# 2nd law of thermodynamics question

1. Oct 17, 2016

### ssor

The reaction CO2 + 4 H2 <==> CH4 + 2 H2O (liq) at 25oC is downhill with a gibbs energy of -31 kcal/mol. It has a negative entropy of -98 cal/K. Thus an isolated system of the two starting gases should go to equilibrium on the basis of the Gibbs energy value, but the process would then violate the 2nd law. What am I missing here?

Thanks

2. Oct 17, 2016

### Bystander

What is equilibrium?

3. Oct 17, 2016

### James Pelezo

This is a case where ΔH < 0 and ΔS < 0 with ΔG either + or - => Spontaneous at 'low' Temps but Non-spontaneous at 'high' temps. You need to determine the Thermodynamic equilibrium temperature from ΔH - TΔS = 0. ΔG above this temp will be + and non-spontaneous where as ΔG below this temp will be negative and spontaneous.

4. Oct 17, 2016

### ssor

Thanks for your response, but the question remains. (Note: What appeared to be 250C was meant to be 25 degrees C.) The reaction with a negative delta G at 25 degrees C will be spontaneous, correct? However its negative entropy and the second law dictate it cannot be spontaneous. The 2nd law as I understand it deals with entropy, but Gibbs energies direct equilibria. Thus my confusion. Any additional comment is welcome, and thanks again.

5. Oct 17, 2016

### Ygggdrasil

The second law states that the entropy of the universe is always increasing. ΔS tells you how the entropy of the system changes. What's missing is the change in the entropy of the surroundings. You should review your thermodynamics text to figure out how to calculate the change in the entropy of the surroundings.

6. Oct 17, 2016

### James Pelezo

Yes, the reaction is spontaneous at 25oC as well as any temperature below the Thermodynamic Equilibrium Temperature. Again, solve ΔGo = ΔHo - TeqΔSo = 0 for Teq. Then calculate ΔGo for any temperature above Teq => ΔGo > 0 => non-spontaneous. Then calculate ΔGo for any temperature below Teq (including 25oC) => ΔGo < 0 => spontaneous. Nothing is violated relative to the 2nd Law.

Perhaps you need to understand that ΔHo and ΔSo are calculated using thermodynamic constants that are listed at 25o. However, these values when applied to ΔHo - TeqΔSo = 0 for Teq, the temperature value obtained (Teq) is the transition temperature separating spontaneous from non-spontaneous. Use Hess's Law Equation to determine ΔHo and ΔSo and test what I'm trying to say. ΔHo and ΔSo are both negative. (Hint: Teq = 612K for the listed rxn).

7. Oct 17, 2016

### ssor

I am still puzzled and appreciate your patience. I fully understand the notion of an equilibrium temperature, and clearly as I have written the equation its equilibrium position at 25oC falls well to the right based on the negative Gibbs energy. It thus should move spontaneously to the right to the equilibrium position (kinetics aside of course). My question comes down to this: At 25oC the reaction has both a negative entropy AND is spontaneous. My understanding of the 2nd law is that a process in an isolated system will take place spontaneously only if the entropy of the process is positive. The contradiction is clear and for me remains a puzzle.

8. Oct 17, 2016

### ssor

As I understand it the second law applies to isolated systems in general, not just the universe. My question presumes an isolated system with a CO2/H2 mixture at 25 degrees C.

9. Oct 17, 2016

### James Pelezo

OK, try this ... Given ΔHo = - 60 kcal; ΔSo = -98 cal/K = -0.098Kcal/K. => ΔHo - TeqΔSo = 0 => Teq = ΔHo/ ΔSo = -60 Kcal/(-0.098Kcal/K) = 612K = Teq

Test for 620K > Teq => ΔGo = ΔHo - TeqΔSo = -60 Kcal - 620K(-0.095 Kcal/K) = -60 Kcal + 60.76 Kcal = +0.76 Kcal > 0 => (non-spontaneous)
Test for 600K < Teq => ΔGo = ΔHo - TeqΔSo = -60 Kcal - 600K(-0.095 Kcal/K) = -60 Kcal + 57.00 Kcal = -3.00 Kcal < 0 => (spontaneous)

Both ΔHo and ΔSo are negative => Rxn is non-spontaneous above 612K (ΔGo > 0) and below 612K, spontaneous (ΔGo < 0).

10. Oct 17, 2016

### Ygggdrasil

The reaction releases energy. If the system is isolated, what happens to the energy that gets released? How does that affect the entropy of the system?

11. Oct 18, 2016

### James Pelezo

Define 'isolated system'.