Hess's law and enthalpy cycles

Result

Hess's law and enthalpy cycles

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Statement

For a closed system at constant pressure doing no work other than expansion work, the heat absorbed equals the change in enthalpy \(H=U+pV\); since \(U\), \(p\) and \(V\) are state functions, so is \(H\), and \(\mathrm{d}H\) is an exact differential. Consequently the reaction enthalpy \(\Delta_{\mathrm{r}}H=(\partial H/\partial\xi)_{T,p}=\sum_i \nu_i\overline{H}_i\) is a linear functional of the stoichiometric vector \(\boldsymbol{\nu}\): if a target reaction satisfies \(\boldsymbol{\nu}_{\text{t}}=\sum_k c_k\boldsymbol{\nu}^{(k)}\) for reactions \(k\) whose enthalpies are known at the same temperature, pressure and set of species states, then \(\Delta_{\mathrm{r}}H_{\text{t}}=\sum_k c_k\,\Delta_{\mathrm{r}}H_k\), and every closed cycle of reactions carries zero net enthalpy, \(\oint\mathrm{d}H=0\). This is Hess's law; reversal (\(c_k=-1\)), scaling (\(c_k=n\)) and addition are its three elementary cases.

Why it matters

Most reaction enthalpies of interest are not directly measurable. A reaction may be too slow to give a calorimetric signal, may run to a mixture of products, may be violently unsafe, or — as with a lattice enthalpy or an electron affinity — may involve species that cannot be prepared in isolation at all. Hess's law converts every one of these into arithmetic on quantities that can be measured, because it says the enthalpy of a transformation is fixed by its endpoints and nothing else.

Read as linear algebra rather than as a bookkeeping trick, it also explains why one table replaces every cycle: the balanced reactions among \(N\) species built from \(E\) elements form a vector space of dimension \(N-\operatorname{rank}A\) (with \(A\) the element-count matrix), the formation reactions of the non-elemental species are a basis for it, and a linear functional on a vector space is determined by its values on a basis. That single sentence contains the entire content of the standard formation-enthalpy formula, the combustion-enthalpy formula, the Born–Haber cycle and the bond-enthalpy estimate; each is one choice of basis.

Hypotheses

The system is closed and the process is at constant pressure with expansion work only.Then and only then does the measured heat equal \(\Delta H\). Discharge the same chemical change through a galvanic cell and part of the energy leaves as electrical work: \(q_p=\Delta H-w_{\text{elec}}\ne\Delta H\), so a calorimeter wired to a load reports the wrong number. At constant volume the measured heat is \(\Delta U\) instead, and \(\Delta H-\Delta U=\Delta n_{\text{gas}}RT\) — for methane combustion to liquid water that gap is \(4.96\,\text{kJ}\,\text{mol}^{-1}\), far larger than calorimetric precision.

Every species carries a fully specified thermodynamic state: chemical identity, phase, temperature, pressure, and (in solution) composition.Cancellation in a cycle is cancellation of states, not of formulae. Cancelling \(\text{H}_2\text{O}(l)\) in one leg against \(\text{H}_2\text{O}(g)\) in another silently discards \(\Delta_{\text{vap}}H^\circ=+44.0\,\text{kJ}\,\text{mol}^{-1}\) at \(298\,\text{K}\), and the resulting cycle does not close.

All legs of the cycle are referred to one common temperature and pressure.\(H\) is path-independent for fixed endpoints, but \(\Delta_{\mathrm{r}}H\) itself depends on \(T\) and \(p\). Water's enthalpy of vaporisation is \(44.0\,\text{kJ}\,\text{mol}^{-1}\) at \(298\,\text{K}\) and \(40.65\,\text{kJ}\,\text{mol}^{-1}\) at its normal boiling point \(373\,\text{K}\); mixing the two into one cycle is a \(3.4\,\text{kJ}\,\text{mol}^{-1}\) error. The standard convention \(p^\circ=1\,\text{bar}\), \(T=298.15\,\text{K}\) exists precisely to make tabulated legs combinable.

The partial molar enthalpies \(\overline{H}_i\) are constant over the extent of reaction considered.\(\Delta_{\mathrm{r}}H\) is defined as a derivative at fixed composition, \((\partial H/\partial\xi)_{T,p}\); equating it to the enthalpy released over a full mole of reaction requires \(\overline{H}_i\) to be composition-independent. This is exact for pure separate phases and for ideal solutions, and fails measurably for concentrated electrolytes, which is why aqueous data is tabulated at infinite dilution and why enthalpies of dilution are tabulated separately.

The elements' standard states are fixed once, globally, and used consistently.The formation basis is a basis only if every table refers to the same origin. Graphite, not diamond, is carbon's standard state; diamond's own \(\Delta_{\mathrm{f}}H^\circ=+1.9\,\text{kJ}\,\text{mol}^{-1}\). Two tables built on different allotropes cannot be added.

Proof

The argument runs in three movements: first that \(H\) is a state function (steps 1–4), then that reaction enthalpy is a linear functional of stoichiometry (steps 5–7), then that cycles and bases follow as pure linear algebra (steps 8–10). Only the first movement is thermodynamics.

\[ \mathrm{d}U = \delta q + \delta w \]

The first law for a closed system. \(U\) is a state function — the statement that \(\oint(\delta q+\delta w)=0\) for every cycle is exactly what energy conservation asserts — whereas \(q\) and \(w\) separately are path functions, written with \(\delta\) rather than \(\mathrm{d}\). A

\[ \delta w = -p\,\mathrm{d}V \quad\Longrightarrow\quad \delta q_p = \mathrm{d}U + p\,\mathrm{d}V \]

Impose the hypothesis of expansion work only against a constant external pressure equal to the system pressure. Any additional mode of work — electrical, surface, magnetic — would add a term here and break the identification made in step 4. A

\[ H \equiv U + pV, \qquad \mathrm{d}H = \mathrm{d}U + p\,\mathrm{d}V + V\,\mathrm{d}p \]

Definition, then the product rule. No physics is used: \(H\) is a construction from \(U\), \(p\) and \(V\), each of which is a property of the state alone. A

\[ \mathrm{d}p=0 \;\Longrightarrow\; \mathrm{d}H = \delta q_p, \qquad \int_{A}^{B}\mathrm{d}H = H_B-H_A, \qquad \oint \mathrm{d}H = 0 \]

Setting \(\mathrm{d}p=0\) in step 3 and substituting step 2 identifies \(\mathrm{d}H\) with the constant-pressure heat. Because \(H\) is a function of state, \(\mathrm{d}H\) is an exact differential, so its line integral depends only on the endpoints and vanishes on any closed path. This is the whole of Hess's law in one line; everything after it is organisation. B

\[ n_i(\xi) = n_i(0) + \nu_i\,\xi, \qquad H = \sum_i n_i \overline{H}_i, \quad \overline{H}_i \equiv \left(\frac{\partial H}{\partial n_i}\right)_{T,p,n_{j\ne i}} \]

Coordinatise composition by the extent of reaction \(\xi\) (units: mol), whose defining property is that all species advance in the fixed ratio \(\nu_i\), negative for reactants. The enthalpy of the mixture is the sum of partial molar enthalpies weighted by amount — Euler's theorem for the homogeneous first-order function \(H\). B

\[ \Delta_{\mathrm{r}}H \equiv \left(\frac{\partial H}{\partial \xi}\right)_{T,p} = \sum_i \nu_i \overline{H}_i \;=\; \boldsymbol{\nu}\cdot\overline{\boldsymbol{H}} \]

Differentiate step 5 with respect to \(\xi\) at fixed \(T,p\), using \(\mathrm{d}n_i=\nu_i\,\mathrm{d}\xi\) and the Gibbs–Duhem relation \(\sum_i n_i\,\mathrm{d}\overline{H}_i=0\) at constant \(T,p\), which kills the term in \(\mathrm{d}\overline{H}_i\). This is the central identity: with the vector \(\overline{\boldsymbol{H}}\) fixed by \(T\), \(p\) and the species states, \(\Delta_{\mathrm{r}}H\) is a linear functional of \(\boldsymbol{\nu}\) alone. Units check: \(\text{J}\,\text{mol}^{-1}\), enthalpy per mole of reaction as written. B

\[ \boldsymbol{\nu}_{\text{t}} = \sum_k c_k\,\boldsymbol{\nu}^{(k)} \quad\Longrightarrow\quad \Delta_{\mathrm{r}}H_{\text{t}} = \boldsymbol{\nu}_{\text{t}}\cdot\overline{\boldsymbol{H}} = \sum_k c_k\left(\boldsymbol{\nu}^{(k)}\cdot\overline{\boldsymbol{H}}\right) = \sum_k c_k\,\Delta_{\mathrm{r}}H_k \]

Apply linearity of the dot product. The three schoolbook rules are the cases \(c=-1\) (reverse: the sign flips), \(c=n\) (scale: the value scales) and \(c_k=1\) (add: the enthalpies add). Note what the derivation does not require: no claim that the intermediate reactions occur, and no claim that they are elementary. A

\[ \sum_k c_k\,\boldsymbol{\nu}^{(k)} = \boldsymbol{0} \quad\Longrightarrow\quad \sum_k c_k\,\Delta_{\mathrm{r}}H_k = 0 \]

Set \(\boldsymbol{\nu}_{\text{t}}=\boldsymbol{0}\) in step 7. An enthalpy cycle is precisely an element of the kernel of the matrix whose columns are the \(\boldsymbol{\nu}^{(k)}\); the closure condition on a Born–Haber diagram is this equation and nothing more. One unknown leg in a closing cycle is therefore determined uniquely. B

\[ \mathcal{R} = \ker A \subseteq \mathbb{R}^{N}, \qquad \dim\mathcal{R} = N - \operatorname{rank}A, \qquad A\in\mathbb{R}^{E\times N} \]

A stoichiometric vector is balanced exactly when it conserves every element, i.e. lies in the kernel of the element-count matrix \(A\) (entry \(A_{ei}\) = atoms of element \(e\) in species \(i\)). For \(N\) species built from \(E\) elements with \(\operatorname{rank}A=E\), the reaction space has dimension \(N-E\), and the \(N-E\) formation reactions of the non-elemental species are linearly independent members of it — each introduces exactly one species the others lack — hence a basis. A linear functional is fixed by its values on a basis, so the \(N-E\) numbers \(\Delta_{\mathrm{f}}H^\circ\) determine \(\Delta_{\mathrm{r}}H^\circ\) for every balanced reaction: \(\Delta_{\mathrm{r}}H^\circ=\sum_i \nu_i\,\Delta_{\mathrm{f}}H_i^\circ\), with elements contributing zero by the choice of origin. C

\[ \left(\frac{\partial\,\Delta_{\mathrm{r}}H}{\partial T}\right)_{p} = \Delta_{\mathrm{r}}C_p \quad\Longrightarrow\quad \Delta_{\mathrm{r}}H(T_2) = \Delta_{\mathrm{r}}H(T_1) + \int_{T_1}^{T_2}\Delta_{\mathrm{r}}C_p\,\mathrm{d}T \]

Kirchhoff's relation, obtained by differentiating step 6 with respect to \(T\) and using \(\overline{C}_{p,i}=(\partial\overline{H}_i/\partial T)_p\). It is the repair kit for the third hypothesis: legs measured at different temperatures are not additive as they stand, but each can be transported to a common \(T\) along a leg of the cycle whose enthalpy is \(\int C_p\,\mathrm{d}T\). Cycles close in a plane of constant \(T\); Kirchhoff is how you change plane. C

Result

\[ \boldsymbol{\nu}_{\text{t}}=\sum_k c_k\boldsymbol{\nu}^{(k)} \;\Longrightarrow\; \Delta_{\mathrm{r}}H_{\text{t}}=\sum_k c_k\,\Delta_{\mathrm{r}}H_k, \qquad \oint\mathrm{d}H = 0, \qquad \Delta_{\mathrm{r}}H^\circ=\sum_i \nu_i\,\Delta_{\mathrm{f}}H_i^\circ \]

Reading. Reaction enthalpy is a linear functional on the space of balanced reactions. Combine equations however you like — reverse, scale, add — and the enthalpies undergo the identical combination; any set of reactions that returns to its starting state sums to zero; and because formation reactions are a basis of that space, one table of \(\Delta_{\mathrm{f}}H^\circ\) values evaluates every cycle you will ever draw.

Scope. Constant pressure, expansion work only, one common \((T,p)\), and species matched by full thermodynamic state and not merely by formula. The same algebra holds verbatim for \(S\), \(G\) and any other state function, and fails for \(q\) and \(w\) taken separately.

Corollaries & converses

Formation basis. \(\Delta_{\mathrm{r}}H^\circ=\sum n\,\Delta_{\mathrm{f}}H^\circ(\text{products})-\sum n\,\Delta_{\mathrm{f}}H^\circ(\text{reactants})\); the standard formation-enthalpy result is step 9 with the canonical basis, and \(\Delta_{\mathrm{f}}H^\circ=0\) for an element in its standard state is the choice of origin, not a measurement.
Combustion basis. Combustion reactions of the organic species also span, giving \(\Delta_{\mathrm{r}}H^\circ=\sum n\,\Delta_{\mathrm{c}}H^\circ(\text{reactants})-\sum n\,\Delta_{\mathrm{c}}H^\circ(\text{products})\) — the reversed sign pattern is a change of basis, not a different law (Example 2).
Born–Haber. A closing cycle with exactly one unknown leg determines that leg uniquely (step 8); this is how lattice enthalpies and electron affinities, neither directly measurable, are obtained (Example 1, Problem 3).
Bond enthalpies. Taking mean bond enthalpies as an approximate dual basis over an atomisation cycle gives \(\Delta_{\mathrm{r}}H\approx\sum D(\text{bonds broken})-\sum D(\text{bonds made})\); it is approximate only because a mean \(D\) is not environment-independent, never because Hess's law is in doubt.
Converse (existence). A target enthalpy is obtainable from a data set if and only if \(\boldsymbol{\nu}_{\text{t}}\) lies in the span of the known \(\boldsymbol{\nu}^{(k)}\). Outside the span no cycle exists and no amount of algebra produces one — a rank test on the stoichiometric matrix decides the question before any arithmetic is attempted.
Converse (uniqueness). If \(\boldsymbol{\nu}_{\text{t}}\) is in the span, the resulting \(\Delta_{\mathrm{r}}H_{\text{t}}\) is independent of which combination \(\{c_k\}\) is used, even when the \(\boldsymbol{\nu}^{(k)}\) are linearly dependent: two combinations differ by a kernel element, which contributes zero by step 8. Redundant data therefore gives a genuine consistency check on the measurements.

Fails without

Expansion work only, dropped: run \(\text{Zn}(s)+\text{Cu}^{2+}(aq)\to\text{Zn}^{2+}(aq)+\text{Cu}(s)\) in a beaker and the calorimeter reports the full \(\Delta_{\mathrm{r}}H\); run it as a Daniell cell delivering current and the heat evolved is smaller by the electrical work extracted. \(\Delta H\) is unchanged — it is a state function — but it is no longer the measured heat, so a cycle built from calorimetric readings taken under non-zero electrical work does not close.
Constant pressure, dropped: a bomb calorimeter measures \(\Delta U\), not \(\Delta H\). For \(\text{CH}_4(g)+2\text{O}_2(g)\to\text{CO}_2(g)+2\text{H}_2\text{O}(l)\), \(\Delta n_{\text{gas}}=-2\) and \(\Delta H-\Delta U=\Delta n_{\text{gas}}RT=-4.96\,\text{kJ}\,\text{mol}^{-1}\) at \(298.15\,\text{K}\). Feeding raw bomb values into an enthalpy cycle mis-states every leg whose gas count changes, and the errors do not cancel unless \(\Delta n_{\text{gas}}\) happens to cancel too.
One common temperature, dropped: combine \(\Delta_{\text{vap}}H(\text{H}_2\text{O})=40.65\,\text{kJ}\,\text{mol}^{-1}\) at \(373\,\text{K}\) with formation enthalpies at \(298\,\text{K}\) and the cycle is out by \(3.4\,\text{kJ}\,\text{mol}^{-1}\), because the two legs are drawn in different constant-\(T\) planes. Kirchhoff's relation (step 10) is the only legitimate bridge.
State-matched cancellation, dropped: cancelling \(\text{H}_2\text{O}(g)\) against \(\text{H}_2\text{O}(l)\), or \(\text{HCl}(g)\) against \(\text{HCl}(aq)\), deletes a real leg — \(+44.0\) and \(-74.8\,\text{kJ}\,\text{mol}^{-1}\) respectively at \(298\,\text{K}\) for the enthalpies of vaporisation and of solution. The formulae match; the states do not, and the algebra is over states.
Path independence itself, dropped: the identical manipulations applied to \(q\) or \(w\) alone are simply false. A Carnot cycle returns to its initial state with \(\oint\mathrm{d}U=0\) but \(\oint\delta q=-\oint\delta w=w_{\text{out}}\ne0\), where \(w_{\text{out}}\) is the net work delivered to the surroundings (the sign convention of step 1 counts work done on the system). Hess's law is a statement about \(H\), and it is exactly as strong as \(H\)'s status as a state function.

Common errors

“I reversed the equation and kept \(\Delta H\).” Reversal is \(c=-1\) in step 7; the equation and the number must change together, always.
“I doubled the coefficients, so \(\Delta H\) is unchanged.” \(\Delta_{\mathrm{r}}H\) is per mole of reaction as written; doubling \(\boldsymbol{\nu}\) doubles \(\boldsymbol{\nu}\cdot\overline{\boldsymbol{H}}\). The unit \(\text{kJ}\,\text{mol}^{-1}\) is meaningless until the balanced equation it refers to is quoted.
“\(\text{H}_2\text{O}\) appears on both sides, so it cancels.” Only if the phase, temperature and (for solutions) concentration match. Formula-level cancellation is the single most common source of a cycle that fails to close.
“The combustion formula must have the same sign pattern as the formation formula.” It is inverted: \(\Delta_{\mathrm{c}}H\) measures distance from the products of combustion, not from the elements, so reactants and products swap roles.
“The intermediate steps have to be what really happens.” Step 7 uses no mechanistic input at all. Thermodynamics fixes the endpoints; kinetics owns the path, and the two questions never meet here.
“Elements have \(\Delta_{\mathrm{f}}H^\circ=0\), so any form of an element does.” Only the designated standard state: diamond is \(+1.9\), ozone \(+142.7\), and atomic \(\text{H}(g)\) is \(+218.0\,\text{kJ}\,\text{mol}^{-1}\).
“A negative \(\Delta_{\mathrm{r}}H\) means the reaction goes.” Spontaneity is governed by \(\Delta_{\mathrm{r}}G=\Delta_{\mathrm{r}}H-T\Delta_{\mathrm{r}}S\); Hess's law is silent about it, though it applies to \(G\) and \(S\) with the same algebra.

Discussion

Germain Henri Hess published the law of constant heat summation in 1840, on the strength of his own calorimetry on the hydration of sulfuric acid, several years before Joule, Mayer and Helmholtz put the first law of thermodynamics on its feet. It is one of the clearest cases in physical science of an empirical regularity that later turns out to be a corollary: what Hess found by measurement is, in the modern reading, nothing more than the exactness of \(\mathrm{d}H\), which is in turn nothing more than energy conservation plus the definition \(H=U+pV\). The historical order also explains the law's traditional presentation as three rules about manipulating equations, which conceals how little content it has once state functions are available.

The linear-algebra reading repays the effort. It tells you in advance whether a problem is solvable (is the target in the span?), it explains why redundant thermochemical data is worth collecting (inconsistencies are non-zero cycle sums, so the residuals of an over-determined system measure experimental error), and it identifies what a thermochemical table actually is: a linear functional recorded in a chosen basis. Modern reference compilations such as those maintained by NIST and CODATA are exactly that, and the periodic re-adjustment of a "key value" propagates through every dependent entry because the basis vectors are shared.

The subtlest hypothesis is not the constant pressure but the constancy of \(\overline{H}_i\). Strictly, \(\Delta_{\mathrm{r}}H=(\partial H/\partial\xi)_{T,p}\) is a derivative at a composition, and the "enthalpy of reaction" quoted per mole is that derivative only when the partial molar enthalpies do not drift as \(\xi\) advances. For pure condensed phases and ideal gases they do not; for real electrolytes they do, sometimes strongly, which is why aqueous ionic data is referred to the hypothetical infinite-dilution state and why enthalpies of dilution appear as separate legs. A second delicacy: the conventional ionic table fixes \(\Delta_{\mathrm{f}}H^\circ(\text{H}^+,aq)\equiv0\), a second, independent origin choice, so single-ion enthalpies are meaningful only in electroneutral combinations — the functional is defined on the reaction space, not on species.

Computational thermochemistry exploits the same structure deliberately. Isodesmic and homodesmotic reaction schemes are cycles chosen so that the same bond types appear on both sides; the systematic errors of an approximate electronic-structure method then lie almost entirely in the kernel and cancel, delivering a reaction enthalpy far more accurate than either total energy that entered it. The design principle is Hess's law used backwards: pick the cycle so the errors, not the species, cancel.

Common misconceptions. That Hess's law is a separate empirical law needing separate justification — it is a theorem given the first law. That it requires the intermediate steps to be real — it never inspects them. And that it is peculiar to enthalpy — every state function obeys the identical algebra, which is why entropy and Gibbs-energy cycles are drawn the same way and read the same way.

Worked examples

Example 1. Determine the lattice enthalpy of sodium chloride, \(\text{NaCl}(s)\to\text{Na}^{+}(g)+\text{Cl}^{-}(g)\), which cannot be measured directly, from the standard tabulated \(298.15\,\text{K}\) data \(\Delta_{\mathrm{f}}H^\circ(\text{NaCl},s)=-411.2\), \(\Delta_{\text{sub}}H^\circ(\text{Na})=+107.3\), \(I_1(\text{Na})=+495.8\), \(D(\text{Cl}\!-\!\text{Cl})=+242.6\) and the electron-gain enthalpy \(\Delta_{\text{eg}}H^\circ(\text{Cl})=-349.0\), all in \(\text{kJ}\,\text{mol}^{-1}\).

\[ \text{Na}(s)+\tfrac12\text{Cl}_2(g)\;\longrightarrow\;\text{NaCl}(s) \qquad \Delta_{\mathrm{f}}H^\circ = -411.2 \]

The direct leg. The cycle will connect the same two endpoints through gaseous atoms and ions, so by step 8 the two paths must carry equal enthalpy. A

\[ \Delta_{\mathrm{f}}H^\circ = \Delta_{\text{sub}}H^\circ(\text{Na}) + I_1(\text{Na}) + \tfrac12 D(\text{Cl}_2) + \Delta_{\text{eg}}H^\circ(\text{Cl}) + \Delta_{\text{latt,form}}H^\circ \]

The indirect leg, assembled species by species: atomise the metal, ionise it, atomise half a mole of chlorine, attach the electron, then condense the gaseous ion pair into the lattice. Note \(\tfrac12 D\), because the balanced equation consumes half a mole of \(\text{Cl}_2\) — the scaling rule of step 7. A

\[ \Delta_{\text{latt,form}}H^\circ = \Delta_{\mathrm{f}}H^\circ - \left[\Delta_{\text{sub}}H^\circ + I_1 + \tfrac12 D + \Delta_{\text{eg}}H^\circ\right] \]

Rearrange for the single unknown leg before any number is inserted; the cycle has one unknown, so step 8 fixes it uniquely. A

\[ \tfrac12 D = 121.3, \qquad 107.3+495.8+121.3-349.0 = 375.4 \]

Evaluate the bracket. Every term is in \(\text{kJ}\,\text{mol}^{-1}\) and refers to \(298.15\,\text{K}\), so the legs are additive (third hypothesis). A

\[ \Delta_{\text{latt,form}}H^\circ = -411.2 - 375.4 = -786.6\,\text{kJ}\,\text{mol}^{-1} \]

Substitute. This is the enthalpy of forming the lattice from its gaseous ions; the lattice enthalpy as usually tabulated is the reverse process, so its sign flips by the \(c=-1\) case of step 7. A

\[ \Delta_{\text{latt}}H^\circ\!\left(\text{NaCl}\right) = +786.6\;\text{kJ}\,\text{mol}^{-1} \]

Reading. A quantity with no direct experimental route — you cannot weigh out a mole of free \(\text{Na}^{+}(g)\) and \(\text{Cl}^{-}(g)\) — is fixed to better than \(1\,\text{kJ}\,\text{mol}^{-1}\) by five measurable legs and the closure condition \(\sum_k c_k\Delta_{\mathrm{r}}H_k=0\). The value agrees with the accepted Born–Haber lattice enthalpy of about \(787\,\text{kJ}\,\text{mol}^{-1}\).

Scope. The same five-leg construction works for any binary ionic solid; for \(\text{MX}_2\) salts the second ionisation energy and a doubled electron-gain term enter, and the scaling factors on \(D\) change accordingly.

Example 2. Find the standard enthalpy of hydrogenation of ethene, \(\text{C}_2\text{H}_4(g)+\text{H}_2(g)\to\text{C}_2\text{H}_6(g)\), from combustion data alone: \(\Delta_{\mathrm{c}}H^\circ(\text{C}_2\text{H}_4)=-1411.0\), \(\Delta_{\mathrm{c}}H^\circ(\text{H}_2)=-285.8\), \(\Delta_{\mathrm{c}}H^\circ(\text{C}_2\text{H}_6)=-1559.7\;\text{kJ}\,\text{mol}^{-1}\), all at \(298.15\,\text{K}\) with \(\text{H}_2\text{O}(l)\) as the product.

\[ \begin{aligned} R_1&: \; \text{C}_2\text{H}_4 + 3\,\text{O}_2 \to 2\,\text{CO}_2 + 2\,\text{H}_2\text{O}(l) \\ R_2&: \; \text{H}_2 + \tfrac12\text{O}_2 \to \text{H}_2\text{O}(l) \\ R_3&: \; \text{C}_2\text{H}_6 + \tfrac72\text{O}_2 \to 2\,\text{CO}_2 + 3\,\text{H}_2\text{O}(l) \end{aligned} \]

Write the three basis reactions explicitly, with the water phase stated. All three share the products \(\text{CO}_2(g)\) and \(\text{H}_2\text{O}(l)\), which is what makes them combinable at all. A

\[ \boldsymbol{\nu}_{\text{t}} = c_1\boldsymbol{\nu}^{(1)} + c_2\boldsymbol{\nu}^{(2)} + c_3\boldsymbol{\nu}^{(3)} \quad\text{with}\quad c_1=1,\; c_2=1,\; c_3=-1 \]

Solve for the coefficients rather than guessing: \(\text{C}_2\text{H}_4\) appears only in \(R_1\) and must be consumed once, so \(c_1=1\); \(\text{H}_2\) only in \(R_2\), so \(c_2=1\); \(\text{C}_2\text{H}_6\) only in \(R_3\) and must be produced, so \(c_3=-1\). B

\[ \text{O}_2:\; 3+\tfrac12-\tfrac72 = 0, \qquad \text{CO}_2:\; 2+0-2 = 0, \qquad \text{H}_2\text{O}(l):\; 2+1-3 = 0 \]

Verify that every spectator species cancels — component by component, and with the phase carried along. What survives is exactly \(\text{C}_2\text{H}_4+\text{H}_2\to\text{C}_2\text{H}_6\), so the combination is legitimate. A

\[ \Delta_{\mathrm{r}}H^\circ = \Delta_{\mathrm{c}}H^\circ(\text{C}_2\text{H}_4) + \Delta_{\mathrm{c}}H^\circ(\text{H}_2) - \Delta_{\mathrm{c}}H^\circ(\text{C}_2\text{H}_6) \]

Apply step 7 with the coefficients of step 2. In general this is \(\sum n\,\Delta_{\mathrm{c}}H^\circ(\text{reactants})-\sum n\,\Delta_{\mathrm{c}}H^\circ(\text{products})\): the combustion basis inverts the sign pattern of the formation basis. B

\[ \Delta_{\mathrm{r}}H^\circ = (-1411.0) + (-285.8) - (-1559.7) = -1696.8 + 1559.7 = -137.1\,\text{kJ}\,\text{mol}^{-1} \]

Insert the numbers only now. The sign is negative, as expected for converting a \(\pi\) bond and an \(\text{H}\!-\!\text{H}\) bond into two \(\text{C}\!-\!\text{H}\) bonds. A

\[ \Delta_{\mathrm{r}}H^\circ\!\left(\text{C}_2\text{H}_4 + \text{H}_2 \to \text{C}_2\text{H}_6\right) = -137.1\;\text{kJ}\,\text{mol}^{-1} \]

Reading. Three easily measured combustion enthalpies, combined with coefficients \((1,1,-1)\), give a hydrogenation enthalpy without hydrogenating anything; the value reproduces the accepted \(\approx-137\,\text{kJ}\,\text{mol}^{-1}\).

Scope. Valid for any target lying in the span of the available combustion reactions, provided every leg is quoted with the same water phase; switching one leg to \(\text{H}_2\text{O}(g)\) would shift the result by multiples of \(44.0\,\text{kJ}\,\text{mol}^{-1}\).

Problems

Given \(\text{N}_2(g)+\text{O}_2(g)\to2\,\text{NO}(g)\) with \(\Delta_{\mathrm{r}}H^\circ=+180.5\,\text{kJ}\,\text{mol}^{-1}\), find \(\Delta_{\mathrm{r}}H^\circ\) for \(\text{NO}(g)\to\tfrac12\text{N}_2(g)+\tfrac12\text{O}_2(g)\), and state what this second number is conventionally called.
Solution
The target is \(c=-\tfrac12\) times the given reaction: reversing it exchanges reactants and products, and halving it divides all coefficients by two. By step 7, \(\Delta_{\mathrm{r}}H^\circ_{\text{t}} = -\tfrac12\,(+180.5) = -90.25\,\text{kJ}\,\text{mol}^{-1}\).
The reverse of this target — \(\tfrac12\text{N}_2+\tfrac12\text{O}_2\to\text{NO}\), \(+90.25\,\text{kJ}\,\text{mol}^{-1}\) — forms one mole of \(\text{NO}\) from its elements in their standard states, so it is the standard enthalpy of formation of nitric oxide, \(\Delta_{\mathrm{f}}H^\circ(\text{NO},g)=+90.25\,\text{kJ}\,\text{mol}^{-1}\), matching the tabulated value. The target itself is therefore \(-\Delta_{\mathrm{f}}H^\circ(\text{NO},g)\).
From \(\text{C}(\text{graphite})+\text{O}_2(g)\to\text{CO}_2(g)\), \(\Delta H_1=-393.5\); \(\text{H}_2(g)+\tfrac12\text{O}_2(g)\to\text{H}_2\text{O}(l)\), \(\Delta H_2=-285.8\); and \(\text{C}_2\text{H}_6(g)+\tfrac72\text{O}_2(g)\to2\,\text{CO}_2(g)+3\,\text{H}_2\text{O}(l)\), \(\Delta H_3=-1559.7\;\text{kJ}\,\text{mol}^{-1}\), determine \(\Delta_{\mathrm{f}}H^\circ(\text{C}_2\text{H}_6,g)\).
Solution
Target: \(2\,\text{C}(\text{graphite})+3\,\text{H}_2(g)\to\text{C}_2\text{H}_6(g)\). Coefficients: \(\text{C}_2\text{H}_6\) occurs only in \(R_3\) and must be produced, so \(c_3=-1\); graphite only in \(R_1\), consumed twice, \(c_1=2\); \(\text{H}_2\) only in \(R_2\), consumed three times, \(c_2=3\).
Check the spectators: \(\text{O}_2:\;2(1)+3(\tfrac12)-\tfrac72=0\); \(\text{CO}_2:\;2(1)+0-2=0\); \(\text{H}_2\text{O}(l):\;0+3(1)-3=0\). The combination is valid.
Symbolically \(\Delta_{\mathrm{f}}H^\circ = 2\Delta H_1 + 3\Delta H_2 - \Delta H_3\); numerically \(2(-393.5)+3(-285.8)-(-1559.7) = -787.0-857.4+1559.7 = -84.7\,\text{kJ}\,\text{mol}^{-1}\), which is the tabulated formation enthalpy of ethane.
For potassium chloride the standard \(298.15\,\text{K}\) data are \(\Delta_{\mathrm{f}}H^\circ(\text{KCl},s)=-436.5\), \(\Delta_{\text{sub}}H^\circ(\text{K})=+89.0\), \(I_1(\text{K})=+418.8\), \(D(\text{Cl}\!-\!\text{Cl})=+242.6\), and the tabulated lattice enthalpy \(\Delta_{\text{latt}}H^\circ(\text{KCl})=+717.0\;\text{kJ}\,\text{mol}^{-1}\). Recover the electron-gain enthalpy of chlorine and compare it with the accepted \(-349.0\,\text{kJ}\,\text{mol}^{-1}\).
Solution
The closing cycle (Example 1, step 2) with \(\Delta_{\text{latt,form}}H^\circ=-\Delta_{\text{latt}}H^\circ\) reads \(\Delta_{\mathrm{f}}H^\circ = \Delta_{\text{sub}}H^\circ + I_1 + \tfrac12 D + \Delta_{\text{eg}}H^\circ - \Delta_{\text{latt}}H^\circ\). Rearranging for the unknown before substituting: \(\Delta_{\text{eg}}H^\circ = \Delta_{\mathrm{f}}H^\circ - \Delta_{\text{sub}}H^\circ - I_1 - \tfrac12 D + \Delta_{\text{latt}}H^\circ\).
With \(\tfrac12 D = 121.3\): \(\Delta_{\text{eg}}H^\circ = -436.5 - 89.0 - 418.8 - 121.3 + 717.0 = -348.6\,\text{kJ}\,\text{mol}^{-1}\).
The recovered value differs from the accepted \(-349.0\) by \(0.4\,\text{kJ}\,\text{mol}^{-1}\), about \(0.06\%\) of the largest leg. Since the cycle is over-determined once an independent electron-affinity measurement exists, this residual is a non-zero cycle sum and therefore a direct estimate of the combined experimental uncertainty (Corollaries, converse on uniqueness) — not an error in Hess's law.
A bomb calorimeter gives \(-885.3\,\text{kJ}\,\text{mol}^{-1}\) for the combustion of methane at \(298.15\,\text{K}\), \(\text{CH}_4(g)+2\,\text{O}_2(g)\to\text{CO}_2(g)+2\,\text{H}_2\text{O}(l)\). Convert it to a constant-pressure enthalpy and explain why this conversion must precede any use of the value in an enthalpy cycle. Take \(R=8.314\;\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}\).
Solution
A bomb is a constant-volume vessel, so the measured heat is \(\Delta_{\mathrm{r}}U\), not \(\Delta_{\mathrm{r}}H\). From \(H=U+pV\) and the ideal-gas relation \(pV=n_{\text{gas}}RT\) applied to the gaseous species only (the volumes of the condensed phases are negligible), \(\Delta_{\mathrm{r}}H = \Delta_{\mathrm{r}}U + \Delta n_{\text{gas}}RT\).
Here \(\Delta n_{\text{gas}} = 1 - 3 = -2\) (liquid water is not counted). So \(\Delta n_{\text{gas}}RT = -2 \times 8.314 \times 298.15\;\text{J}\,\text{mol}^{-1} = -4958\;\text{J}\,\text{mol}^{-1} = -4.96\;\text{kJ}\,\text{mol}^{-1}\).
Hence \(\Delta_{\mathrm{r}}H^\circ = -885.3 + (-4.96) = -890.3\;\text{kJ}\,\text{mol}^{-1}\), the standard combustion enthalpy of methane.
The conversion is mandatory because Hess's law as proved holds for \(H\) at constant pressure (steps 2 and 4); \(\Delta U\) obeys its own additive algebra, but the two must not be mixed within one cycle. The correction is leg-specific, being proportional to that leg's \(\Delta n_{\text{gas}}\), so it does not cancel in general — here it is \(4.96\,\text{kJ}\,\text{mol}^{-1}\), a hundred times a good calorimeter's precision.
Ammonia synthesis, \(\text{N}_2(g)+3\,\text{H}_2(g)\to2\,\text{NH}_3(g)\), has \(\Delta_{\mathrm{f}}H^\circ(\text{NH}_3,g)=-45.9\;\text{kJ}\,\text{mol}^{-1}\) at \(298.15\,\text{K}\). Using the standard molar heat capacities \(C_{p,\mathrm{m}}=29.12\;(\text{N}_2)\), \(28.82\;(\text{H}_2)\), \(35.06\;(\text{NH}_3)\;\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}\), estimate \(\Delta_{\mathrm{r}}H^\circ\) at \(500\,\text{K}\), and say precisely which hypothesis of the derivation this calculation is repairing and which extra approximation it introduces.
Solution
At \(298.15\,\text{K}\), by the formation basis, \(\Delta_{\mathrm{r}}H^\circ = 2(-45.9) - 0 - 3(0) = -91.8\;\text{kJ}\,\text{mol}^{-1}\) (both reactants are elements in their standard states).
The heat-capacity change of reaction is \(\Delta_{\mathrm{r}}C_p = 2(35.06) - 29.12 - 3(28.82) = 70.12 - 29.12 - 86.46 = -45.46\;\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}\).
Kirchhoff's relation (step 10) with \(\Delta_{\mathrm{r}}C_p\) taken as constant gives \(\Delta_{\mathrm{r}}H^\circ(T_2) = \Delta_{\mathrm{r}}H^\circ(T_1) + \Delta_{\mathrm{r}}C_p\,(T_2-T_1)\). With \(T_2-T_1 = 500 - 298.15 = 201.85\,\text{K}\): correction \(= -45.46 \times 201.85 = -9176\;\text{J}\,\text{mol}^{-1} = -9.18\;\text{kJ}\,\text{mol}^{-1}\).
So \(\Delta_{\mathrm{r}}H^\circ(500\,\text{K}) \approx -91.8 - 9.2 = -101.0\;\text{kJ}\,\text{mol}^{-1}\): the reaction becomes about \(10\%\) more exothermic on heating, because the products' heat capacity is smaller than the reactants'.
The hypothesis being repaired is the third one, that all legs share a common temperature: the \(298.15\,\text{K}\) formation data cannot simply be asserted at \(500\,\text{K}\), and Kirchhoff supplies the two heating legs that close the rectangle. The extra approximation is that \(C_{p,\mathrm{m}}\) is temperature-independent over \(200\,\text{K}\); real \(C_{p,\mathrm{m}}\) rises with \(T\), so an exact treatment integrates a fitted \(C_p(T)\) instead, and the estimate above is good to roughly a kilojoule per mole rather than exact.