The van 't Hoff equation

Result

The van 't Hoff equation

T-142Home CU-203Threads equilibrium

Statement

For a reaction whose thermodynamic equilibrium constant \(K\) is defined from activities referred to standard states fixed at a temperature-independent standard pressure \(P^\circ\), the temperature dependence of \(K\) is governed entirely by the standard reaction enthalpy: \(\dfrac{d\ln K}{dT}=\dfrac{\Delta H^\circ}{RT^{2}}\), equivalently \(\dfrac{d\ln K}{d(1/T)}=-\dfrac{\Delta H^\circ}{R}\). This differential statement is exact. If, in addition, \(\Delta H^\circ\) may be treated as constant over the interval \([T_1,T_2]\) — that is, if the standard reaction heat capacity \(\Delta C_P^\circ\) is negligible over that range — integration gives the two-point form \(\ln\dfrac{K_2}{K_1}=-\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)\) and the linear form \(\ln K=-\dfrac{\Delta H^\circ}{RT}+\dfrac{\Delta S^\circ}{R}\).

Why it matters

gibbs-equilibrium-constant fixes the position of an equilibrium at one temperature, through \(\Delta G^\circ=-RT\ln K\). It says nothing about what happens when the temperature is changed, and the naive reading of that relation — that \(\ln K\) simply scales as \(1/T\) — is wrong, because \(\Delta G^\circ\) is itself a function of temperature. The van ’t Hoff equation supplies the missing law, and the answer is strikingly clean: of the two thermodynamic quantities that set an equilibrium, only the enthalpy controls how it moves with temperature. The entropy term is what fixes the intercept, not the slope.

Practically, this is the single most-used equation in experimental equilibrium chemistry. It converts a set of equilibrium measurements at different temperatures — solubilities, dissociation constants, binding constants, vapour pressures, partition coefficients — into \(\Delta H^\circ\) and \(\Delta S^\circ\) without any calorimeter, by the slope and intercept of a plot of \(\ln K\) against \(1/T\). It also turns the qualitative statement of le-chatelier — that heating shifts an endothermic equilibrium towards products — into a number, and it contains clausius-clapeyron as the special case in which the “reaction” is vaporisation.

Hypotheses

\(K\) is the thermodynamic equilibrium constant: a dimensionless product of activities, each referred to a standard state.A concentration quotient carrying units, or one written with bare pressures in atmospheres, changes value when the unit convention changes, so \(\ln K\) acquires an additive constant that is arbitrary. The constant is temperature-independent and so cancels from the differential form, but only if the same convention is used at every temperature; mixing \(K\) in \(\text{bar}\) at one temperature with \(K\) in \(\text{atm}\) at another corrupts the slope directly.

The standard states themselves do not depend on temperature.The IUPAC standard pressure \(P^\circ=1\,\text{bar}\) is a fixed number, which is what allows \(\Delta G^\circ\), \(\Delta H^\circ\) and \(\Delta S^\circ\) to be functions of \(T\) alone and lets partial derivatives be replaced by ordinary ones in Step 2. Had the standard state been defined as, say, the equilibrium pressure of the substance at each temperature, the derivation would carry an extra term and the result would be false.

\(\Delta H^\circ\) is constant over the temperature interval — needed only for the integrated forms.Since \(d\Delta H^\circ/dT=\Delta C_P^\circ\) (Kirchhoff’s law), this is the statement that the products and reactants have nearly equal total heat capacities. Where they do not, the van ’t Hoff plot curves and the two-point formula returns an enthalpy averaged over the range rather than the value at either endpoint (Fails without; Step 8 gives the correction).

No species changes phase, and no change of speciation or of the dominant reaction, occurs within the interval.Crossing a melting or boiling point of any participant changes which standard state \(K\) refers to and produces a genuine discontinuity in the slope; a weak acid whose degree of association changes with temperature, or an equilibrium whose dominant complex changes stoichiometry, produces an apparent \(\Delta H^\circ\) that is a composite of several processes, not a reaction enthalpy at all.

Proof

The whole derivation is one line of thermodynamics — the Gibbs–Helmholtz identity — applied to the logarithm of the equilibrium constant. Symbols are rearranged completely before any number appears.

\[ \ln K(T) = -\frac{\Delta G^\circ(T)}{RT} = -\frac{1}{R}\cdot\frac{\Delta G^\circ(T)}{T} \]

Rearranging gibbs-equilibrium-constant’s relation \(\Delta G^\circ=-RT\ln K\). Writing it with \(\Delta G^\circ/T\) grouped as a single object is the whole trick: it is that combination, not \(\Delta G^\circ\) itself, whose temperature derivative is simple. A

\[ \left(\frac{\partial G}{\partial T}\right)_{P}=-S \quad\Longrightarrow\quad \frac{d\,\Delta G^\circ}{dT} = -\Delta S^\circ \]

The fundamental relation \(dG=-S\,dT+V\,dP\) (from \(G=H-TS\) with \(dU=T\,dS-P\,dV\), as assembled in gibbs-free-energy) evaluated at the fixed standard pressure, then summed over the reaction with stoichiometric signs. Because the standard state is a fixed pressure (Hypotheses), the partial derivative at constant \(P\) is an ordinary derivative in \(T\). B

\[ \begin{aligned} \frac{d}{dT}\!\left(\frac{\Delta G^\circ}{T}\right) &= \frac{1}{T}\frac{d\,\Delta G^\circ}{dT}-\frac{\Delta G^\circ}{T^{2}} = -\frac{\Delta S^\circ}{T}-\frac{\Delta H^\circ-T\Delta S^\circ}{T^{2}} \\ &= -\frac{\Delta S^\circ}{T}-\frac{\Delta H^\circ}{T^{2}}+\frac{\Delta S^\circ}{T} \;=\; -\frac{\Delta H^\circ}{T^{2}} \end{aligned} \]

The quotient rule, then Step 2 for the first term and the definition \(\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ\) (gibbs-free-energy) for the second. The two entropy terms cancel identically — this is the Gibbs–Helmholtz equation, and the cancellation is exact, with no assumption yet that \(\Delta H^\circ\) or \(\Delta S^\circ\) is constant. B

\[ \frac{d\ln K}{dT} = -\frac{1}{R}\,\frac{d}{dT}\!\left(\frac{\Delta G^\circ}{T}\right) = \frac{\Delta H^\circ}{RT^{2}} \]

Differentiating Step 1 and substituting Step 3. This is the van ’t Hoff equation in its primitive form; note that \(\Delta S^\circ\) has disappeared entirely, so the direction in which an equilibrium moves on heating is decided by enthalpy alone. B

\[ d\!\left(\frac{1}{T}\right)=-\frac{dT}{T^{2}} \quad\Longrightarrow\quad \frac{d\ln K}{d(1/T)} = -\frac{\Delta H^\circ}{R} \]

A change of independent variable from \(T\) to \(1/T\). This is the form that matters experimentally: it says that \(\ln K\) plotted against reciprocal absolute temperature has local slope \(-\Delta H^\circ/R\), whatever else is true. A

\[ \int_{\ln K_1}^{\ln K_2}\! d\ln K = -\frac{\Delta H^\circ}{R}\int_{1/T_1}^{1/T_2}\! d\!\left(\frac{1}{T}\right) \quad\Longrightarrow\quad \ln\frac{K_2}{K_1}=-\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \]

Integrating Step 5 between two temperatures. Taking \(\Delta H^\circ\) outside the integral is the one approximation in the argument (Hypotheses, third): it is legitimate exactly when \(\Delta C_P^\circ\) is negligible over \([T_1,T_2]\). A

\[ \ln K = -\frac{\Delta H^\circ}{RT}+\frac{\Delta S^\circ}{R} \]

The indefinite integral of Step 5, with the constant of integration identified by comparing with Step 1 written as \(\ln K=-(\Delta H^\circ-T\Delta S^\circ)/(RT)\). Slope \(-\Delta H^\circ/R\), intercept \(\Delta S^\circ/R\): the van ’t Hoff plot. A

\[ \frac{d}{dT}\!\left[-\frac{\Delta H^\circ}{RT}+\frac{\Delta S^\circ}{R}\right] = \frac{\Delta H^\circ}{RT^{2}}-\frac{1}{RT}\frac{d\,\Delta H^\circ}{dT}+\frac{1}{R}\frac{d\,\Delta S^\circ}{dT} = \frac{\Delta H^\circ}{RT^{2}}-\frac{\Delta C_P^\circ}{RT}+\frac{\Delta C_P^\circ}{RT} \]

A consistency check that explains why Step 4 is exact while Step 7 is not. Differentiating the linear form and using Kirchhoff’s law in both its enthalpy and entropy versions, \(d\Delta H^\circ/dT=\Delta C_P^\circ\) and \(d\Delta S^\circ/dT=\Delta C_P^\circ/T\), the two heat-capacity terms cancel exactly and the differential law is recovered whether or not \(\Delta C_P^\circ\) vanishes. What fails without constant \(\Delta H^\circ\) is not the van ’t Hoff equation but the claim that the plot is a straight line. C

\[ \ln\frac{K(T)}{K(T_0)} = -\frac{\Delta H^\circ(T_0)}{R}\left(\frac{1}{T}-\frac{1}{T_0}\right)+\frac{\Delta C_P^\circ}{R}\left[\ln\frac{T}{T_0}+\frac{T_0}{T}-1\right] \]

Substituting \(\Delta H^\circ(T)=\Delta H^\circ(T_0)+\Delta C_P^\circ (T-T_0)\) into Step 4 and integrating term by term, using \(\int dT/T^{2}=-1/T\) and \(\int dT/T=\ln T\), with \(\Delta C_P^\circ\) taken constant. The bracket vanishes to first order in \((T-T_0)/T_0\), which is why the uncorrected form survives over modest ranges and fails over wide ones. C

Result

\[ \frac{d\ln K}{dT}=\frac{\Delta H^\circ}{RT^{2}}, \qquad \ln\frac{K_2}{K_1}=-\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right), \qquad \ln K=-\frac{\Delta H^\circ}{RT}+\frac{\Delta S^\circ}{R} \]

Reading. An equilibrium constant responds to temperature in proportion to the standard reaction enthalpy and inversely to the square of the absolute temperature. Endothermic reactions (\(\Delta H^\circ>0\)) have \(K\) increasing with \(T\); exothermic reactions (\(\Delta H^\circ\lt 0\)) have \(K\) decreasing with \(T\); a thermoneutral reaction has an equilibrium constant that is temperature-independent even though \(\Delta G^\circ=-RT\ln K\) is not.

Units check. \(\Delta H^\circ/R\) has units of \(\text{J}\,\text{mol}^{-1}/(\text{J}\,\text{mol}^{-1}\text{K}^{-1})=\text{K}\), which is exactly what the reciprocal-temperature factor \((1/T_2-1/T_1)\) needs to leave \(\ln(K_2/K_1)\) dimensionless. \(R=8.314\,\text{J}\,\text{mol}^{-1}\text{K}^{-1}\) is the molar gas constant, so \(\Delta H^\circ\) must be entered in \(\text{J}\,\text{mol}^{-1}\), not \(\text{kJ}\,\text{mol}^{-1}\).

Scope. The differential form is exact for any thermodynamic \(K\) with temperature-independent standard states. The integrated forms require \(\Delta C_P^\circ\approx 0\) across the interval; Step 9 gives the leading correction when it is not.

Corollaries & converses

Le Chatelier, quantified. The sign of \(dK/dT\) equals the sign of \(\Delta H^\circ\), and the size of the shift over a given interval is fixed, not merely qualitative: le-chatelier’s temperature clause is a corollary of this result rather than an independent principle.
The van ’t Hoff plot. Measuring \(K\) at several temperatures and fitting \(\ln K\) against \(1/T\) yields \(\Delta H^\circ\) from the slope and \(\Delta S^\circ\) from the intercept, using no calorimetric data whatever. This is the standard “second-law” determination of a reaction enthalpy.
Clausius–Clapeyron as a special case. For liquid \(\rightleftharpoons\) vapour with the pure liquid’s activity set to \(1\), \(K=P/P^\circ\) and \(\Delta H^\circ=\Delta H_{\text{vap}}\), and the Result reduces line for line to clausius-clapeyron’s integrated equation.
The isochoric form. For ideal gases, \(K_P=K_c\,(c^\circ RT/P^\circ)^{\Delta n}\) and \(\Delta H^\circ=\Delta U^\circ+\Delta n\,RT\), from which \(d\ln K_c/dT=\Delta U^\circ/(RT^{2})\) (Problems, 4). This is the relation to which the name “van ’t Hoff isochore” strictly belongs; the constant-pressure form is sometimes called the isobar.
Link to kinetics. For an elementary reversible step, \(K=k_{\text{f}}/k_{\text{r}}\), so combining with arrhenius-equation gives \(E_{a,\text{f}}-E_{a,\text{r}}=\Delta U^\circ\), and \(\approx\Delta H^\circ\) in condensed phases. Thermodynamics thus constrains the two activation energies of a mechanism, though it fixes neither alone.
Converse. Curvature in a measured \(\ln K\) against \(1/T\) plot is evidence of non-zero \(\Delta C_P^\circ\), of a changing speciation, or of a phase change — never of a failure of thermodynamics, since Step 8 shows the differential law survives all of these.

Fails without

Constant \(\Delta H^\circ\) dropped (large \(\Delta C_P^\circ\)): protein unfolding and protein–ligand binding have standard heat capacities of order \(\text{kJ}\,\text{mol}^{-1}\text{K}^{-1}\) — one to three orders of magnitude larger than the few tens of \(\text{J}\,\text{mol}^{-1}\text{K}^{-1}\) typical of small-molecule gas reactions, such as the \(-45.5\,\text{J}\,\text{mol}^{-1}\text{K}^{-1}\) of Worked example 1 — because burying or exposing hydrophobic surface changes the heat capacity of the surrounding water. The van ’t Hoff plot is then strongly curved, \(\Delta H^\circ\) passes through zero at some temperature, and \(K\) has a genuine maximum there (Problems, 5) — the thermodynamic origin of cold denaturation, which the straight-line form cannot represent at all. See protein-folding-thermodynamics.
A phase change inside the interval: for a solid whose sublimation is studied across the triple point, the slope changes abruptly from \(-\Delta H_{\text{sub}}/R\) below to \(-\Delta H_{\text{vap}}/R\) above, because the condensed standard state changes identity. Fitting one straight line through the kink returns an enthalpy that is a meaningless average of two different processes; the correct treatment is two separate fits.
A non-thermodynamic \(K\) at varying ionic strength: a concentration quotient measured in electrolyte solution absorbs the activity coefficients, whose own temperature dependence (debye-huckel) is not part of \(\Delta H^\circ\). The slope then delivers an apparent enthalpy that mixes the reaction enthalpy with an enthalpy of the ionic atmosphere, and the discrepancy grows with ionic strength.
Equilibrium not actually attained: if the measurement time is short compared with the relaxation time at low temperature but long compared with it at high temperature, the “\(K\)” values are kinetically limited at one end of the range. The resulting plot has a slope contaminated by activation energies (arrhenius-equation), not by \(\Delta H^\circ\).

Common errors

“The slope of \(\ln K\) against \(1/T\) is \(-\Delta G^\circ/R\).” Tempting, because \(\ln K=-\Delta G^\circ/(RT)\) looks like a proportionality to \(1/T\). It is not: \(\Delta G^\circ\) is itself temperature-dependent, and the whole content of Steps 2–4 is that the \(T\)-dependence of \(\Delta G^\circ\) removes the entropy contribution from the slope, leaving \(-\Delta H^\circ/R\).
“Use \(E_a\) in the two-point formula.” The identically-shaped arrhenius-equation two-point relation governs a rate constant with \(E_a\); this one governs an equilibrium constant with \(\Delta H^\circ\). The two are different quantities and, for an elementary step, differ by the reverse activation energy.
Confusing this result with the van ’t Hoff factor. The symbol \(i\) in the colligative-property expressions, and the osmotic-pressure relation \(\Pi=icRT\), are separate results of the same chemist. Neither has anything to do with the temperature dependence of \(K\).
Entering \(\Delta H^\circ\) in \(\text{kJ}\,\text{mol}^{-1}\) alongside \(R\) in \(\text{J}\,\text{mol}^{-1}\text{K}^{-1}\). A factor of \(1000\) in an exponent; the answer is wrong by many orders of magnitude and usually looks superficially plausible.
Celsius temperatures. \(1/T\) demands the absolute scale — the same error already flagged in ideal-gas-law.
“Heating always speeds a reaction up, so it must push the equilibrium forward.” Rate and position are independent: heating raises both \(k_{\text{f}}\) and \(k_{\text{r}}\), and it is only their ratio, governed by \(\Delta H^\circ\), that decides which way \(K\) moves. An exothermic reaction reaches a less product-rich equilibrium faster.

Discussion

Jacobus Henricus van ’t Hoff published the relation in his 1884 Études de dynamique chimique, the work that founded chemical dynamics as a quantitative subject and that earned him the first Nobel Prize in Chemistry in 1901. He is the same chemist who, a decade earlier, had proposed the tetrahedral carbon atom underlying chirality-optical-activity, and whose osmotic-pressure work supplies the van ’t Hoff factor. The equation reached him by a route different from the derivation above: he argued from the temperature dependence of the equilibrium of a reversible cycle, before Gibbs’s formalism was widely absorbed. The modern derivation, three lines from Gibbs–Helmholtz, shows the result to be a corollary of the second law rather than an independent empirical law.

The structural content of the result is worth dwelling on. Two quantities, \(\Delta H^\circ\) and \(\Delta S^\circ\), determine \(K\) at any one temperature, yet only one of them determines how \(K\) changes. The reason is visible in Step 3: the entropy enters \(\Delta G^\circ/T\) as a pure constant, \(-\Delta S^\circ\), and constants have zero derivative. Geometrically, the entropy sets where the van ’t Hoff line sits and the enthalpy sets how steeply it tilts. This is also why extrapolating a van ’t Hoff line to \(1/T\to 0\) to read off \(\Delta S^\circ\) is so unreliable in practice: the intercept lies at infinite temperature, far outside any measured range, so a small error in slope is levered into a large error in intercept.

There is a well-known and instructive discrepancy in biophysical chemistry between the “van ’t Hoff enthalpy” obtained from the slope of an equilibrium-versus-temperature curve and the calorimetric enthalpy measured directly by isothermal titration calorimetry. Step 8 shows that the two must agree for a genuine two-state process with the model correctly specified, so a robust disagreement is diagnostic: it indicates either that \(\Delta C_P^\circ\) is large enough that the fitted straight line is reporting a range-average, or that the assumed reaction is not the one occurring — a coupled protonation, an aggregation step, or a partly populated intermediate. The equation is therefore useful not only when it works but also, as a null hypothesis, when it fails.

Common misconceptions. That the equation predicts what a reaction will do when heated: it predicts only where equilibrium lies, and says nothing about whether the system can get there in the available time. That \(K\) always rises with temperature because “everything goes faster when hot”: about half of all reactions are exothermic, and for every one of them \(K\) falls on heating. And that a large \(\Delta H^\circ\) means a large \(K\): \(\Delta H^\circ\) sets the sensitivity of \(K\) to temperature, while \(\Delta G^\circ\) sets its value.

Worked examples

Example 1. For the Haber–Bosch synthesis \(\text{N}_2(g)+3\,\text{H}_2(g)\rightleftharpoons 2\,\text{NH}_3(g)\), estimate \(K\) at \(700\,\text{K}\) from standard thermodynamic data at \(298.15\,\text{K}\), and then assess how much the constant-\(\Delta H^\circ\) approximation costs over so wide a range. Standard values used, all at \(298.15\,\text{K}\): \(\Delta_{\text{f}}H^\circ(\text{NH}_3,g)=-45.9\,\text{kJ}\,\text{mol}^{-1}\); molar entropies \(S^\circ(\text{N}_2)=191.6\), \(S^\circ(\text{H}_2)=130.7\), \(S^\circ(\text{NH}_3)=192.8\,\text{J}\,\text{mol}^{-1}\text{K}^{-1}\); molar heat capacities \(C_{P}^\circ(\text{N}_2)=29.12\), \(C_{P}^\circ(\text{H}_2)=28.82\), \(C_{P}^\circ(\text{NH}_3)=35.06\,\text{J}\,\text{mol}^{-1}\text{K}^{-1}\).

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

Hess’s-law sums over the balanced equation (hess-law); the elements have zero formation enthalpy by convention. The entropy is strongly negative because four moles of gas become two. A

\[ \Delta G^\circ_{298} = \Delta H^\circ - T\Delta S^\circ = -91\,800-(298.15)(-198.1) = -32\,740\,\text{J}\,\text{mol}^{-1} \]

gibbs-free-energy, with \(\Delta H^\circ\) converted to \(\text{J}\,\text{mol}^{-1}\) first so the two terms are commensurate. A

\[ K_{298} = \exp\!\left(-\frac{\Delta G^\circ_{298}}{RT}\right) = \exp\!\left(\frac{32\,740}{(8.314)(298.15)}\right) = e^{13.21} = 5.4\times 10^{5} \]

gibbs-equilibrium-constant, giving the reference point the van ’t Hoff equation propagates. \(K\) is dimensionless, defined as \((p_{\text{NH}_3}/P^\circ)^2/[(p_{\text{N}_2}/P^\circ)(p_{\text{H}_2}/P^\circ)^3]\) with \(P^\circ=1\,\text{bar}\). A

\[ \ln\frac{K_{700}}{K_{298}} = -\frac{-91\,800}{8.314}\left(\frac{1}{700}-\frac{1}{298.15}\right) = (11\,042)(-1.9254\times10^{-3}) = -21.26 \]

The two-point form, Step 6 of the Proof. The bracket is negative because \(T_2>T_1\), and \(\Delta H^\circ\) is negative, so the product is negative: an exothermic equilibrium is driven backwards by heating. A

\[ K_{700} = (5.4\times10^{5})\,e^{-21.26} = 3.2\times10^{-4} \]

Nine orders of magnitude lost over four hundred kelvin — the quantitative statement of why ammonia synthesis is run at high pressure (\(\Delta n=-2\), so pressure pushes the other way) and with a catalyst to make the low temperature kinetically usable. A

\[ \Delta C_P^\circ = 2(35.06)-29.12-3(28.82) = -45.5\,\text{J}\,\text{mol}^{-1}\text{K}^{-1} \quad\Longrightarrow\quad \Delta H^\circ(700)\approx -91\,800+(-45.5)(401.85) = -110.1\,\text{kJ}\,\text{mol}^{-1} \]

Kirchhoff’s law. The reaction becomes appreciably more exothermic at \(700\,\text{K}\), by about twenty per cent, so the constant-\(\Delta H^\circ\) assumption is visibly strained here. C

\[ \ln\frac{K_{700}}{K_{298}} = -21.26+\frac{-45.5}{8.314}\left[\ln\frac{700}{298.15}+\frac{298.15}{700}-1\right] = -21.26+(-5.473)(0.2794) = -22.79 \]

Step 9 of the Proof, with \(\Delta C_P^\circ\) held constant across the range (itself an approximation, since heat capacities rise with temperature). C

\[ K_{700}^{\text{corr}} = (5.4\times10^{5})\,e^{-22.79} = 6.9\times10^{-5} \]

The correction divides the estimate by about \(4.6\). That is the honest measure of the error incurred by the straight-line form over a \(400\,\text{K}\) extrapolation: not catastrophic on a logarithmic scale, but far too large to quote three significant figures from the uncorrected number. C

\[ K_{298}=5.4\times10^{5}; \qquad K_{700}\approx 3.2\times10^{-4}\ \ (\Delta C_P^\circ=0), \qquad K_{700}\approx 6.9\times10^{-5}\ \ (\Delta C_P^\circ\ \text{corrected}) \]

Reading. Ammonia synthesis is thermodynamically excellent at room temperature and thermodynamically poor at the temperature where its rate is acceptable. The van ’t Hoff equation quantifies exactly this conflict, which is the reason the industrial process exists in the form it does.

Scope. Ideal-gas standard states throughout; the corrected value assumes only that \(\Delta C_P^\circ\) is constant, which is better than assuming \(\Delta H^\circ\) is, but still approximate.

Example 2. For \(\text{N}_2\text{O}_4(g)\rightleftharpoons 2\,\text{NO}_2(g)\), compute \(K\) at \(298.15\,\text{K}\) and at \(350\,\text{K}\), and convert both into the degree of dissociation at a total pressure of \(1.000\,\text{bar}\). Standard values used at \(298.15\,\text{K}\): \(\Delta_{\text{f}}H^\circ(\text{NO}_2)=+33.2\), \(\Delta_{\text{f}}H^\circ(\text{N}_2\text{O}_4)=+9.16\,\text{kJ}\,\text{mol}^{-1}\); \(S^\circ(\text{NO}_2)=240.1\), \(S^\circ(\text{N}_2\text{O}_4)=304.3\,\text{J}\,\text{mol}^{-1}\text{K}^{-1}\).

\[ \Delta H^\circ = 2(33.2)-9.16 = +57.24\,\text{kJ}\,\text{mol}^{-1}, \qquad \Delta S^\circ = 2(240.1)-304.3 = +175.9\,\text{J}\,\text{mol}^{-1}\text{K}^{-1} \]

Breaking the N–N bond is endothermic and increases the number of gas molecules, so both quantities are positive — the classic entropy-driven dissociation. A

\[ \Delta G^\circ_{298} = 57\,240-(298.15)(175.9) = +4795\,\text{J}\,\text{mol}^{-1} \quad\Longrightarrow\quad K_{298}=e^{-4795/2478.8}=0.1445 \]

\(RT=(8.314)(298.15)=2478.8\,\text{J}\,\text{mol}^{-1}\). Quoted to three figures this is \(K_{298}=0.145\); the fourth figure is carried into Step 4 so that the rounding does not propagate. A \(K\) below one: at standard pressure and room temperature the dimer is favoured. A

\[ \ln\frac{K_{350}}{K_{298}} = -\frac{57\,240}{8.314}\left(\frac{1}{350}-\frac{1}{298.15}\right) = (-6884.8)(-4.9687\times10^{-4}) = +3.421 \]

The two-point form again. Here \(\Delta H^\circ>0\) and the reciprocal-temperature bracket is negative, so the logarithm is positive: heating an endothermic equilibrium drives it forward. A

\[ K_{350} = 0.1445\,e^{3.421} = 4.42 \]

A thirty-fold rise in \(K\) for a temperature rise of only \(52\,\text{K}\); the equilibrium crosses from dimer-favoured to monomer-favoured. A

\[ K = \frac{4\alpha^{2}}{1-\alpha^{2}}\cdot\frac{P}{P^\circ} \quad\Longrightarrow\quad \alpha=\sqrt{\frac{K}{4(P/P^\circ)+K}} \]

Starting from one mole of \(\text{N}_2\text{O}_4\) dissociated to extent \(\alpha\), the amounts are \(1-\alpha\) and \(2\alpha\) in a total of \(1+\alpha\) moles, so \(K=[2\alpha/(1+\alpha)]^{2}(P/P^\circ)^{2}\big/\{[(1-\alpha)/(1+\alpha)](P/P^\circ)\}\), which simplifies as shown; the algebra is rearranged for \(\alpha\) before any number is inserted. B

\[ \alpha_{298}=\sqrt{\frac{0.145}{4.145}}=0.187, \qquad \alpha_{350}=\sqrt{\frac{4.42}{8.42}}=0.725 \]

Setting \(P=P^\circ=1.000\,\text{bar}\). Dissociation rises from about nineteen per cent to about seventy-two per cent. Since \(\text{N}_2\text{O}_4\) is colourless and \(\text{NO}_2\) is brown, this is the familiar lecture demonstration in which a sealed tube darkens on warming and pales on cooling. A

\[ K_{298}=0.145,\ \alpha_{298}=0.187 \qquad\longrightarrow\qquad K_{350}=4.42,\ \alpha_{350}=0.725 \quad (P=1.000\,\text{bar}) \]

Reading. A modest \(52\,\text{K}\) warming converts a mostly-dimerised gas into a mostly-dissociated one, and the visible colour change is a direct read-out of the van ’t Hoff equation with \(\Delta H^\circ=+57.24\,\text{kJ}\,\text{mol}^{-1}\).

Scope. Ideal-gas behaviour and constant \(\Delta H^\circ\) over a \(52\,\text{K}\) range, where the neglected \(\Delta C_P^\circ\) term contributes a correction of well under one per cent to \(\ln K\).

Problems

An exothermic reaction has \(\Delta H^\circ=-50.0\,\text{kJ}\,\text{mol}^{-1}\) and \(K=25.0\) at \(298.15\,\text{K}\). Assuming \(\Delta H^\circ\) is constant, find \(K\) at \(350\,\text{K}\).
Solution
Rearrange first: \(K_2=K_1\exp\!\left[-\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)\right]\).
\(-\Delta H^\circ/R = +50\,000/8.314 = +6014\,\text{K}\). The bracket is \(1/350-1/298.15 = 2.85714\times10^{-3}-3.35402\times10^{-3} = -4.9688\times10^{-4}\,\text{K}^{-1}\).
Hence \(\ln(K_2/K_1) = (6014)(-4.9688\times10^{-4}) = -2.988\), so \(K_2 = 25.0\,e^{-2.988} = 25.0\times0.0504 = 1.26\).
The equilibrium constant falls by a factor of about twenty: heating an exothermic reaction by \(52\,\text{K}\) has taken it from strongly product-favoured to barely product-favoured, exactly as the sign rule requires.
A van ’t Hoff plot of \(\ln K\) against \(1/T\) for a reaction is a straight line of slope \(-3.75\times10^{3}\,\text{K}\) and intercept \(12.4\). Find \(\Delta H^\circ\) and \(\Delta S^\circ\), then compute \(K\) and \(\Delta G^\circ\) at \(298.15\,\text{K}\) and check the two routes to \(\Delta G^\circ\) agree.
Solution
From \(\ln K=-\dfrac{\Delta H^\circ}{RT}+\dfrac{\Delta S^\circ}{R}\), the slope is \(-\Delta H^\circ/R\) and the intercept is \(\Delta S^\circ/R\).
\(\Delta H^\circ = -R\times(\text{slope}) = -(8.314)(-3750) = +3.118\times10^{4}\,\text{J}\,\text{mol}^{-1} = +31.2\,\text{kJ}\,\text{mol}^{-1}\), so the reaction is endothermic and \(K\) rises with temperature.
\(\Delta S^\circ = R\times(\text{intercept}) = (8.314)(12.4) = 103.1\,\text{J}\,\text{mol}^{-1}\text{K}^{-1}\).
At \(298.15\,\text{K}\): \(\ln K = -3750/298.15+12.4 = -12.578+12.4 = -0.1776\), so \(K=e^{-0.1776}=0.837\).
Route one: \(\Delta G^\circ=-RT\ln K = -(8.314)(298.15)(-0.1776) = +440\,\text{J}\,\text{mol}^{-1}\). Route two: \(\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ = 31\,177.5-(298.15)(103.094) = 31\,177.5-30\,737.4 = +440\,\text{J}\,\text{mol}^{-1}\). The two agree, as they must, because the linear form of the van ’t Hoff equation is nothing but \(\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ\) divided by \(-RT\).
Using the ammonia data of Worked example 1 (\(\Delta H^\circ=-91.8\,\text{kJ}\,\text{mol}^{-1}\), \(\Delta S^\circ=-198.1\,\text{J}\,\text{mol}^{-1}\text{K}^{-1}\), both treated as temperature-independent), find the temperature at which \(K=1\), and comment on what this implies for the choice of industrial conditions.
Solution
\(K=1\) means \(\ln K=0\), which by \(\Delta G^\circ=-RT\ln K\) means \(\Delta G^\circ=0\), i.e. \(\Delta H^\circ=T\Delta S^\circ\). Rearranging, \(T=\Delta H^\circ/\Delta S^\circ\).
\(T = \dfrac{-91\,800\,\text{J}\,\text{mol}^{-1}}{-198.1\,\text{J}\,\text{mol}^{-1}\text{K}^{-1}} = 463\,\text{K}\).
Above \(463\,\text{K}\) the standard-state equilibrium lies on the reactant side (\(K\lt 1\)); below it, on the product side. Both \(\Delta H^\circ\) and \(\Delta S^\circ\) are negative, so this is the temperature at which the entropic penalty of consuming four moles of gas overtakes the enthalpic reward.
Since acceptable rates require roughly \(650\)–\(750\,\text{K}\), the process is necessarily run where \(K\ll1\). The compensation is pressure: \(\Delta n=-2\), so raising the total pressure to hundreds of bar shifts the equilibrium composition forward without touching \(K\) itself, and an iron catalyst supplies the rate at the lowest temperature that remains workable. Note that the \(463\,\text{K}\) figure is itself approximate, since Worked example 1 showed \(\Delta H^\circ\) drifting by about twenty per cent over this range.
Derive the isochoric form \(\dfrac{d\ln K_c}{dT}=\dfrac{\Delta U^\circ}{RT^{2}}\) for a gas-phase reaction with \(\Delta n\) moles of gas produced, given the ideal-gas relations \(K_P=K_c\left(\dfrac{c^\circ RT}{P^\circ}\right)^{\Delta n}\) and \(\Delta H^\circ=\Delta U^\circ+\Delta n\,RT\).
Solution
Take logarithms of the first relation: \(\ln K_P=\ln K_c+\Delta n\left[\ln T+\ln\dfrac{c^\circ R}{P^\circ}\right]\). The second bracketed term is a temperature-independent constant.
Differentiate with respect to \(T\): \(\dfrac{d\ln K_P}{dT}=\dfrac{d\ln K_c}{dT}+\dfrac{\Delta n}{T}\).
Substitute the van ’t Hoff equation on the left and rearrange for the unknown: \(\dfrac{d\ln K_c}{dT}=\dfrac{\Delta H^\circ}{RT^{2}}-\dfrac{\Delta n}{T}\).
Now insert \(\Delta H^\circ=\Delta U^\circ+\Delta n RT\): \(\dfrac{d\ln K_c}{dT}=\dfrac{\Delta U^\circ+\Delta n RT}{RT^{2}}-\dfrac{\Delta n}{T}=\dfrac{\Delta U^\circ}{RT^{2}}+\dfrac{\Delta n}{T}-\dfrac{\Delta n}{T}=\dfrac{\Delta U^\circ}{RT^{2}}\).
The \(\Delta n\) terms cancel identically. So the constant-pressure equilibrium constant tracks the enthalpy and the constant-volume one tracks the internal energy — each pairs with the state function appropriate to its own constraint. For \(\Delta n=0\) the two constants and the two forms coincide.
A protein–ligand association has \(\Delta H^\circ=-30.0\,\text{kJ}\,\text{mol}^{-1}\) at \(T_0=298.15\,\text{K}\) and \(\Delta C_P^\circ=-1.50\,\text{kJ}\,\text{mol}^{-1}\text{K}^{-1}\), the latter typical of the burial of hydrophobic surface. Show that the binding constant passes through a maximum, locate it, and find by what factor \(K\) at body temperature \(310\,\text{K}\) falls below that maximum. Treat \(\Delta C_P^\circ\) as constant.
Solution
Existence and location of the maximum. By the van ’t Hoff equation \(d\ln K/dT=\Delta H^\circ(T)/(RT^{2})\), and \(RT^{2}>0\), so \(\ln K\) is stationary exactly where \(\Delta H^\circ(T)=0\). Kirchhoff’s law gives \(\Delta H^\circ(T)=\Delta H^\circ(T_0)+\Delta C_P^\circ(T-T_0)\), so
\(T_{\max}=T_0-\dfrac{\Delta H^\circ(T_0)}{\Delta C_P^\circ}=298.15-\dfrac{-30\,000}{-1500}=298.15-20.0=278.15\,\text{K}\) (about \(5\,{}^\circ\text{C}\)).
Below \(T_{\max}\) the binding is endothermic (\(\Delta H^\circ>0\)) so \(K\) rises with \(T\); above it the binding is exothermic so \(K\) falls. The stationary point is therefore a maximum, and the sign change is real, not an artefact.
Magnitudes. Use the corrected integral, \(\ln\dfrac{K(T)}{K(T_0)}=-\dfrac{\Delta H^\circ(T_0)}{R}\left(\dfrac{1}{T}-\dfrac{1}{T_0}\right)+\dfrac{\Delta C_P^\circ}{R}\left[\ln\dfrac{T}{T_0}+\dfrac{T_0}{T}-1\right]\), with \(-\Delta H^\circ(T_0)/R=+3608\,\text{K}\) and \(\Delta C_P^\circ/R=-180.4\).
At \(T=278.15\,\text{K}\): the first term is \((3608)(3.59518\times10^{-3}-3.35402\times10^{-3})=(3608)(2.4116\times10^{-4})=+0.870\); the bracket is \(\ln(0.93292)+1.07190-1=-0.06944+0.07190=2.47\times10^{-3}\), giving \((-180.4)(2.47\times10^{-3})=-0.445\). Total \(+0.425\), so \(K(278.15)/K(T_0)=e^{0.425}=1.53\).
At \(T=310\,\text{K}\): the first term is \((3608)(3.22581\times10^{-3}-3.35402\times10^{-3})=-0.463\); the bracket is \(\ln(1.03974)+0.96177-1=0.03898-0.03823=7.50\times10^{-4}\), giving \((-180.4)(7.50\times10^{-4})=-0.135\). Total \(-0.598\), so \(K(310)/K(T_0)=e^{-0.598}=0.550\).
Hence \(\dfrac{K(310)}{K(278.15)}=\dfrac{0.550}{1.53}=0.36\): binding at body temperature is only about thirty-six per cent as strong as at its optimum near \(5\,{}^\circ\text{C}\).
A straight-line van ’t Hoff fit could not have produced any of this, since a straight line is monotonic in \(1/T\) and can never show a maximum. A curved plot is the signature of a large \(\Delta C_P^\circ\), and here the curvature carries the entire chemical story.