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.
The 2nd law of thermodynamics states that the total entropy of a closed system will always increase over time. It means that the amount of disorder or randomness in a closed system will always increase, and energy will inevitably be lost as heat.
The 2nd law of thermodynamics is responsible for various natural phenomena that we experience in our daily lives, such as the cooling of hot objects, the rusting of metal, and the decay of organic matter. It also explains why it is impossible to create a perfectly efficient machine.
Entropy and energy are related but distinct concepts. Energy is the ability to do work, while entropy is a measure of the disorder or randomness in a system. In simple terms, energy is the quantity of matter or radiation, while entropy is the quality of that matter or radiation.
No, the 2nd law of thermodynamics is a fundamental law of nature and cannot be violated. It has been experimentally verified countless times and is a cornerstone of modern physics. However, it is possible to manipulate and control energy and entropy to our advantage, such as in the production of electricity.
The 2nd law of thermodynamics is closely related to the concept of time. It states that in a closed system, entropy will always increase over time. This means that time is asymmetrical, and things tend to move from a state of order to a state of disorder, rather than the other way around. This concept is also known as the "arrow of time."