Mortality

Law 2 of thermodynamics states that \(dΩ_{isolated} \geq 0, dS_{\text{isolated}} \geq 0\)

\(ΔS_{isolated} = ΔS_{total} = ΔS_{sys} + ΔS_{env} ≥ 0\)
\(ΔS_{env} = -ΔH_{sys} / T\) constant pressure
ΔS_total = ΔS_sys - ΔH_sys / T ≥ 0
ΔS_sys - ΔH_sys / T ≥ 0
\(ΔH_{sys} / T - ΔS_{sys} \leq 0\)
\(Gibbs\ energy = ΔH_{sys} - TΔS_{sys} \leq 0\)

is sys is human, a kind of organism:

\(ΔH_{sys}=ΔH_{human}​<0\), oxidation reaction is Exothermic.

glucose: \(C_6H_{12}O_6 + 6O_2 + \color{red}{36~ADP + 36~Pi} -> 6CO_2 + 6H_2O + \color{red}{36~ATP} + \text{多余热能}\)
\(\Delta_cH\) enthalpy of combustion reaction 1: \(\Delta_cH_{glucose} = Δc​H(葡萄糖)=[6Δf​H∘(CO2)+6Δf​H∘(H2O)]−Δ_f​H∘(C_6H_12O_6) = -2810kJ/mol, \Delta_cH_{simple\ substance} = \Delta_cH_{O_2} = 0\)
full \(\Delta_fH\) enthalpy of formation reaction 2: \(ΔH_{ADP->ATP}=∑Δf​H∘(产物)−∑Δf​H∘(反应物)=36 \times \left[ \Delta_fH^\circ(ATP) - \Delta_fH^\circ(ADP) - \Delta_fH^\circ(Pi) \right] = +1098kJ/mol\)
-1769 kJ/mol
sucrose: \(C_{12} H_{22} O_{11}(aq) + 12O_2 (g) -> 12CO_2 (g) + 11H_2 O(l)\)
oleic acid: \(C_{18} H_{34} O_2 (aq) + \frac{25}{2}O_2 (g) -> 18CO_2 (g) + 17H_2 O(l)\)

photosynthesis can be \(ΔH>0\). So for human, \(ΔH_{human} << 0\).

b. 熵变 ΔS 计算

为了计算熵变,我们需要标准摩尔熵值(S°)。参考热力学数据(在298 K):

  • C₆H₁₂O₆(s): 212 J/mol·K
  • O₂(g): 205 J/mol·K
  • CO₂(g): 214 J/mol·K
  • H₂O(l): 70 J/mol·K
  • ADP(aq) 和 ATP(aq) 的熵值较为复杂,但可以从ATP水解的熵变推断。

首先,计算葡萄糖氧化部分的熵变:

  • 反应物熵:S°(C₆H₁₂O₆) + 6 × S°(O₂) = 212 + 6 × 205 = 212 + 1230 = 1442 J/mol·K
  • 产物熵:6 × S°(CO₂) + 6 × S°(H₂O) = 6 × 214 + 6 × 70 = 1284 + 420 = 1704 J/mol·K
  • ΔS_oxidation = 1704 - 1442 = +262 J/mol·K

现在,考虑ATP合成部分。ATP水解反应(ATP → ADP + Pi)的熵变通常为正值:ΔS_hydrolysis ≈ +35 J/mol·K(基于标准数据:ΔG°_hydrolysis = -30.5 kJ/mol, ΔH°_hydrolysis ≈ -20 kJ/mol, 因此 ΔS_hydrolysis = (ΔH° - ΔG°)/T = (-20 - (-30.5))/298 ≈ 0.0352 kJ/mol·K = 35.2 J/mol·K)。
对于ATP合成(逆反应),熵变为负值:ΔS_synthesis = -35.2 J/mol·K per ATP。对于36个ATP:
\(\Delta S_{\text{synthesis}} = -35.2 \times 36 = -1267.2 \text{ J/mol·K} \approx -1.267 \text{ kJ/mol·K}\)

整体反应的熵变:
\(\Delta S_{\text{oxidation}} + \Delta S_{\text{synthesis}} = +262 - 1267.2 = -1005.2 \text{ J/mol·K} \approx -1.005 \text{ kJ/mol·K}\)

This is just an example, \(\Delta S_{\text{human}} <= 0\)

c. 吉布斯自由能 ΔG 计算

在人体温度约310 K(37°C)下计算:
\(T \Delta S_{\text{human}} = 310 \times (-1.005) = -311.55 \text{ kJ/mol} \Delta G_{\text{human}} = \Delta H_{\text{human}} - T \Delta S_{\text{human}} = -1712 - (-311.55) = -1400.45 \text{ kJ/mol} < 0\)

因此,即使熵减少(ΔS_human < 0),由于焓减少的幅度很大(ΔH_human << 0),ΔG 仍然为负。这证明了对于人类代谢反应,ΔG < 0。

3. 整体人类的热力学论证

人类是一个开放的、非平衡的系统,处于稳态。从热力学第二定律:

  • 系统熵变 ΔS_human 可能为负(由于有序结构的形成),但环境熵变 ΔS_env = -ΔH_human / T > 0(因为 ΔH_sys < 0),且总有 ΔS_total ≥ 0。
  • 因此,ΔG_human = ΔH_human - TΔS_human ≤ 0 必然成立。

ΔG_human ≤ 0。人类作为一个生物系统,其过程是自发的,符合热力学原理。尽管熵可能局部减少,但通过放热和环境熵增,总熵增加,维持了生命的持续。

\(dS_{human}​>TδQ_{irrev}\)​​

\(dS=d_i​S+d_e​S\)

\(ΔS=∫dS=∫(d_i​S+d_e​S)\)

\(d_i​S≥0\)

So that's fundamental definition of entropy change and only holds exactly for irreversible paths by Clausius Inequality.

Life is essentially a system that maintains an ordered state far from equilibrium by continuously inputting negative entropy flow (dₑS<0) to counteract internal entropy production (dᵢS>0).

Using energy (\(ΔG_{human} ≤ 0, ΔS_{total} ≥ 0\) in constant temperature and pressure) to counteract \(d_i​S≥0\), locally creating and maintaining \(dS_{human}≤0​\).

The energy can maintain the \(ΔS_{human} ≤ 0\) forever until all of the irreversible possibilities(\(d_i​S≥0\)).

posted @ 2025-10-07 18:55  千心  阅读(10)  评论(0)    收藏  举报