Hess's Law & Reaction Enthalpy

Pick a reaction nobody can measure directly. Scale, reverse and add the measured reactions below it until everything that is not in the target cancels, and the enthalpy comes out of the sum.

Worked example — Hess's Law & Reaction Enthalpy

Reaction 1, ×1	-393.5 kJ/mol
Reaction 2, ×1	-283.0 kJ/mol
Your combination	-676.5 kJ/mol

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Hess's Law & Reaction Enthalpy

The enthalpy change of a reaction, ΔH, is the heat it exchanges with its surroundings at constant pressure: negative when it releases heat, positive when it absorbs it. Hess's law says ΔH depends only on where the reaction starts and where it ends, so reactions add like equations. Reverse one and its ΔH changes sign; multiply one and its ΔH is multiplied. The working form uses standard enthalpies of formation, ΔfH°, the ΔH of making one mole of a substance from its elements in their standard states, which makes ΔfH° of an element zero: ΔH° = Σ ν ΔfH°(products) − Σ ν ΔfH°(reactants).

What each symbol means

ΔH: the enthalpy change of the reaction as written, in kJ per mole of reaction. Double the equation and it doubles
ΔfH°: standard enthalpy of formation, in kJ/mol: the ΔH of forming one mole of the substance from its elements at 25 °C and 1 bar
ν: the stoichiometric coefficient of each species in the balanced equation

Assumes: Everything is at 25 °C and 1 bar. The formation enthalpies are rounded to 0.1 kJ/mol, which is finer than some of them are known: benzene's carries ±0.9 and cyclohexene's ±1.0, so the 150.4 is good to about two kilojoules and not to its last digit. The phase is part of the species. H₂O(l) and H₂O(g) are different substances on this page, 44.0 kJ/mol apart, and the three C₆ rings are all taken as gases so that no heat of vaporisation hides inside the benzene comparison.
Breaks when: Away from 25 °C, strictly. ΔH itself drifts with temperature by the difference between the heat capacities of the products and the reactants (Kirchhoff's law of thermochemistry), so a 298 K figure used for a furnace at 1,000 °C carries an error this page does not show. The law itself does not break: a state function is path-independent at every temperature, and what changes is only the number.

The heat nobody can measure is known to a tenth of a kilojoule 🖖

Carbon burned in a calorimeter gives CO₂, or a mixture, and never pure carbon monoxide, so the target reaction cannot be run by itself. Press Show a solution: reaction 1 forwards, reaction 2 backwards. The stairs drop 393.5 kJ/mol to CO₂ and climb 283.0 back up to CO, and the target is what is left, −110.5 kJ/mol. The direct arrow on the right lands exactly where the stairs do, because enthalpy belongs to the substances in the flask and not to the route by which they got there. For organic compounds nearly every formation enthalpy in the data books was found like this: burn the compound, burn its elements, subtract.

Benzene is 150 kJ/mol more stable than its own drawing 🖖

Cyclohexene has one C=C double bond, and adding hydrogen across it releases 118.8 kJ/mol. Benzene is drawn with three, so it ought to release three times as much, −356.4 kJ/mol. Solve the benzene problem and the combustions say −206.0. On the second panel the dotted line is where the arrow would end if benzene had three ordinary double bonds, and benzene's arrow stops 150.4 kJ/mol short of it: the ring started that much lower than its drawing implies. George Kistiakowsky measured these hydrogenations at Harvard in the 1930s, and the gap is still the standard measure of the idea that benzene's six π electrons belong to the whole ring rather than to three bonds.

Bond enthalpies work exactly as well as the molecule is ordinary 🖖

Counting bonds is the shortcut when a compound has no data: add the bonds broken, subtract the bonds made. For methane burning to steam it gives −802 kJ/mol against −802.2 from the formation enthalpies. For benzene it gives −372 against −206.0, off by 166, because the table's C=C is an average over ordinary double bonds and benzene does not have any. The shortcut is also blind to phase. It counts the same O–H bonds in steam and in liquid water, which are 44.0 kJ/mol apart, and graphite, where both carbon problems start, has no row in a bond table at all.

Problem solved in full

Benzene's enthalpy of formation, from one combustion 4 steps
Burning one mole of gaseous benzene to CO₂ and liquid water releases 3,301.3 kJ. With ΔfH° of CO₂(g) = −393.5 and of H₂O(l) = −285.8 kJ/mol, find ΔfH° of benzene. Then say how its sign sits beside the 150.4 kJ/mol this page finds.

Open this problem in the tool
1. Write the combustion in formation enthalpies. O₂ is an element and contributes nothing, so only the two products and the benzene appear.
  
  $Δ_{c} H^{\circ} = 6 Δ_{f} H^{\circ} (C O_{2}) + 3 Δ_{f} H^{\circ} (H_{2} O) - Δ_{f} H^{\circ} (C_{6} H_{6})$
2. The products: six moles of CO₂ and three of water, together −3,218.4 kJ.
  
  $6 \cdot (- 393.5) + 3 \cdot (- 285.8) = - 2, 361.0 - 857.4 = - 3, 218.4 kJ$
3. Rearrange for benzene. The combustion released 3,301.3 kJ and the products account for only 3,218.4 of it, so benzene started 82.9 kJ above its elements.
  
  $Δ_{f} H^{\circ} (C_{6} H_{6}) = - 3, 218.4 - (- 3, 301.3) = + 82.9 kJ/mol$
4. For comparison, a benzene that really had three ordinary double bonds would sit three cyclohexene hydrogenations above cyclohexane.
  
  $- 123.1 - 3 \cdot (- 118.8) = + 233.3 kJ/mol$
Answer

ΔfH°(C₆H₆, g) = +82.9 kJ/mol. It is positive: benzene holds more enthalpy than the graphite and hydrogen it is made of, and forming it from them would absorb heat. That does not contradict the 150.4. Stability is always measured against something, and benzene is unstable next to its elements and unusually stable next to the ring of three double bonds it is drawn as, which on the same table would sit at +233.3.

$Δ_{f} H^{\circ} (C_{6} H_{6}, g) = + 82.9 kJ/mol 233.3 - 82.9 = 150.4 kJ/mol$

Learning path

Why a reaction goes

References (3)

The measurements behind insight block 2: the heats of hydrogenation of cyclohexene, the cyclohexadienes and benzene, taken in the gas phase at Harvard: G. B. Kistiakowsky, J. R. Ruhoff, H. A. Smith and W. E. Vaughan, "Heats of Organic Reactions. IV. Hydrogenation of Some Dienes and of Benzene." Journal of the American Chemical Society 58(1), 1936, pp. 146-153. DOI 10.1021/ja01292a043
Benzene's gas-phase enthalpy of formation, +82.9 ± 0.9 kJ/mol, is this review's evaluated value: M. V. Roux, M. Temprado, J. S. Chickos and Y. Nagano, "Critically Evaluated Thermochemical Properties of Polycyclic Aromatic Hydrocarbons." Journal of Physical and Chemical Reference Data 37(4), 2008, pp. 1855-1996. DOI 10.1063/1.2955570
Where the three C6 rings' gas-phase formation enthalpies were read: benzene +82.9, cyclohexene -4.32 (Steele, Chirico et al. 1996) and cyclohexane -123.1 (Prosen, Johnson et al. 1946), each with the measurement it comes from. Rounded to 0.1 kJ/mol on this page: P. J. Linstrom and W. G. Mallard (eds.), NIST Chemistry WebBook, NIST Standard Reference Database Number 69. National Institute of Standards and Technology, Gaithersburg MD. Benzene, gas-phase thermochemistry.