But, how will the universe end? https://www.researchgate.net/publication/373717413_White_Privilege_Antiracist_Razor

Dear Bo Miao ,

I understand your concern, as at first glance, it seems like there are more equations than variables, which could imply that there is no solution. However, upon a more rigorous mathematical analysis, I believe the second law still holds in this case.

In your system, you have CO, O2, CO2 gases, and solid carbon (C) with three chemical equilibrium reactions: 2CO + O2 ⇌ CO2, C + O2 ⇌ CO2, and C + CO2 ⇌ 2CO. The corresponding equilibrium equations are ρ_co^2 / (ρ_co2 * ρ_o2) = K1, ρ_co2 / ρ_o2 = K2, and ρ_co^2 / ρ_co2 = K3. Additionally, you have an element conservation equation, 0.5ρ_co + ρ_co2 + ρ_o2 = A, where A is the average concentration of O2.

The equilibrium constants K1, K2, and K3 are not independent variables. They are related to the Gibbs free energy change (ΔG) of each reaction at a given temperature T by the equation ΔG = -RT ln(K), where R is the gas constant. Therefore, only two of the three K values are independent.

We can rewrite the equilibrium equations in terms of ΔG: ΔG1 = -RT ln(K1) = -RT ln(ρ_co^2 / (ρ_co2 * ρ_o2)), ΔG2 = -RT ln(K2) = -RT ln(ρ_co2 / ρ_o2), and ΔG3 = -RT ln(K3) = -RT ln(ρ_co^2 / ρ_co2). The sum of the ΔG values for the three reactions must be zero, as the overall process is at equilibrium: ΔG1 + ΔG2 + ΔG3 = 0. Substituting the expressions for ΔG1, ΔG2, and ΔG3, we get -RT ln(ρ_co^2 / (ρ_co2 * ρ_o2)) - RT ln(ρ_co2 / ρ_o2) - RT ln(ρ_co^2 / ρ_co2) = 0.

This equation, along with the element conservation equation, forms a system of two equations with three variables (ρ_co, ρ_co2, ρ_o2). While this system is underdetermined, it does not necessarily mean there is no solution or that the second law of thermodynamics is violated. The second law states that the entropy of an isolated system at equilibrium is maximum, and the change in Gibbs free energy is zero. The equilibrium state of this system, if it exists, would satisfy these conditions.

The apparent contradiction between the number of equations and variables in this chemical system does not inherently violate the second law of thermodynamics. The equilibrium constants are related through the Gibbs free energy, reducing the number of independent equations. The system may have an equilibrium state consistent with the second law, even if it is not uniquely determined by the given equations alone. Additional constraints, such as the positivity of concentrations or the minimization of Gibbs free energy, could potentially lead to a unique solution.

The second law of thermodynamics is a fundamental principle that has been extensively verified experimentally. I think it is important to critically examine such principles, but the presented argument does not provide sufficient evidence to overturn this well-established law.

Best,

Alessandro Rizzo

A great response, I will carefully consider it. Thank you! But the following example.

The actual decomposition voltage is less than the reversible decomposition voltage, which violates the second law of thermodynamics.

1）Experiments show that there exists an actual decomposition voltage which is less than the reversible decomposition voltage. ==》

2）The change of Gibbs free energy is related to the path.==》

3）The cyclic integral of entropy is not zero.==》

4）The second law of thermodynamics becomes invalid.

5） Scientists can only pretend to be deaf and dumb to such a fact.

Please see the picture for details.

I think you should wrok with the rate equations for the 3 processes. Then (4) follows from them as a conservation law. I do not see any contradiction