# Homework Help: Can a reaction be neither exo- nor endothermic?

1. Oct 19, 2016

### kingkong23

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

Did I do it right? is it possible?

2. Oct 19, 2016

### Bystander

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

3. Oct 19, 2016

### Staff: Mentor

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

4. Oct 19, 2016

### Staff: Mentor

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

5. Oct 19, 2016

### kingkong23

enthalpy

6. Oct 19, 2016

### kingkong23

oh yeah, I totally missed that.

does this number seem reasonable ?

7. Oct 19, 2016

### kingkong23

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8. Oct 19, 2016

### Staff: Mentor

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

9. Oct 20, 2016

### kingkong23

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. Oct 20, 2016

### Staff: Mentor

As best I can tell, you didn't make a mistake.

11. Oct 20, 2016

### kingkong23

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. Oct 20, 2016

### Staff: Mentor

To my knowledge, there is no such thing as a negative absolute temperature.