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Can a reaction be neither exo- nor endothermic?

  1. Oct 19, 2016 #1
    1. The problem statement, all variables and given/known data
    Organic life is based on complex organic molecules formed from smaller ones during a long evolution. Using data in Appendix 2, investigate one such reaction:
    2 C2H6 (g) → C4H10 (g) + H2 (g)

    (a) Calculate ∆rH and ∆rS for such a reaction under standard conditions (Po = 1 bar, To = 298 K).
    Is there a temperature, TH, for which the reaction is neither endo- nor exothermic – calculate it?
    - I did this already - The ∆rH = 43.7kJ/mol which is needed to answer the next question

    b) Is there a temperature, TH, for which the reaction is neither endo- nor exothermic – calculate it?

    CPm(C2H6) = 52.5 J/(mol K)

    CPm(C4H10) = 97.5 J/(mol K)

    Smo(C4H10) = 310.2 J/(mol K)

    ∆fHmo(C4H10) = -125.7 kJ/mol


    2. Relevant equations

    Δr H (T2) = Δr H (T1) + ΔCp (T2-T1)
    3. The attempt at a solution

    upload_2016-10-19_14-22-53.png

    Did I do it right? is it possible?
     
  2. jcsd
  3. Oct 19, 2016 #2

    Bystander

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    You've used "H" for "hot," for "hydrogen," for "enthalpy," ... ?
     
  4. Oct 19, 2016 #3
    Your algebra for the heat capacity term doesn't seem correct, and you omitted the effect of the heat capacity of hydrogen.
     
  5. Oct 19, 2016 #4

    Borek

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    But - in general - the answer to the main question is "yes".
     
  6. Oct 19, 2016 #5
    enthalpy
     
  7. Oct 19, 2016 #6
    oh yeah, I totally missed that.

    upload_2016-10-19_16-38-2.png
    does this number seem reasonable ?
     
  8. Oct 19, 2016 #7

    Attached Files:

  9. Oct 19, 2016 #8
    I think so. So, for this reaction, there apparently isn't a cross-over.
     
  10. Oct 20, 2016 #9
    in a follow up question I was asked "obtain the reaction entropy, ∆rS, for such a temperature" (which I calculated to be -1725.15k)

    So I tried using this:
    Δr S0 Tf= Δr S0 Ti + Δr CP ln(Tf / Ti)

    but I can't do ln for a negative number so is my temperature calculation wrong?
     
  11. Oct 20, 2016 #10
    As best I can tell, you didn't make a mistake.
     
  12. Oct 20, 2016 #11
    okay, thanks for replying again. Just one more thing, at negative temperature is the entropy 0?

    Because I can't do ln for a negative number.

    I used this equation : Δr S Tf= Δr S Ti + Δr CP ln(Tf / Ti)
     
  13. Oct 20, 2016 #12
    To my knowledge, there is no such thing as a negative absolute temperature.
     
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