The discussion centers on the thermodynamics of the reaction CO2 + 4 H2 <==> CH4 + 2 H2O (liq) at 25°C, which has a Gibbs energy of -31 kcal/mol and a negative entropy of -98 cal/K. Participants clarify that while the reaction is spontaneous at 25°C, it does not violate the second law of thermodynamics because the entropy change of the surroundings must also be considered. The equilibrium temperature (Teq) is calculated to be 612K, indicating that the reaction is non-spontaneous above this temperature and spontaneous below it. The confusion arises from the misunderstanding of entropy changes in isolated systems.
PREREQUISITESChemists, chemical engineers, and students studying thermodynamics who seek to understand the principles governing reaction spontaneity and the second law of thermodynamics.
What is equilibrium?ssor said: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?
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.James Pelezo said: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).
Ygggdrasil said: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.
ssor said: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.