Ten Orders of Magnitude from Equilibrium: What a Cell's Steady State Costs — Epoche C2
The claim that living matter maintains order in defiance of the second law is repeated often enough to have acquired the status of a fact. It is not one, and the reason it is not is worth setting out step by step, because the correct account replaces a metaphor with a number. Units are strict SI throughout, as in the biophysical literature, with molar free energies in kJ mol$^{-1}$ and single-molecule energies in units of $k_BT$; at the body temperature $T = 310$ K, $RT = 8.314\times310 = 2577$ J mol$^{-1}$, so one $k_BT$ corresponds to 2.577 kJ mol$^{-1}$. The premise that dissolves the paradox The second law constrains the total entropy of a system together with its surroundings, not the entropy of a part. A system may lower its own entropy indefinitely provided it exports at least as much to the environment. This is not a special provision for biology; a domestic refrigerator does it. So the mere existence of internal order is no puzzle, and Schrödinger's 1944 formulation — that an organism feeds on negative entropy — is already the answer rather than a paradox. The real question is what kind of thermodynamic condition a cell occupies, and here there are three possibilities, not two. Equilibrium requires detailed balance: every elementary reaction proceeds at the same rate in both directions, so all net fluxes vanish. A transient relaxation has time-varying concentrations and non-zero fluxes. A cell has neither. Its concentrations are constant over hours while its fluxes are large. That combination — $\mathrm{d}c_i/\mathrm{d}t = 0$ for every species $i$ with net flux $J \neq 0$ — defines a nonequilibrium steady state, and it is a distinct condition that can be sustained only by a continuous supply of free energy. Measuring the displacement The single most informative number is the distance of ATP hydrolysis from its own equilibrium. The free energy change is $$\Delta G = \Delta G^{\circ\prime} + RT\ln Q, \qquad Q = \frac{[\mathrm{ADP}][\mathrm{P_i}]}{[\mathrm{ATP}]},$$ where $\Delta G^{\circ\prime} = -30.5$ kJ mol$^{-1}$ is the standard value at pH 7 and $Q$ is the reaction quotient, referred to a 1 M standard state. Typical mammalian cytosolic concentrations are $[\mathrm{ATP}] = 3$ mM, $[\mathrm{ADP}] = 0.03$ mM and $[\mathrm{P_i}] = 5$ mM, giving $$Q = \frac{(3\times10^{-5})(5\times10^{-3})}{3\times10^{-3}} = 5.0\times10^{-5}.$$ Then $RT\ln Q = 2.577\times(-9.90) = -25.5$ kJ mol$^{-1}$, so $\Delta G = -30.5 - 25.5 = -56.0$ kJ mol$^{-1}$, which is $56.0/2.577 = 21.7\,k_BT$ per molecule hydrolysed. Now the displacement itself. At equilibrium $\Delta G = 0$, so $Q_{\mathrm{eq}} = \exp(-\Delta G^{\circ\prime}/RT) = \exp(30500/2577) = \exp(11.83) = 1.37\times10^{5}$. The cell holds $Q$ at $5.0\times10^{-5}$. The ratio is $$\frac{Q_{\mathrm{eq}}}{Q} = \frac{1.37\times10^{5}}{5.0\times10^{-5}} = 2.7\times10^{9}.$$ Nearly ten orders of magnitude. This is not a transient; it is a value held constant while ATP is consumed at enormous rates. Note the consistency check: $\Delta G = RT\ln(Q/Q_{\mathrm{eq}}) = 2.577\times\ln(3.6\times10^{-10}) = -56.0$ kJ mol$^{-1}$, the same figure by a different route. What the displacement buys, and what it costs Consider the sodium-potassium pump, which exports three Na$^+$ and imports two K$^+$ per ATP. Exporting one Na$^+$ against a concentration ratio of $145/12 = 12.1$ costs $RT\ln 12.1 = 6.4$ kJ mol$^{-1}$, and against a membrane potential of $-70$ mV inside costs a further $zF\Delta\psi = 1\times96485\times0.070 = 6.8$ kJ mol$^{-1}$, where $F$ is the Faraday constant — a total of 13.2. Importing one K$^+$ against a ratio of $140/4 = 35$ costs $RT\ln 35 = 9.2$ kJ mol$^{-1}$, from which the potential returns 6.8, leaving 2.4. One cycle therefore stores $3\times13.2 + 2\times2.4 = 44.4$ kJ mol$^{-1}$ out of the 56.0 available: an efficiency of 79%. The remaining 11.6 kJ mol$^{-1}$ is dissipated, which at 310 K is an entropy production of $11600/310 = 37$ J mol$^{-1}$ K$^{-1}$, or $37/8.314 = 4.5\,k_B$ per pump cycle. Scale that up. A resting adult dissipates about 100 W, which over a day is $100\times86400 = 8.6$ MJ. Had all of it passed through ATP hydrolysis at 56.0 kJ mol$^{-1}$, the turnover would be $8.6\times10^{6}/5.6\times10^{4} = 154$ mol per day; at 507 g mol$^{-1}$ that is 78 kg. The familiar remark that one turns over roughly one's own body mass in ATP each day is not folklore but arithmetic, and the true figure is somewhat smaller because not every joule passes through ATP. Meanwhile the entropy production rate is $100/310 = 0.32$ W K$^{-1}$, or $0.32/1.381\times10^{-23} = 2.3\times10^{22}\,k_B$ per second. Against this, the entropy reduction represented by a cell's internal organisation is negligible. The second law is satisfied by an overwhelming margin, and the interesting quantity is not whether it is satisfied but by how much. Turning the condition into an observable The deduction so far requires knowing the chemistry. The last step removes that requirement, and it is what makes the subject quantitative rather than rhetorical. Equilibrium is equivalent to detailed balance, and detailed balance in a cyclic process has a signature: around any closed cycle, the product of forward rate constants equals the product of backward ones. Equivalently, in the space of a system's coordinates there is no circulating probability current. Any observed circulation is proof of a non-zero cycle affinity and hence of ongoing dissipation. Battle and colleagues exploited exactly this in 2016, tracking the bending modes of beating cilia and finding closed probability-flux loops in the plane of the first two modes — a direct, model-free demonstration that the system is not at equilibrium, obtained without identifying a single chemical reaction. The thermodynamic uncertainty relation, established by Barato and Seifert in 2015, goes further and converts fluctuation statistics into a lower bound on the dissipation. In the form convenient for a stepping motor it r