Can a reaction be neither exo- nor endothermic?

[kingkong23]

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

Did I do it right? is it possible?

 

[Bystander]

You've used "H" for "hot," for "hydrogen," for "enthalpy," ... ?

 

[Chestermiller]

Your algebra for the heat capacity term doesn't seem correct, and you omitted the effect of the heat capacity of hydrogen.

 

[Borek]

But - in general - the answer to the main question is "yes".

 

[kingkong23]

You've used "H" for "hot," for "hydrogen," for "enthalpy," ... ?

enthalpy

 

[kingkong23]

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.

does this number seem reasonable ?

 

[kingkong23]

oh yeah, I totally missed that.

View attachment 107717
does this number seem reasonable ?

 

[Chestermiller]

I think so. So, for this reaction, there apparently isn't a cross-over.

 

[kingkong23]

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?

 

[Chestermiller]

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.

 

[kingkong23]

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)

 

[Chestermiller]

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.