena772
    • one year ago
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    0.196

  31. Lena772
    • one year ago
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    0.05 + (2*0.073) = 0.196

  32. Photon336
    • one year ago
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    \[0.196^{2} = 0.038416 \] \[(0.10-0.073)^{2} = 0.000729\] Now \[\frac{ 0.038416 }{ 0.000729 } = 52.6~ \]

  33. Photon336
    • one year ago
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    Now my justification to put this all together

  34. Lena772
    • one year ago
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    So is the answer 0.000729? im confused

  35. Photon336
    • one year ago
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    \[[HI] = 0.05M + 2(0.73) = 0.196M\] \[I_{2} = 0.10M - 0.073 = 0.027 M \]

  36. Photon336
    • one year ago
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    \[H_{2} = 0.10M - 0.073 = 0.027 M \]

  37. Photon336
    • one year ago
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    @Lena772 I plugged in the value 0.073 into this remember so plug in 0.073 into this and tell me what you get \[\frac{ [2x+0.05]^{2} }{ [0.1-x]^{2} }\] I did this to see if the value was correct and it is because it equals our Kc roughly 50

  38. Lena772
    • one year ago
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    52.7

  39. Lena772
    • one year ago
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    So 0.196?

  40. Photon336
    • one year ago
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    I realized the last answer didn't make sense because we started with 0.10 for each. |dw:1442110682306:dw|

  41. Photon336
    • one year ago
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    @lena772 do they mark it wrong if you don't get sig figures right?

  42. Lena772
    • one year ago
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    not really, there's always a tolerance. that was correct. Thanks so much!

  43. Photon336
    • one year ago
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    OMG YES!!! |dw:1442111149801:dw|

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