Le Chatelier's principle, quantified

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Le Chatelier's principle, quantified

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Statement

Let a single reaction run in one homogeneous phase, its composition described by one extent of reaction, and let the mixture sit at a stable equilibrium. Write \(\nu_i\) for the stoichiometric numbers (negative for reactants, positive for products), \(\xi\) for the extent, so that \(dn_i=\nu_i\,d\xi\), and \(\Delta_{\mathrm r}G\equiv(\partial G/\partial\xi)_{T,P}\), \(G_{\xi\xi}\equiv(\partial^2G/\partial\xi^2)_{T,P}\) for the reaction Gibbs energy and its curvature; stability is \(\Delta_{\mathrm r}G=0\) with \(G_{\xi\xi}\gt 0\). Then for any control variable \(\lambda\) the equilibrium extent moves by \((\partial\xi_{\mathrm{eq}}/\partial\lambda)=-G_{\xi\xi}^{-1}(\partial\Delta_{\mathrm r}G/\partial\lambda)_{\xi}\); specialising \(\lambda\) gives \((\partial\xi/\partial T)_P=\Delta_{\mathrm r}H/(T\,G_{\xi\xi})\), \((\partial\xi/\partial P)_T=-\Delta_{\mathrm r}V/G_{\xi\xi}\), and, for adding \(dn_j\) of any species \(j\) to an ideal mixture at fixed \(T\) and \(P\), \((\partial\xi/\partial n_j)_{T,P}=(\Delta\nu-\nu_j/x_j)\big/\big(\sum_k\nu_k^2/x_k-\Delta\nu^2\big)\). The first two are Le Chatelier's principle made exact; the third shows that the popular verbal form of the principle is false as stated, because adding a reactant at constant pressure can drive the reaction backwards.

Why it matters

Le Chatelier's principle is usually taught as a slogan — a system at equilibrium responds so as to oppose an imposed change — and slogans do not survive contact with a real process design. Every question that actually matters industrially is quantitative: how much does raising the pressure from 100 to 200 bar buy in ammonia yield, and is that worth the compressor duty; how far does a 30 K temperature swing move a selectivity; is it cheaper to recycle unconverted feed or to add more of it. The derivative form derived here answers all of them from thermodynamic data alone, with no new postulate.

The quantitative form also exposes where the slogan breaks. It is not a law but a first-order sign rule, and it is only guaranteed for the two disturbances — temperature and pressure — that couple to \(\xi\) through a single, sign-definite second derivative. For the third common disturbance, adding a substance, the coupling depends on composition, and the sign genuinely flips inside an accessible region of composition space. Knowing exactly where that region is (a mole-fraction threshold, derived below) is the difference between a mnemonic and a result.

Hypotheses

The system starts at a stable equilibrium: \(\Delta_{\mathrm r}G=0\) and \(G_{\xi\xi}\gt 0\).The first condition makes \(\xi_{\mathrm{eq}}\) a stationary point of \(G\); the second makes it a minimum, and supplies the positive denominator that fixes every sign in the Result. Applied to a mixture that is merely kinetically frozen — a hydrogen–oxygen mixture at room temperature, say — the formulas return the response of an equilibrium the system is not at, and predict nothing about the real mixture.

One independent reaction coordinate.With \(r\) simultaneous independent reactions the equilibrium condition is a vector equation and the response is \(\partial\boldsymbol\xi/\partial\lambda=-\mathbf{G}_{\xi\xi}^{-1}\,\partial\boldsymbol{\Delta_{\mathrm r}G}/\partial\lambda\), an inverse Hessian rather than a division. The Hessian is positive definite, but its inverse has off-diagonal entries of either sign, so one reaction can be dragged the “wrong” way by its coupling to another — the routine situation in reforming and in coupled acid–base equilibria.

Ideal mixture, for the explicit formulas only.The master formula in the Result is exact for any equation of state; only the evaluations \(\Delta_{\mathrm r}V=\Delta\nu\,RT/P\) and \(G_{\xi\xi}=RT\big(\sum_i\nu_i^2/n_i-\Delta\nu^2/n\big)\) assume ideality. At the \(160\ \mathrm{bar}\) of Worked example 2 the fugacity coefficients of nitrogen and hydrogen sit a few per cent above unity and that of ammonia a few per cent below, so the combination \(K_{\varphi}=\varphi_{\mathrm{NH_3}}^{2}\big/\big(\varphi_{\mathrm{N_2}}\varphi_{\mathrm{H_2}}^{3}\big)\) departs from unity by tens of per cent: the ideal numbers below have the right structure and the wrong second digit.

The disturbance is differential, and the constraint set is named.“Add nitrogen” is not a thermodynamic instruction until one says what is held fixed while it is added. The response at constant \((T,P)\) and the response at constant \((T,V)\) have, for the ammonia synthesis, opposite signs in part of composition space; a finite addition must be integrated, and the linear coefficient over-predicts the shift by several per cent for changes of order ten per cent (Worked example 1).

Proof

The whole argument is the implicit function theorem applied to the equilibrium condition, plus one inequality from stability. Nothing about chemistry enters until Step 7.

\[ n_i(\xi) = n_i^{0} + \nu_i\,\xi, \qquad n(\xi)=n^{0}+\Delta\nu\,\xi, \qquad \Delta\nu\equiv\sum_i\nu_i \]

Definition of the extent of reaction: stoichiometry ties every amount to one scalar, so the composition of a closed reacting mixture is a curve in composition space parameterised by \(\xi\), and \(G\) restricted to that curve is a function of \((T,P,\xi)\) alone. A

\[ \Delta_{\mathrm r}G \equiv \left(\frac{\partial G}{\partial \xi}\right)_{T,P} = \sum_i \nu_i\,\mu_i, \qquad \mathcal{A} \equiv -\Delta_{\mathrm r}G \]

Chain rule on \(G=\sum_i n_i\mu_i\) with \(dn_i=\nu_i\,d\xi\); the Gibbs–Duhem relation \(\sum_i n_i\,d\mu_i=0\) at fixed \(T,P\) kills the remaining term. The negative of this slope is De Donder's affinity \(\mathcal{A}\), the generalised force conjugate to \(\xi\). A

\[ \Delta_{\mathrm r}G(T,P,\xi_{\mathrm{eq}})=0, \qquad G_{\xi\xi}\equiv\left(\frac{\partial^2 G}{\partial\xi^2}\right)_{T,P}\gt 0 \]

First- and second-order conditions for \(G\) to be minimised along the reaction curve at fixed \(T,P\). The strict inequality is the statement that the equilibrium is stable rather than critical, and it is the only inequality the whole result needs. A

\[ 0 = \frac{d}{d\lambda}\,\Delta_{\mathrm r}G\big(\lambda,\xi_{\mathrm{eq}}(\lambda)\big) = \left(\frac{\partial \Delta_{\mathrm r}G}{\partial \lambda}\right)_{\xi} + G_{\xi\xi}\,\frac{d\xi_{\mathrm{eq}}}{d\lambda} \;\Longrightarrow\; \frac{d\xi_{\mathrm{eq}}}{d\lambda} = -\frac{1}{G_{\xi\xi}}\left(\frac{\partial \Delta_{\mathrm r}G}{\partial \lambda}\right)_{\xi} \]

The implicit function theorem: \(G_{\xi\xi}\neq0\) by Step 3, so the equilibrium condition defines \(\xi_{\mathrm{eq}}(\lambda)\) as a differentiable function near the equilibrium, and its derivative follows by differentiating the constraint. Because \(G_{\xi\xi}\gt 0\), the sign of every response is minus the sign of how the disturbance tilts the affinity — this single line is Le Chatelier's principle. B

\[ \left(\frac{\partial \Delta_{\mathrm r}G}{\partial T}\right)_{P,\xi} = \frac{\partial}{\partial\xi}\left(\frac{\partial G}{\partial T}\right)_{P} = -\Delta_{\mathrm r}S \;\Longrightarrow\; \left(\frac{\partial \xi_{\mathrm{eq}}}{\partial T}\right)_{P} = \frac{\Delta_{\mathrm r}S}{G_{\xi\xi}} = \frac{\Delta_{\mathrm r}H}{T\,G_{\xi\xi}} \]

Mixed partials commute for the smooth potential \(G\), and \((\partial G/\partial T)_P=-S\). The last equality uses the equilibrium condition itself: \(\Delta_{\mathrm r}G=\Delta_{\mathrm r}H-T\Delta_{\mathrm r}S=0\) forces \(\Delta_{\mathrm r}S=\Delta_{\mathrm r}H/T\) at equilibrium, which is what converts an entropy criterion into the familiar enthalpy one. B

\[ \left(\frac{\partial \Delta_{\mathrm r}G}{\partial P}\right)_{T,\xi} = \frac{\partial}{\partial\xi}\left(\frac{\partial G}{\partial P}\right)_{T} = \Delta_{\mathrm r}V \;\Longrightarrow\; \left(\frac{\partial \xi_{\mathrm{eq}}}{\partial P}\right)_{T} = -\frac{\Delta_{\mathrm r}V}{G_{\xi\xi}} \]

The same commutation with \((\partial G/\partial P)_T=V\). Note what the exact statement is: compression shifts the reaction toward the side of smaller volume, not of fewer moles; the two coincide only when the mixture is a gas obeying \(PV=nRT\). A

\[ \mu_i = \mu_i^{\circ}(T) + RT\ln\frac{x_i P}{P^{\circ}} \;\Longrightarrow\; \Delta_{\mathrm r}G = \Delta_{\mathrm r}G^{\circ}(T) + RT\ln Q_x + RT\,\Delta\nu\,\ln\frac{P}{P^{\circ}}, \qquad Q_x=\prod_i x_i^{\nu_i} \]

Chemical potential of a component of an ideal gas mixture, summed over the reaction with weights \(\nu_i\) — the standard route to the equilibrium constant, developed in free energy and the equilibrium constant. Separating the mole-fraction quotient \(Q_x\) from the total-pressure factor is the step that makes the pressure dependence explicit. A

\[ G_{\xi\xi} = RT\,\frac{\partial}{\partial\xi}\Big(\sum_i\nu_i\ln n_i-\Delta\nu\ln n\Big) = RT\left(\sum_i\frac{\nu_i^{2}}{n_i}-\frac{\Delta\nu^{2}}{n}\right) \gt 0 \]

Differentiate Step 7 using Step 1 (\(dn_i/d\xi=\nu_i\), \(dn/d\xi=\Delta\nu\)); the \(\Delta_{\mathrm r}G^{\circ}\) and pressure terms are \(\xi\)-independent and drop. Positivity is Cauchy–Schwarz: \(\Delta\nu^{2}=\big(\sum_i(\nu_i/\sqrt{n_i})\sqrt{n_i}\big)^{2}\le\big(\sum_i\nu_i^{2}/n_i\big)\big(\sum_i n_i\big)\), with equality only if every \(\nu_i/n_i\) is the same constant — impossible when some \(\nu_i\) are negative and all \(n_i\gt 0\). So an ideal reacting mixture is never unstable against reaction, and Step 3's hypothesis is automatic here rather than assumed. C

\[ \left(\frac{\partial \Delta_{\mathrm r}G}{\partial n_j}\right)_{T,P,\xi} = RT\left(\frac{\nu_j}{n_j}-\frac{\Delta\nu}{n}\right) \;\Longrightarrow\; \left(\frac{\partial \xi_{\mathrm{eq}}}{\partial n_j}\right)_{T,P} = \frac{\Delta\nu-\nu_j/x_j}{\displaystyle\sum_k\frac{\nu_k^{2}}{x_k}-\Delta\nu^{2}} \]

Injecting \(dn_j\) from outside raises \(n_j\) and \(n\) each by \(dn_j\) at fixed \(\xi\); differentiate Step 7 accordingly, insert into Step 4 with \(G_{\xi\xi}\) from Step 8, and clear the common factor \(RT/n\) by writing \(n/n_k=1/x_k\). The denominator is positive by Step 8, so the entire sign question is the numerator \(\Delta\nu-\nu_j/x_j\). Setting \(\nu_j=0\) covers addition of an inert diluent. B

\[ \text{forward shift} \iff \frac{\nu_j}{x_j}\lt\Delta\nu; \qquad \text{for a reactant } (\nu_j\lt 0):\ \text{reverse shift} \iff x_j \gt \frac{|\nu_j|}{|\Delta\nu|},\ \ \Delta\nu\lt -|\nu_j| \]

Read the numerator of Step 9. For a reactant, \(\nu_j/x_j=-|\nu_j|/x_j\), so the naive prediction reverses when \(-|\nu_j|/x_j\gt\Delta\nu\), i.e. when \(x_j\gt|\nu_j|/|\Delta\nu|\) with \(\Delta\nu\) negative. Since \(x_j\le1\), such a threshold lies in the physical range only if \(|\Delta\nu|\gt|\nu_j|\): the reaction must destroy more moles overall than the added species itself contributes, so that the dilution of the other reactants outweighs the enrichment in \(j\). For \(\mathrm{N_2}+3\mathrm{H_2}\rightleftharpoons2\mathrm{NH_3}\) this gives the threshold \(x_{\mathrm{N_2}}=1/2\), and no threshold at all for \(\mathrm{H_2}\) (\(|\nu|=3\gt|\Delta\nu|=2\)). C

Result

\[ \left(\frac{\partial \xi_{\mathrm{eq}}}{\partial \lambda}\right) = -\frac{1}{G_{\xi\xi}}\left(\frac{\partial \Delta_{\mathrm r}G}{\partial \lambda}\right)_{\xi}, \qquad G_{\xi\xi}\gt 0 \] \[ \left(\frac{\partial \xi}{\partial T}\right)_{P}=\frac{\Delta_{\mathrm r}H}{T\,G_{\xi\xi}}, \qquad \left(\frac{\partial \xi}{\partial P}\right)_{T}=-\frac{\Delta_{\mathrm r}V}{G_{\xi\xi}}, \qquad \left(\frac{\partial \xi}{\partial n_j}\right)_{T,P}=\frac{\Delta\nu-\nu_j/x_j}{\sum_k \nu_k^{2}/x_k-\Delta\nu^{2}} \]

Reading. Stability supplies a positive denominator once and for all, so every equilibrium response has the sign of minus the disturbance's tilt on the affinity. Heating shifts an endothermic reaction forward because \(\Delta_{\mathrm r}H\gt 0\); compression shifts toward smaller volume because of the minus sign on \(\Delta_{\mathrm r}V\); adding a species does whatever the composition-dependent numerator \(\Delta\nu-\nu_j/x_j\) says, which is not always what the slogan says.

Dimensions. \(G_{\xi\xi}\) is J mol\(^{-2}\) (energy per squared extent), \(\Delta_{\mathrm r}H\) is J mol\(^{-1}\), so \((\partial\xi/\partial T)_P\) is mol K\(^{-1}\); \(\Delta_{\mathrm r}V\) is m\(^3\) mol\(^{-1}\), giving \((\partial\xi/\partial P)_T\) in mol Pa\(^{-1}\). The addition coefficient is a pure number, as it must be: moles of reaction per mole added.

Scope. The master formula holds for any single reaction in any stable equilibrium, with any equation of state. The three specialisations hold as written for an ideal mixture; for real fluids replace \(x_i\) by activities and \(\Delta_{\mathrm r}V\) by the true partial-molar volume difference.

Corollaries & converses

Van 't Hoff. With \(\Delta_{\mathrm r}G^{\circ}=-RT\ln K\), the same Gibbs–Helmholtz manipulation that produced Step 5 gives \(d\ln K/dT=\Delta_{\mathrm r}H^{\circ}/(RT^{2})\); integrating with \(\Delta_{\mathrm r}H^{\circ}\) treated as constant gives \(\ln(K_2/K_1)=-(\Delta_{\mathrm r}H^{\circ}/R)(1/T_2-1/T_1)\).
Temperature is the only disturbance that moves \(K\). \(K\) is built from \(\Delta_{\mathrm r}G^{\circ}(T)\) alone, so pressure and composition changes move \(Q\) to meet a fixed \(K\); heating moves \(K\) itself. The mole-fraction constant does depend on pressure: \(K_x=K\,(P/P^{\circ})^{-\Delta\nu}\).
Exact null case. If \(\Delta\nu=0\) then \(\Delta_{\mathrm r}V=0\) for an ideal gas mixture and \((\partial\xi/\partial P)_T=0\) exactly — not approximately, and at every pressure.
Inert gas. Step 9 with \(\nu_j=0\) gives \((\partial\xi/\partial n_{\text{inert}})_{T,P}=\Delta\nu\big/\big(\sum_k\nu_k^{2}/x_k-\Delta\nu^{2}\big)\): at constant pressure a diluent shifts the reaction toward the side with more moles of gas, exactly as an expansion would. At constant volume the same addition changes no partial pressure and produces no shift at all.
Moderation, not cancellation. Because \(G_{\xi\xi}\) is finite and positive, the shift is never a restoration: the disturbed amount always ends up displaced in the direction it was pushed, never returned exactly to its old value and never over-shot past it. At constant \(T,V\) the fraction of an added reactant that the shift consumes is \(\big(\nu_j^{2}/n_j\big)\big/\big(\sum_k\nu_k^{2}/n_k\big)\), strictly between \(0\) and \(1\) by the same Cauchy–Schwarz bound as Step 8 (Problem 5 puts a number on it: \(93.9\%\) consumed, not \(100\%\)). At constant \(T,P\) even that bound is lost: in the anomalous regime of Step 10 the consumed fraction is negative — injecting nitrogen into the mixture of Worked example 2 leaves \(0.5\%\) more nitrogen than was injected.
Converse (sign inference). Measuring the direction of the shift under a known, single-variable disturbance determines the sign of \(\Delta_{\mathrm r}H\) (from a temperature step) or of \(\Delta_{\mathrm r}V\) (from a pressure step) without ever measuring \(K\); the magnitude additionally yields \(G_{\xi\xi}\), hence the curvature of \(G\) at the equilibrium.

Fails without

Constant pressure instead of constant volume, when adding a reactant: for \(\mathrm{N_2}+3\mathrm{H_2}\rightleftharpoons2\mathrm{NH_3}\) at fixed \(T,P\) with \(x_{\mathrm{N_2}}\gt1/2\), Step 10 gives a negative \((\partial\xi/\partial n_{\mathrm{N_2}})\): adding nitrogen decomposes ammonia. At \(x_{\mathrm{N_2}}=0.60,\ x_{\mathrm{H_2}}=0.20,\ x_{\mathrm{NH_3}}=0.20\) the coefficient is \(-5.3\times10^{-3}\), while the same addition at constant \(T,V\) gives \(+2.5\times10^{-2}\) — opposite signs, same chemistry, different constraint. The slogan “add a reactant, the equilibrium shifts forward” is simply false at constant pressure.
Stability, when \(G_{\xi\xi}\to0\): near a critical point, a spinodal, or an incipient phase split, the denominator collapses and every response coefficient diverges; the linearisation of Step 4 is then useless (the true shift is governed by the cubic term), and beyond the spinodal, where \(G_{\xi\xi}\lt0\), the stationary point is a maximum, the sign rule inverts, and the “equilibrium” is not one. Le Chatelier's principle is a corollary of stability, so it cannot survive the loss of stability.
A single reaction coordinate, in a coupled system: with two simultaneous equilibria the response is \(-\mathbf{G}_{\xi\xi}^{-1}\) applied to the disturbance vector, and although \(\mathbf{G}_{\xi\xi}\) is positive definite its inverse has off-diagonal elements of either sign. A disturbance that pushes reaction 1 forward can therefore pull reaction 2 backwards — routine in steam reforming, where methane conversion and the water–gas shift compete for the same species.
Ideality, at process pressures: the operative quantity in Step 6 is \(\Delta_{\mathrm r}V\), not \(\Delta\nu\). For reactions in condensed phases \(\Delta_{\mathrm r}V\) is a partial-molar-volume difference of either sign and no mole-counting rule applies; for gases above roughly 100 bar the fugacity corrections shift the numbers by tens of per cent, which is why real ammonia plants are designed from fugacity-based equilibria rather than from \(\Delta\nu=-2\).

Common errors

“Adding more of a reactant always pushes the reaction forward.” True at constant \(T,V\), where the added species is the only concentration that changes; false in general at constant \(T,P\), where the addition also dilutes every other species (Step 10, and the first entry under Fails without).
“High pressure favours the side with fewer molecules.” Only if the pressure is raised by compression. Raising the total pressure by pumping in argon at constant volume does nothing at all, and adding argon at constant pressure shifts the reaction the other way.
“\(K\) changes with pressure because the equilibrium shifts.” \(K\) is a function of \(T\) only, being defined from \(\Delta_{\mathrm r}G^{\circ}\) at the standard pressure. What moves is \(Q_x\), or equivalently \(K_x=K(P/P^{\circ})^{-\Delta\nu}\).
“\(\ln K\) is linear in \(1/T\).” Only while \(\Delta_{\mathrm r}H^{\circ}\) is constant. Over a 400 K span \(\Delta_{\mathrm r}C_P^{\circ}\) matters; the integrated van 't Hoff numbers quoted below are therefore good to a factor of order unity, not to three figures, and should be labelled as such.
“A catalyst shifts the equilibrium.” Nothing in Steps 1–10 refers to a rate. A catalyst changes neither \(G(T,P,\xi)\) nor its derivatives, so it changes neither \(\xi_{\mathrm{eq}}\) nor any response coefficient; it changes only the time taken to arrive.
Confusing \(\Delta_{\mathrm r}H\) with \(\Delta_{\mathrm r}H^{\circ}\) in Step 5. The exact response coefficient carries the reaction enthalpy at the actual composition; for an ideal mixture the enthalpy of mixing vanishes and the two coincide, which is precisely the licence being used.

Discussion

Le Chatelier put forward the qualitative principle in 1884, and Braun published a related statement shortly afterwards; the version that survives is neither author's prose but the derivative form above, which belongs to the thermodynamics of the affinity developed by De Donder and his school and later systematised by Prigogine and Defay. The historical point worth keeping is that the principle was proposed as a general law of moderation applying to any disturbance whatever, and that this generality is exactly what fails: the theorem is about second derivatives of a potential, and only disturbances that enter through a well-signed mixed second derivative are covered.

The structure is completely general, which is why the same algebra recurs across physics. Anything of the form “minimise a convex potential subject to a parameter” obeys \(\partial(\text{response})/\partial(\text{parameter})=-(\text{curvature})^{-1}\times(\text{coupling})\), with the sign fixed by convexity. Mechanical stability (\(\kappa_T\gt0\)) and thermal stability (\(C_P\gt0\)) are the same theorem applied to volume and entropy rather than to extent; that is why a stable substance always expands when the pressure on it is reduced, and why no equilibrium system spontaneously runs away from a small nudge.

There is a sharper version of the moderation idea, sometimes called the Le Chatelier–Braun principle, concerning secondary responses. In the present language it is the Schur-complement inequality for a positive-definite Hessian: releasing a constraint on a second variable can only lower the effective curvature seen by the first, so the released system is the softer one and the constrained response is always the smaller. Read from the other end, that is the moderation statement — clamping the second variable lets its conjugate force build up, and the build-up always pushes back against the primary shift. This is the rigorous content behind \(\kappa_T\gt\kappa_S\), the isothermal compressibility exceeding the adiabatic one because at fixed entropy the temperature rise on compression opposes the compression, and behind the matching observation that a reaction compressed adiabatically shifts less than the same reaction held isothermal: the heat it releases warms it, and for an exothermic reaction warming undoes part of the shift.

Common misconceptions. The persistent one is that Le Chatelier's principle is a law about systems “wanting” to undo disturbances. It is a statement about the curvature of a thermodynamic potential, and the teleological reading gives no way to predict the failures: the constant-pressure reactant-addition anomaly is not a curiosity of an exotic system but occurs in the single most-taught industrial equilibrium, at compositions a plant operator could plausibly choose. A second misconception is that the response is a restoration — the shift is always partial, so the disturbed quantity ends up displaced in the direction it was pushed. Nothing is ever put back.

Worked examples

Example 1. For \(\mathrm{N_2O_4}(g)\rightleftharpoons2\mathrm{NO_2}(g)\) at \(298.15\,\mathrm{K}\), standard tabulated formation data give \(\Delta_{\mathrm f}G^{\circ}(\mathrm{NO_2},g)=+51.3\ \mathrm{kJ\,mol^{-1}}\) and \(\Delta_{\mathrm f}G^{\circ}(\mathrm{N_2O_4},g)=+97.9\ \mathrm{kJ\,mol^{-1}}\). Starting from \(1.000\) mol of \(\mathrm{N_2O_4}\) at \(P=1.000\) bar, find the equilibrium extent, the exact extent at \(2.000\) bar, and check the linear-response coefficient \((\partial\xi/\partial P)_T\) against an exact calculation at \(1.100\) bar. Take \(R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}\) and \(P^{\circ}=1\) bar.

\[ \Delta_{\mathrm r}G^{\circ}=2(51.3)-97.9=+4.7\ \mathrm{kJ\,mol^{-1}}, \qquad K=\exp\!\left(-\frac{\Delta_{\mathrm r}G^{\circ}}{RT}\right)=\exp\!\left(-\frac{4700}{2478.8}\right)=0.150 \]

Hess-law combination of formation energies, then \(\Delta_{\mathrm r}G^{\circ}=-RT\ln K\) with \(RT=8.314\times298.15=2478.8\ \mathrm{J\,mol^{-1}}\). A

\[ n_{\mathrm{N_2O_4}}=1-\xi,\quad n_{\mathrm{NO_2}}=2\xi,\quad n=1+\xi \;\Longrightarrow\; K=\frac{(2\xi/(1+\xi))^{2}}{(1-\xi)/(1+\xi)}\cdot\frac{P}{P^{\circ}} = \frac{4\xi^{2}}{1-\xi^{2}}\cdot\frac{P}{P^{\circ}} \]

Step 1 of the Proof with \(\nu=(-1,+2)\), then Step 7's \(K=Q_x(P/P^{\circ})^{\Delta\nu}\) with \(\Delta\nu=+1\). Symbols are rearranged before any number is inserted. A

\[ \xi^{2}=\frac{K P^{\circ}/P}{4+K P^{\circ}/P} \;\Longrightarrow\; \xi(1.000\ \mathrm{bar})=\sqrt{\frac{0.150}{4.150}}=0.1901, \qquad \xi(2.000\ \mathrm{bar})=\sqrt{\frac{0.0750}{4.0750}}=0.1357 \]

Solving for \(\xi^{2}\) — the equilibrium condition is linear in it — then substituting. Doubling the pressure cuts the degree of dissociation from \(19.0\%\) to \(13.6\%\), the direction demanded by \(\Delta_{\mathrm r}V\gt0\) in the Result. A

\[ \frac{G_{\xi\xi}}{RT}=\sum_i\frac{\nu_i^{2}}{n_i}-\frac{\Delta\nu^{2}}{n}=\frac{1}{1-\xi}+\frac{4}{2\xi}-\frac{1}{1+\xi} = 1.2347+10.5198-0.8403=10.914\ \mathrm{mol^{-1}} \]

Proof Step 8 evaluated at \(\xi=0.1901\). Positive, as Cauchy–Schwarz guarantees, and dominated by the \(2/\xi\) term because \(\mathrm{NO_2}\) is the scarce species — scarcity of any participant stiffens the equilibrium against being moved. B

\[ \left(\frac{\partial \xi}{\partial P}\right)_{T}=-\frac{\Delta_{\mathrm r}V}{G_{\xi\xi}}=-\frac{\Delta\nu\,RT/P}{RT(10.914)}=-\frac{1}{(1.000\ \mathrm{bar})(10.914)}=-0.0916\ \mathrm{mol\ bar^{-1}} \]

Result, second specialisation, with \(\Delta_{\mathrm r}V=\Delta\nu RT/P\) for an ideal gas mixture; the factor \(RT\) cancels, so the coefficient depends only on composition and pressure, not on temperature. A

\[ \Delta\xi_{\text{lin}}(+0.100\ \mathrm{bar})=-0.00916, \qquad \Delta\xi_{\text{exact}}=\sqrt{\frac{0.13636}{4.13636}}-0.19012=0.18157-0.19012=-0.00855 \]

The linear coefficient over-predicts the shift by \(7.2\%\) for a \(10\%\) pressure change, because \(G_{\xi\xi}\) itself rises as \(\xi\) falls (Step 4's \(2/\xi\) term). This is the quantitative content of “the disturbance must be differential” in the Hypotheses. B

\[ \xi(1\ \mathrm{bar})=0.190,\qquad \xi(2\ \mathrm{bar})=0.136,\qquad \left(\frac{\partial\xi}{\partial P}\right)_{T}=-0.0916\ \mathrm{mol\ bar^{-1}} \]

Reading. The equilibrium is soft: a one per cent pressure rise moves the extent by \(9\times10^{-4}\ \mathrm{mol}\), about half a per cent of \(\xi\) itself, and the linear coefficient stays trustworthy to about seven per cent over a ten per cent perturbation.

Scope. Ideal-gas mixture at \(298.15\ \mathrm{K}\); \(K=0.150\) carries the uncertainty of the two formation energies, roughly \(\pm0.1\ \mathrm{kJ\,mol^{-1}}\) each, i.e. about \(\pm10\%\) in \(K\) and \(\pm5\%\) in \(\xi\).

Example 2. For the ammonia synthesis \(\mathrm{N_2}(g)+3\mathrm{H_2}(g)\rightleftharpoons2\mathrm{NH_3}(g)\), standard tabulated data at \(298.15\ \mathrm{K}\) give \(\Delta_{\mathrm f}H^{\circ}(\mathrm{NH_3},g)=-45.9\ \mathrm{kJ\,mol^{-1}}\) and third-law entropies \(S^{\circ}=192.8,\ 191.6,\ 130.7\ \mathrm{J\,K^{-1}\,mol^{-1}}\) for \(\mathrm{NH_3}\), \(\mathrm{N_2}\), \(\mathrm{H_2}\). Find \(K\) at \(700\ \mathrm{K}\); find the total pressure at which the mixture \(x_{\mathrm{N_2}}=0.60,\ x_{\mathrm{H_2}}=0.20,\ x_{\mathrm{NH_3}}=0.20\) is an equilibrium mixture at that temperature; and compute the response to adding \(0.0100\ \mathrm{mol}\) of \(\mathrm{N_2}\) to \(1.000\ \mathrm{mol}\) of it, at constant \((T,P)\) and at constant \((T,V)\).

\[ \Delta_{\mathrm r}H^{\circ}=2(-45.9)=-91.8\ \mathrm{kJ\,mol^{-1}}, \qquad \Delta_{\mathrm r}S^{\circ}=2(192.8)-191.6-3(130.7)=-198.1\ \mathrm{J\,K^{-1}\,mol^{-1}} \]

Elements in their standard states have zero formation enthalpy; the entropy is an absolute third-law sum weighted by \(\nu_i\). A

\[ \Delta_{\mathrm r}G^{\circ}(298)=-91800-298.15(-198.1)=-32.74\ \mathrm{kJ\,mol^{-1}} \;\Longrightarrow\; K(298)=e^{32740/2478.8}=5.4\times10^{5} \]

\(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\), then \(K=e^{-\Delta_{\mathrm r}G^{\circ}/RT}\). Strongly product-favoured at room temperature — and useless there, because the rate is immeasurably slow. A

\[ \ln K(700)=\ln K(298)-\frac{\Delta_{\mathrm r}H^{\circ}}{R}\left(\frac{1}{700}-\frac{1}{298}\right)=13.21-21.28=-8.07 \;\Longrightarrow\; K(700)=3.1\times10^{-4} \]

Integrated van 't Hoff (Corollaries) with \(\Delta_{\mathrm r}H^{\circ}\) held constant; \(-\Delta_{\mathrm r}H^{\circ}/R=+11042\ \mathrm{K}\) and \(1/700-1/298=-1.927\times10^{-3}\ \mathrm{K^{-1}}\). Nine orders of magnitude of \(K\) are destroyed by the heating — the exothermic penalty of Step 5 of the Proof, integrated. Two significant figures at most: \(\Delta_{\mathrm r}C_P^{\circ}\) is not zero over \(400\ \mathrm{K}\). B

\[ Q_x=\frac{x_{\mathrm{NH_3}}^{2}}{x_{\mathrm{N_2}}x_{\mathrm{H_2}}^{3}}=\frac{0.20^{2}}{0.60(0.20)^{3}}=8.33 \;\Longrightarrow\; \frac{P}{P^{\circ}}=\left(\frac{Q_x}{K}\right)^{1/2}=\left(\frac{8.33}{3.1\times10^{-4}}\right)^{1/2}=1.6\times10^{2} \]

Equilibrium requires \(K=Q_x(P/P^{\circ})^{\Delta\nu}\) with \(\Delta\nu=-2\), so \(Q_x=K(P/P^{\circ})^{2}\); solve for \(P\). About \(160\ \mathrm{bar}\) — squarely inside the operating range of real ammonia plants, which is the point: the composition chosen here is a physically reachable one, not a contrivance. B

\[ \sum_k\frac{\nu_k^{2}}{x_k}-\Delta\nu^{2}=\frac{1}{0.60}+\frac{9}{0.20}+\frac{4}{0.20}-4=1.667+45+20-4=62.67 \]

Denominator of the addition coefficient, from Proof Step 9. Large because hydrogen carries \(\nu^{2}=9\) at a small mole fraction: the equilibrium is stiff. A

\[ \left(\frac{\partial\xi}{\partial n_{\mathrm{N_2}}}\right)_{T,P}=\frac{\Delta\nu-\nu_{\mathrm{N_2}}/x_{\mathrm{N_2}}}{62.67}=\frac{-2-(-1/0.60)}{62.67}=\frac{-0.333}{62.67}=-5.32\times10^{-3} \]

Numerator negative because \(x_{\mathrm{N_2}}=0.60\) exceeds the threshold \(|\nu_j|/|\Delta\nu|=1/2\) of Step 10. Adding \(0.0100\ \mathrm{mol}\) of nitrogen therefore gives \(\Delta\xi=-5.3\times10^{-5}\ \mathrm{mol}\): about \(1.1\times10^{-4}\ \mathrm{mol}\) of ammonia is destroyed by adding a reactant. C

\[ \left(\frac{\partial\xi}{\partial n_{j}}\right)_{T,V}=\frac{-\nu_j/n_j}{\sum_k \nu_k^{2}/n_k}=\frac{1/0.600}{1/0.600+9/0.200+4/0.200}=\frac{1.667}{66.67}=+2.50\times10^{-2} \]

At constant volume \(\Delta_{\mathrm r}G=\Delta_{\mathrm r}G^{\circ}+RT\sum_i\nu_i\ln(n_i RT/VP^{\circ})\), so \(\partial\Delta_{\mathrm r}G/\partial n_j=RT\nu_j/n_j\) with no dilution term, and \(G_{\xi\xi}=RT\sum_k\nu_k^{2}/n_k\). The sign is now \(-\mathrm{sign}(\nu_j)\), always forward for a reactant. B

\[ K(700\ \mathrm{K})\approx3.1\times10^{-4},\quad P\approx1.6\times10^{2}\ \mathrm{bar}; \qquad \left(\frac{\partial\xi}{\partial n_{\mathrm{N_2}}}\right)_{T,P}=-5.3\times10^{-3} \ \ \text{but} \ \ \left(\frac{\partial\xi}{\partial n_{\mathrm{N_2}}}\right)_{T,V}=+2.5\times10^{-2} \]

Reading. The same chemical action — injecting nitrogen into an equilibrium ammonia mixture — makes ammonia at constant volume and destroys it at constant pressure, because at constant pressure the injection also dilutes the hydrogen, and hydrogen appears cubed. The verbal principle cannot distinguish the two cases; the derivative formula does so automatically.

Scope. Ideal-gas mixture at \(700\ \mathrm{K}\); at \(160\ \mathrm{bar}\) fugacity coefficients would change \(K\) and the coefficients by tens of per cent but not the sign, since the sign comes from stoichiometry and composition rather than from \(K\). The anomaly needs \(x_{\mathrm{N_2}}\gt1/2\), which a stoichiometric \(1:3\) feed can never reach (\(x_{\mathrm{N_2}}\le0.25\) along its whole reaction path).

Problems

(Routine.) For \(\mathrm{H_2}(g)+\mathrm{I_2}(g)\rightleftharpoons2\mathrm{HI}(g)\), \(\Delta_{\mathrm r}H^{\circ}=-9.4\ \mathrm{kJ\,mol^{-1}}\). State the exact effect of compression at constant temperature, and compute the fractional change in \(K\) on heating from \(700\ \mathrm{K}\) to \(710\ \mathrm{K}\).
Solution
Here \(\Delta\nu=2-1-1=0\), so \(\Delta_{\mathrm r}V=\Delta\nu RT/P=0\) and the Result gives \((\partial\xi/\partial P)_T=-\Delta_{\mathrm r}V/G_{\xi\xi}=0\) exactly, at every pressure — not merely a weak effect but an identically vanishing one (for an ideal gas mixture).
For temperature, \(d\ln K/dT=\Delta_{\mathrm r}H^{\circ}/(RT^{2})=-9400/(8.314\times700^{2})=-9400/4.074\times10^{6}=-2.31\times10^{-3}\ \mathrm{K^{-1}}\). Over \(\Delta T=10\ \mathrm{K}\), \(\Delta\ln K\approx-2.31\times10^{-2}\), so \(K\) falls by \(1-e^{-0.0231}=2.3\%\). The reaction is mildly exothermic, so heating reduces \(K\), and the effect is small because \(|\Delta_{\mathrm r}H^{\circ}|\) is small compared with \(RT=5.8\ \mathrm{kJ\,mol^{-1}}\) at this temperature.
(Routine.) Using \(K=0.150\) for \(\mathrm{N_2O_4}\rightleftharpoons2\mathrm{NO_2}\) at \(298.15\ \mathrm{K}\), find the total pressure at which \(1.000\) mol of \(\mathrm{N_2O_4}\) is \(10.0\%\) dissociated.
Solution
From Example 1 Step 2, \(K=\dfrac{4\xi^{2}}{1-\xi^{2}}\cdot\dfrac{P}{P^{\circ}}\). Rearranged for pressure before substituting: \(\dfrac{P}{P^{\circ}}=\dfrac{K(1-\xi^{2})}{4\xi^{2}}\).
With \(\xi=0.100\): \(\dfrac{P}{P^{\circ}}=\dfrac{0.150(1-0.0100)}{4(0.0100)}=\dfrac{0.1485}{0.0400}=3.71\), so \(P=3.71\ \mathrm{bar}\).
Check of direction: \(3.71\ \mathrm{bar}\gt1.000\ \mathrm{bar}\) and the dissociation has fallen from \(19.0\%\) to \(10.0\%\), consistent with \(\Delta_{\mathrm r}V\gt0\) and the negative sign in \((\partial\xi/\partial P)_T\). Note that a \(3.7\)-fold pressure rise was needed for a factor-\(1.9\) reduction in \(\xi\): the linear coefficient of Example 1 badly over-predicts a change this large.
(Standard.) Prove that adding \(\mathrm{H_2}\) to an ammonia-synthesis equilibrium at constant \(T,P\) always shifts the reaction forward, whatever the composition, and state the general condition on \(\nu_j\) and \(\Delta\nu\) under which a reactant addition can ever shift a reaction backwards at constant \(T,P\).
Solution
From Proof Step 9 the sign of \((\partial\xi/\partial n_j)_{T,P}\) is the sign of \(\Delta\nu-\nu_j/x_j\), the denominator being positive. For \(\mathrm{H_2}\), \(\nu_j=-3\) and \(\Delta\nu=-2\), so the numerator is \(-2+3/x_{\mathrm{H_2}}\). Since \(0\lt x_{\mathrm{H_2}}\le1\) we have \(3/x_{\mathrm{H_2}}\ge3\gt2\), hence the numerator is at least \(+1\) and always positive: the shift is forward for every composition. At \(x_{\mathrm{H_2}}=0.20\) (Example 2) the numerator is \(-2+15=+13\) and the coefficient is \(13/62.67=+0.207\), forty times the magnitude of the nitrogen coefficient and of the opposite sign.
In general, for a reactant \(\nu_j=-|\nu_j|\), a reverse shift requires \(\Delta\nu-\nu_j/x_j\lt0\), i.e. \(\Delta\nu\lt-|\nu_j|/x_j\). This needs \(\Delta\nu\lt0\), and, since \(|\nu_j|/x_j\ge|\nu_j|\), it needs \(|\Delta\nu|\gt|\nu_j|\). When that holds the threshold is \(x_j\gt|\nu_j|/|\Delta\nu|\). Interpretation: the added species must be a minor stoichiometric participant in a reaction that contracts strongly, so that the dilution it imposes on the other reactants outweighs its own enrichment.
For \(\mathrm{N_2}\): \(|\nu_j|=1\lt|\Delta\nu|=2\), threshold \(x_{\mathrm{N_2}}\gt1/2\) — attainable. For \(\mathrm{H_2}\): \(3\gt2\), so no threshold exists. For \(2\mathrm{SO_2}+\mathrm{O_2}\rightleftharpoons2\mathrm{SO_3}\) (\(\Delta\nu=-1\)) neither \(|\nu_{\mathrm{SO_2}}|=2\) nor \(|\nu_{\mathrm{O_2}}|=1\) is smaller than \(|\Delta\nu|=1\), so no reactant addition can ever reverse that equilibrium at constant \(T,P\).
(Standard.) Using the ammonia data of Example 2, find the temperature at which \(K=1\), and comment on why industrial synthesis is nevertheless run some \(200\ \mathrm{K}\) above it.
Solution
\(K=1\) means \(\Delta_{\mathrm r}G^{\circ}=-RT\ln K=0\), so with \(\Delta_{\mathrm r}G^{\circ}=\Delta_{\mathrm r}H^{\circ}-T\Delta_{\mathrm r}S^{\circ}\) held to constant \(\Delta H^{\circ},\Delta S^{\circ}\), \(T=\Delta_{\mathrm r}H^{\circ}/\Delta_{\mathrm r}S^{\circ}\). Symbols first, then numbers: \(T=(-91800\ \mathrm{J\,mol^{-1}})/(-198.1\ \mathrm{J\,K^{-1}\,mol^{-1}})=463\ \mathrm{K}\).
Above \(463\ \mathrm{K}\) the standard-state equilibrium is reactant-favoured, and by \(700\ \mathrm{K}\), \(K\approx3\times10^{-4}\). Equilibrium alone would therefore say run cold. But nothing in this result mentions rates: the \(\mathrm{N}\equiv\mathrm{N}\) bond makes the uncatalysed reaction immeasurably slow at \(463\ \mathrm{K}\), and even on iron the rate at \(500\ \mathrm{K}\) is uneconomic. The plant buys back the lost \(K\) with pressure: at fixed \(T\), \(Q_x=K(P/P^{\circ})^{2}\), so raising \(P\) from \(1\) to \(160\ \mathrm{bar}\) multiplies the attainable mole-fraction quotient by \(2.6\times10^{4}\) — a factor of \(\sim10^{4}\) recovered against the \(\sim10^{9}\) lost to heating, plus recycle of unconverted feed for the rest. Le Chatelier's two levers are being played against each other, quantitatively.
(Harder — moderation.) A \(1.000\ \mathrm{L}\) vessel at \(298.15\ \mathrm{K}\) is charged with \(0.1000\ \mathrm{mol}\) of \(\mathrm{N_2O_4}\) and reaches equilibrium. Given \(K=0.150\) (pressure basis, \(P^{\circ}=1\ \mathrm{bar}\)), convert to a concentration basis, find the equilibrium concentrations, then add \(0.0100\ \mathrm{mol}\) of \(\mathrm{NO_2}\) at constant \(T,V\) and find the new equilibrium. What fraction of the added \(\mathrm{NO_2}\) is consumed?
Solution
Conversion. With \(p_i=c_iRT\), \(K=K_c\,(c^{\circ}RT/P^{\circ})^{\Delta\nu}\). Here \(\Delta\nu=+1\), \(c^{\circ}=1\ \mathrm{mol\,L^{-1}}\) and \(RT=0.083145\times298.15=24.79\ \mathrm{L\,bar\,mol^{-1}}\), so \(K_c=K/24.79=0.150/24.79=6.05\times10^{-3}\).
First equilibrium. With \(y\) mol L\(^{-1}\) of \(\mathrm{N_2O_4}\) dissociating, \((2y)^{2}/(0.1000-y)=6.05\times10^{-3}\), i.e. \(4y^{2}+6.05\times10^{-3}y-6.05\times10^{-4}=0\). The positive root is \(y=\big(-6.05\times10^{-3}+\sqrt{3.66\times10^{-5}+9.68\times10^{-3}}\big)/8=0.011565\). So \([\mathrm{NO_2}]=0.023131\) and \([\mathrm{N_2O_4}]=0.088435\ \mathrm{mol\,L^{-1}}\); check \(0.023131^{2}/0.088435=6.05\times10^{-3}\).
The disturbance. Adding \(0.0100\ \mathrm{mol}\) to \(1.000\ \mathrm{L}\) gives \([\mathrm{NO_2}]=0.033131\) and \(Q_c=0.033131^{2}/0.088435=1.241\times10^{-2}\gt K_c\), so the reaction runs in reverse. With \(z\) mol L\(^{-1}\) of \(\mathrm{N_2O_4}\) re-formed, \((0.033131-2z)^{2}/(0.088435+z)=6.05\times10^{-3}\). Solving numerically: \(z=0.00460\) gives \(6.156\times10^{-3}\), \(z=0.00470\) gives \(6.047\times10^{-3}\); interpolating, \(z=0.004697\).
Moderation. New \([\mathrm{NO_2}]=0.033131-2(0.004697)=0.023737\ \mathrm{mol\,L^{-1}}\). Of the \(0.0100\ \mathrm{mol\,L^{-1}}\) added, \(2z=0.009394\) was consumed, a fraction \(93.9\%\). The remaining \(6.1\times10^{-4}\ \mathrm{mol\,L^{-1}}\) leaves \(\mathrm{NO_2}\) permanently above its original \(0.023131\).
This is the general moderation statement of the Corollaries made concrete: the consumed fraction is strictly between \(0\) and \(1\) because \(G_{\xi\xi}\) is strictly positive and finite. A system that restored \([\mathrm{NO_2}]\) exactly would need \(G_{\xi\xi}=0\) (a critical equilibrium), and one that over-shot would need \(G_{\xi\xi}\lt0\), i.e. no equilibrium at all. The fraction is close to \(1\) here only because \(\mathrm{N_2O_4}\) is abundant relative to the perturbation; diluting the vessel tenfold would make the equilibrium much softer and the moderation much weaker.