Biological Thermodynamics

Derivation: First Law for Open Biological Systems

For an open system, the energy balance includes enthalpy carried by mass flow:

$$\frac{dU}{dt} = \dot{Q} - \dot{W} + \sum_{\text{in}} \dot{m}_i h_i - \sum_{\text{out}} \dot{m}_j h_j$$

where $\dot{m}_i$ are mass flow rates and $h_i$ are specific enthalpies. For a living organism at steady state ($dU/dt = 0$), taking food in and excreting waste, we can write:

$$\dot{Q} = \dot{W} + \sum_{\text{out}} \dot{m}_j h_j - \sum_{\text{in}} \dot{m}_i h_i$$

For metabolic reactions at constant pressure, we use enthalpy rather than internal energy. The enthalpy of a metabolic reaction is:

$$\Delta H_{\text{rxn}} = \sum \nu_i \Delta H_{f,i}^{\circ}(\text{products}) - \sum \nu_j \Delta H_{f,j}^{\circ}(\text{reactants})$$

For glucose oxidation:

$$\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \to 6\text{CO}_2 + 6\text{H}_2\text{O}$$

$$\Delta H_{\text{combustion}} = 6(-393.5) + 6(-285.8) - (-1274.4) - 6(0)$$

$$= -2361 - 1714.8 + 1274.4 = -2801.4 \text{ kJ/mol}$$

This equals $-2801.4 / 180.16 = -15.55$ kJ/g, or about 3.72 kcal/g.

Derivation: Atwater Factors from Heats of Combustion

The Atwater general factors relate food composition to physiological energy. The key distinction is between bomb calorimetry (complete combustion) and physiological energy (accounting for incomplete digestion and urinary losses).

Carbohydrates: The heat of combustion of glucose is$\Delta H_c = -15.55$ kJ/g. Starch has $\Delta H_c \approx -17.5$ kJ/g. With a digestibility coefficient of $d \approx 0.97$:

$$E_{\text{physiol}} = \Delta H_c \times d = 17.5 \times 0.97 \approx 17.0 \text{ kJ/g} \approx 4.06 \text{ kcal/g} \approx 4 \text{ kcal/g}$$

Fats: Tripalmitin has $\Delta H_c = -39.16$ kJ/g. Fats are more reduced (higher H:O ratio), so they release more energy per gram. With digestibility $d \approx 0.95$:

$$E_{\text{physiol}} = 39.16 \times 0.95 \approx 37.2 \text{ kJ/g} \approx 8.89 \text{ kcal/g} \approx 9 \text{ kcal/g}$$

Proteins: The heat of combustion is $\Delta H_c \approx -23.6$ kJ/g in a bomb calorimeter. However, the body cannot oxidize the nitrogen in amino acids to$\text{N}_2$ or $\text{NO}_x$. Instead, nitrogen is excreted as urea ($\text{CO(NH}_2\text{)}_2$), which still has combustion energy:

$$\Delta H_{\text{urea}} \approx -5.4 \text{ kJ/g protein}$$

$$E_{\text{physiol}} = (\Delta H_c - \Delta H_{\text{urea}}) \times d = (23.6 - 5.4) \times 0.92 \approx 16.7 \text{ kJ/g} \approx 4.0 \text{ kcal/g}$$

Thus the Atwater factors: Carbohydrate 4 kcal/g, Protein 4 kcal/g, Fat 9 kcal/g. The coincidence that carbohydrates and proteins have equal physiological energy values despite different combustion enthalpies arises from the urinary energy loss for proteins, which exactly compensates for the higher combustion energy.

Derivation: Entropy Balance for Living Systems

For an open system, the total entropy change has two contributions:

$$\frac{dS}{dt} = \frac{dS_i}{dt} + \frac{dS_e}{dt}$$

where $dS_i/dt \geq 0$ is the internal entropy production rate (always non-negative by the second law) and $dS_e/dt$ is the entropy exchange rate with the environment (can be positive or negative).

For a steady-state living organism ($dS/dt = 0$):

$$\frac{dS_i}{dt} = -\frac{dS_e}{dt} > 0$$

The organism must export entropy to the environment at a rate equal to its internal entropy production. This is Schrödinger's "negentropy" concept: the organism absorbs low-entropy food (organized chemical energy) and exports high-entropy waste (heat, $\text{CO}_2$, water).

The entropy exchange has contributions from heat flow and mass flow:

$$\frac{dS_e}{dt} = \frac{\dot{Q}}{T_{\text{env}}} + \sum_{\text{in}} \dot{m}_i s_i - \sum_{\text{out}} \dot{m}_j s_j$$

where $s_i$ are specific entropies. For a resting human with metabolic rate$\dot{Q} \approx 80$ W at $T = 310$ K:

$$\frac{dS_i}{dt} \approx \frac{80}{310} \approx 0.26 \text{ W/K} = 0.26 \text{ J/(s·K)}$$

This represents the minimum entropy production rate for maintaining the organism's steady state. According to Prigogine's minimum entropy production theorem, a system near equilibrium in a steady state minimizes its entropy production rate. Living systems, being far from equilibrium, produce considerably more entropy than this minimum.

Derivation: Coupled Reactions and ATP Hydrolysis

Consider an endergonic reaction A $\to$ B with $\Delta G_1 > 0$. This can be coupled to ATP hydrolysis:

$$\text{ATP} + \text{H}_2\text{O} \to \text{ADP} + \text{P}_i \quad \Delta G_2^{\circ} = -30.5 \text{ kJ/mol}$$

The coupled reaction A + ATP $\to$ B + ADP + P$_i$ has:

$$\Delta G_{\text{coupled}} = \Delta G_1 + \Delta G_2$$

The reaction proceeds spontaneously if $\Delta G_{\text{coupled}} < 0$, which requires$|\Delta G_2| > \Delta G_1$.

The actual free energy of ATP hydrolysis under cellular conditions is far more negative than the standard value. Using the mass action ratio:

$$\Delta G = \Delta G^{\circ} + RT \ln Q$$

where the mass action ratio is:

$$Q = \frac{[\text{ADP}][\text{P}_i]}{[\text{ATP}]}$$

At equilibrium, $\Delta G = 0$ and $Q = K_{eq}$, so:

$$\Delta G^{\circ} = -RT \ln K_{eq}$$

Combining these: $\Delta G = RT \ln(Q/K_{eq})$. In the cell, typical concentrations are [ATP] $\approx 5$ mM, [ADP] $\approx 0.5$ mM, [P$_i$] $\approx 5$ mM. Therefore:

$$Q = \frac{(0.5 \times 10^{-3})(5 \times 10^{-3})}{5 \times 10^{-3}} = 5 \times 10^{-4}$$

$$\Delta G = -30.5 + (8.314 \times 10^{-3})(310) \ln(5 \times 10^{-4})$$

$$= -30.5 + 2.577 \times (-7.60) = -30.5 - 19.6 = -50.1 \text{ kJ/mol}$$

The actual free energy of ATP hydrolysis in the cell is about $-50$ to $-55$ kJ/mol — nearly double the standard value! The cell maintains the ATP/ADP ratio far from equilibrium, which is the fundamental thermodynamic driving force of metabolism. This ratio is sometimes called the phosphorylation potential.

Derivation: Carnot Efficiency at Body Temperature

The Carnot efficiency represents the maximum possible efficiency of a heat engine operating between a hot reservoir at temperature $T_{\text{hot}}$ and a cold reservoir at$T_{\text{cold}}$:

$$\eta_{\text{Carnot}} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}}$$

This is derived from the second law. For a reversible Carnot cycle with isothermal and adiabatic steps, the work output is $W = Q_H - Q_C$ and by the entropy constraint $Q_H/T_H = Q_C/T_C$, so:

$$\eta = \frac{W}{Q_H} = \frac{Q_H - Q_C}{Q_H} = 1 - \frac{Q_C}{Q_H} = 1 - \frac{T_C}{T_H}$$

If the human body were a heat engine with $T_{\text{hot}} = 310$ K (body, 37°C) and $T_{\text{cold}} = 298$ K (environment, 25°C):

$$\eta_{\text{Carnot}} = 1 - \frac{298}{310} = 1 - 0.961 = 0.039 \approx 3.9\%$$

This is absurdly low! A human converting glucose to work at 3.9% efficiency would need to eat about 50,000 kcal/day to sustain normal activity. Clearly, metabolism is NOT a heat engine.

Why metabolism is a chemical engine: Instead of converting heat to work (constrained by Carnot), metabolism directly converts chemical free energy to work via coupled chemical reactions. The relevant efficiency is:

$$\eta_{\text{ATP}} = \frac{\Delta G_{\text{ATP}} \times n_{\text{ATP}}}{\Delta G_{\text{glucose}}}$$

For glucose oxidation: $\Delta G^{\circ}_{\text{glucose}} = -2870$ kJ/mol. The theoretical ATP yield is about 32 ATP molecules, each storing $\Delta G_{\text{ATP}} \approx 30.5$ kJ/mol:

$$\eta_{\text{theoretical}} = \frac{32 \times 30.5}{2870} = \frac{976}{2870} \approx 34\%$$

Under cellular conditions where $\Delta G_{\text{ATP}} \approx -50$ kJ/mol:

$$\eta_{\text{actual}} = \frac{32 \times 50}{2870} = \frac{1600}{2870} \approx 55.7\%$$

The actual efficiency of oxidative phosphorylation, accounting for real ATP yield (~30 ATP) and proton leak, is approximately 38–40%. This vastly exceeds the Carnot limit, confirming that metabolism operates as a chemical engine, not a heat engine. The efficiency of muscle contraction (converting ATP to mechanical work) is about 25%, giving an overall metabolic-to-mechanical efficiency of about 25% × 40% = 10%.

Derivation: The Protein Stability Curve and Cold Denaturation

The free energy of unfolding is:

$$\Delta G(T) = \Delta H(T) - T\Delta S(T)$$

The enthalpy and entropy of unfolding are temperature-dependent due to the large heat capacity change $\Delta C_p$ upon unfolding (exposure of hydrophobic groups):

$$\Delta H(T) = \Delta H(T_0) + \Delta C_p(T - T_0)$$

$$\Delta S(T) = \Delta S(T_0) + \Delta C_p \ln\left(\frac{T}{T_0}\right)$$

Substituting into $\Delta G$, using $T_0 = T_m$ (melting temperature where$\Delta G(T_m) = 0$ and $\Delta S(T_m) = \Delta H(T_m)/T_m$):

$$\Delta G(T) = \Delta H_m\left(1 - \frac{T}{T_m}\right) + \Delta C_p\left[(T - T_m) - T\ln\left(\frac{T}{T_m}\right)\right]$$

This is the Gibbs-Helmholtz stability curve. Because$\Delta C_p > 0$ for unfolding (typically 5–15 kJ/(mol·K) for small proteins), this curve has a parabolic shape with a maximum stability at some intermediate temperature.

Cold denaturation: Setting $\Delta G(T) = 0$ yieldstwo solutions: the hot denaturation temperature $T_{\text{hot}}$ (the usual melting temperature) and a cold denaturation temperature $T_{\text{cold}}$. Differentiating$\Delta G(T)$ to find the temperature of maximum stability:

$$\frac{d\Delta G}{dT} = -\Delta S(T) = 0 \implies T^* = T_m \exp\left(-\frac{\Delta H_m}{\Delta C_p \cdot T_m}\right)$$

More precisely, setting $\Delta S(T) = 0$:

$$\frac{\Delta H_m}{T_m} + \Delta C_p \ln\left(\frac{T^*}{T_m}\right) = 0$$

$$T^* = T_m \exp\left(-\frac{\Delta H_m}{\Delta C_p \cdot T_m}\right)$$

For a typical protein with $T_m = 340$ K, $\Delta H_m = 300$ kJ/mol,$\Delta C_p = 8$ kJ/(mol·K): $T^* = 340 \times e^{-300/(8 \times 340)} = 340 \times e^{-0.110} = 340 \times 0.896 = 305$ K (32°C). The cold denaturation temperature is typically below 0°C for most proteins, but can be observed in destabilized mutants or under high pressure.

Derivation: The Folding Funnel and Levinthal's Paradox

Levinthal (1969) noted that a protein with $N$ residues, each with $\phi$backbone conformations, has $\phi^N$ total conformational states. For a 100-residue protein with $\phi = 3$:

$$\Omega = 3^{100} \approx 5 \times 10^{47}$$

Sampling each at the fastest bond rotation time ($\sim 10^{-13}$ s) would require$5 \times 10^{47} \times 10^{-13} = 5 \times 10^{34}$ seconds — vastly longer than the age of the universe ($4 \times 10^{17}$ s). Yet proteins fold in milliseconds to seconds. This is Levinthal's paradox.

The resolution is the folding funnel. The free energy landscape in conformational space is not flat but funnel-shaped, with the native state at the bottom. The configurational entropy at energy level $E$ is:

$$S_{\text{conf}}(E) = k_B \ln \Omega(E)$$

As the protein descends the funnel, the number of accessible states decreases (lower entropy) while the average energy decreases (lower enthalpy). The free energy$G = E - TS_{\text{conf}}$ decreases monotonically if the funnel is smooth enough:

$$\frac{dG}{d\text{(folding progress)}} = \frac{dE}{d\text{progress}} - T\frac{dS_{\text{conf}}}{d\text{progress}} < 0$$

The folding time can be estimated from the funnel steepness. For a minimally frustrated funnel with energy gap $\delta E$ between native and non-native contacts, and roughness $\Delta E$:

$$t_{\text{fold}} \approx t_0 \exp\left(\frac{\Delta E^2}{k_B T \cdot \delta E}\right)$$

where $t_0 \sim 1$ $\mu$s is the elementary folding attempt time. For a smooth funnel ($\Delta E \ll \delta E$), folding is fast. Misfolded states correspond to local minima (traps) on the landscape. Chaperonins like GroEL/GroES help proteins escape these kinetic traps by providing an unfolding chamber that allows the protein to restart its folding trajectory.

The Two-State Model of Protein Folding

Many small proteins fold in a two-state manner: N (native) $\rightleftharpoons$ U (unfolded), with no detectable intermediates. The equilibrium constant is:

$$K = \frac{[U]}{[N]} = e^{-\Delta G / RT}$$

The fraction folded is $f_N = 1/(1 + K)$, giving the sigmoidal melting curve. At$T = T_m$, $K = 1$ and $f_N = 0.5$. The van't Hoff enthalpy can be extracted from the steepness of the melting curve:

$$\Delta H_{\text{vH}} = -R \frac{d\ln K}{d(1/T)} = 4RT_m^2 \left(\frac{df_N}{dT}\right)_{T_m}$$

Agreement between $\Delta H_{\text{vH}}$ and $\Delta H_{\text{cal}}$ (from calorimetry) confirms two-state folding. A ratio $\Delta H_{\text{vH}}/\Delta H_{\text{cal}} < 1$indicates intermediates; a ratio $> 1$ suggests oligomerization during unfolding.