Can a reaction be neither exo- nor endothermic?

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Discussion Overview

The discussion revolves around the thermodynamic properties of a specific chemical reaction involving ethane and butane, particularly focusing on whether there exists a temperature at which the reaction is neither exothermic nor endothermic. Participants explore calculations related to enthalpy (∆rH) and entropy (∆rS) under standard conditions, as well as implications of negative temperatures on entropy.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant calculates ∆rH for the reaction and seeks to determine a temperature, TH, where the reaction is neither endothermic nor exothermic.
  • Another participant questions the use of "H" in various contexts, suggesting potential confusion in notation.
  • Concerns are raised about the algebra used in calculating the heat capacity term, with a specific mention of neglecting the heat capacity of hydrogen.
  • Some participants assert that there is no crossover point for the reaction, implying a consistent thermodynamic behavior.
  • One participant calculates a negative reaction entropy (∆rS) and expresses confusion about the implications of this result, particularly regarding the logarithm of a negative number.
  • Another participant confirms that there is no mistake in the calculations but notes that negative absolute temperatures do not exist.

Areas of Agreement / Disagreement

Participants express differing views on the implications of their calculations, particularly regarding the existence of a temperature where the reaction is neither endothermic nor exothermic. There is no consensus on the interpretation of negative entropy values or the validity of the calculations presented.

Contextual Notes

Participants highlight limitations in their calculations, including missing considerations for the heat capacity of hydrogen and the implications of negative temperatures on entropy calculations. The discussion remains open-ended regarding the correct interpretation of these thermodynamic properties.

kingkong23
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Homework Statement


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

Homework Equations



Δr H (T2) = Δr H (T1) + ΔCp (T2-T1)

The Attempt at a Solution



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


Did I do it right? is it possible?
 
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You've used "H" for "hot," for "hydrogen," for "enthalpy," ... ?
 
Your algebra for the heat capacity term doesn't seem correct, and you omitted the effect of the heat capacity of hydrogen.
 
But - in general - the answer to the main question is "yes".
 
Bystander said:
You've used "H" for "hot," for "hydrogen," for "enthalpy," ... ?

enthalpy
 
Chestermiller said:
Your algebra for the heat capacity term doesn't seem correct, and you omitted the effect of the heat capacity of hydrogen.
oh yeah, I totally missed that.

upload_2016-10-19_16-38-2.png

does this number seem reasonable ?
 
kingkong23 said:
oh yeah, I totally missed that.

View attachment 107717
does this number seem reasonable ?
upload_2016-10-19_16-41-44.png
 

Attachments

  • upload_2016-10-19_16-41-4.png
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I think so. So, for this reaction, there apparently isn't a cross-over.
 
Chestermiller said:
I think so. So, for this reaction, there apparently isn't a cross-over.
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?
 
  • #10
kingkong23 said:
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?
As best I can tell, you didn't make a mistake.
 
  • #11
Chestermiller said:
As best I can tell, you didn't make a mistake.
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)
 
  • #12
kingkong23 said:
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)
To my knowledge, there is no such thing as a negative absolute temperature.
 

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