Thermochemistry II: enthalpy and Hess's law

Calorimetry measures enthalpy changes directly, but many reactions cannot be studied in a calorimeter. Some are too slow, some produce mixtures of products, and some are simply dangerous. Fortunately, because enthalpy is a state function, enthalpy changes can be added and subtracted like the legs of a journey. This lesson develops Hess's law, standard enthalpies of formation and bond enthalpies, the three tools chemists use to calculate ΔH for reactions that have never been measured, and applies them to fuels and energy.

Hess's law

In 1840 the Swiss-Russian chemist Germain Hess observed that the heat released by a reaction was the same whether it happened in one step or several. We now see this as a consequence of enthalpy being a state function: ΔH depends only on where a reaction starts and ends. Hess's law states that if a reaction can be written as the sum of two or more steps, its enthalpy change is the sum of the enthalpy changes of the steps.

A classic example is carbon monoxide. Burning carbon in a limited supply of oxygen always gives a mixture of CO and CO₂, so the enthalpy of C(s) + ½ O₂(g) → CO(g) cannot be measured cleanly. But the complete combustion of carbon and of carbon monoxide can both be measured, and the target reaction is their difference.

The strategy for any Hess's law problem is the same. Arrange the given equations so that the target's reactants are on the left and its products on the right, reversing equations (and changing the signs of their ΔH values) as needed. Multiply equations (and their ΔH values) by factors so that intermediate species cancel. Then add both the equations and the enthalpies.

The bookkeeping can be automated. Each equation becomes a vector of coefficients (negative for reactants, positive for products), and the problem becomes finding the multipliers that combine the given equations into the target, a small linear algebra problem.

import numpy as np

species = ["C", "H2", "O2", "CO2", "H2O", "CH4"]
# rows: given reactions as coefficient vectors (products +, reactants -)
given = np.array([
    [-1,  0, -1,  1, 0, 0],    # C + O2 -> CO2
    [ 0, -1, -0.5, 0, 1, 0],   # H2 + 1/2 O2 -> H2O
    [ 0,  0, -2,  1, 2, -1],   # CH4 + 2 O2 -> CO2 + 2 H2O
])
dH = np.array([-393.5, -285.8, -890.3])
target = np.array([-1, -2, 0, 0, 0, 1])   # C + 2 H2 -> CH4

# solve multipliers x so that x @ given = target (least squares gives the exact answer here)
x, *_ = np.linalg.lstsq(given.T, target, rcond=None)
print("multipliers:", np.round(x, 3))
print("check residual:", np.round(x @ given - target, 10))
print(f"dH(target) = {x @ dH:.1f} kJ")

multipliers: [ 1.  2. -1.]
check residual: [-0.  0. -0.  0. -0.  0.]
dH(target) = -74.8 kJ

The multipliers, 1, 2 and −1, are exactly the steps chosen by hand: keep the first equation, double the second, reverse the third.

Standard states and standard enthalpies of formation

To tabulate enthalpies usefully, chemists agree on reference conditions. The standard state of a substance is its pure form at a pressure of 1 bar (100 kPa) and a specified temperature, usually 25 °C (298.15 K). For a solute it is a concentration of 1 M. A superscript ° marks a standard quantity: ΔH° is the enthalpy change when reactants in their standard states become products in their standard states.

The standard enthalpy of formation, ΔH°f, of a compound is the enthalpy change when one mole of it forms from its elements in their standard states. For an element in its standard state, the most stable form at 1 bar and 25 °C, ΔH°f is zero by definition: O₂(g), N₂(g), C(graphite), Br₂(l), Na(s). Formation equations often need fractional coefficients so that exactly one mole of product forms, for example ½ N₂(g) + 3/2 H₂(g) → NH₃(g).

Formation enthalpies let us calculate the standard enthalpy of any reaction whose reactants and products are tabulated. Imagine taking the reactants apart into their elements, then building the products from those elements. By Hess's law,

where n and m are the coefficients in the balanced equation. Elements in their standard states contribute zero.

# Standard enthalpies of formation, kJ/mol (approximate table values)
Hf = {"CO2(g)": -393.5, "H2O(l)": -285.8, "H2O(g)": -241.8, "O2(g)": 0.0,
      "CH4(g)": -74.8, "C3H8(g)": -103.8, "C8H18(l)": -250.0, "C2H5OH(l)": -277.6,
      "H2(g)": 0.0}

def dH_rxn(reactants, products):
    return (sum(n * Hf[s] for s, n in products.items())
            - sum(n * Hf[s] for s, n in reactants.items()))

fuels = {   # fuel: (molar mass g/mol, reactants, products)
    "hydrogen": (2.016, {"H2(g)": 1, "O2(g)": 0.5}, {"H2O(l)": 1}),
    "methane":  (16.04, {"CH4(g)": 1, "O2(g)": 2}, {"CO2(g)": 1, "H2O(l)": 2}),
    "propane":  (44.10, {"C3H8(g)": 1, "O2(g)": 5}, {"CO2(g)": 3, "H2O(l)": 4}),
    "octane":   (114.23, {"C8H18(l)": 1, "O2(g)": 12.5}, {"CO2(g)": 8, "H2O(l)": 9}),
    "ethanol":  (46.07, {"C2H5OH(l)": 1, "O2(g)": 3}, {"CO2(g)": 2, "H2O(l)": 3}),
}
print(f"{'fuel':10s}{'kJ/mol':>10s}{'kJ/g':>8s}{'g CO2 per MJ':>14s}")
for name, (M, r, p) in fuels.items():
    dH = dH_rxn(r, p)
    co2 = p.get("CO2(g)", 0) * 44.01
    per_MJ = co2 / (-dH / 1000)
    print(f"{name:10s}{dH:10.1f}{-dH / M:8.1f}{per_MJ:14.1f}")

fuel          kJ/mol    kJ/g  g CO2 per MJ
hydrogen      -285.8   141.8           0.0
methane       -890.3    55.5          49.4
propane      -2219.9    50.3          59.5
octane       -5470.2    47.9          64.4
ethanol      -1366.8    29.7          64.4

The table shows two ways of judging a fuel. Hydrogen releases by far the most energy per gram and no carbon dioxide at the point of use, though how the hydrogen is made decides its overall footprint. Among the carbon fuels, methane gives the least carbon dioxide per megajoule because it has the highest ratio of hydrogen to carbon. Ethanol stores less energy per gram than octane because it is already partly oxidized: one of its atoms is oxygen.

Enthalpies of formation and stability

A large negative ΔH°f means a compound is much lower in enthalpy than its elements; it is energetically stable with respect to decomposing into them. Carbon dioxide and water, the end products of combustion, have large negative values, which is why burning releases so much energy. A positive ΔH°f, as for nitrogen monoxide (+91.3 kJ/mol), means the compound is higher in enthalpy than its elements. NO forms only at the high temperatures of engines and lightning, and it slowly reacts further in the atmosphere, contributing to smog and acid rain.

Energetic stability is not the same as reactivity, however. Whether a reaction actually happens depends also on entropy (lesson 3) and on its rate (lessons 5 and 6). Diamond has ΔH°f = +1.9 kJ/mol relative to graphite, yet diamonds do not turn into graphite on any human time scale.

Bond enthalpies

When formation enthalpies are not available, bond enthalpies give an estimate. The bond enthalpy (or bond dissociation energy) is the enthalpy needed to break one mole of a particular bond in gaseous molecules. Breaking bonds always absorbs energy; forming bonds always releases it. A reaction's enthalpy is roughly the energy needed to break the bonds in the reactants minus the energy released in forming the bonds of the products.

Bond enthalpy estimates are less accurate than formation-enthalpy calculations for two reasons. Tabulated bond enthalpies are averages over many molecules, and a C–H bond in methane is not identical to one in ethanol. And bond enthalpies apply only to gases, so reactions involving liquids or solids also require enthalpies of vaporization or sublimation. Their value is insight: they show why a reaction is exothermic. Combustion releases energy because the C=O and O–H bonds formed are stronger than the C–H and O=O bonds broken.

Enthalpies of solution and lattice energy

Dissolving a salt involves two competing steps: breaking up the crystal lattice, which absorbs energy (the lattice energy from CHEM-101's Born–Haber cycles), and hydrating the separated ions, which releases energy as water molecules cluster around them. The enthalpy of solution is the sum. When hydration wins, dissolving is exothermic, as for calcium chloride (about −82.8 kJ/mol), which is why calcium chloride melts ice on roads even in deep cold. When the lattice wins, dissolving is endothermic, as for ammonium nitrate (about +25.7 kJ/mol), the basis of instant cold packs.

Fuels, food and energy density

Formation enthalpies underpin the energy comparisons used in public policy. Gasoline and diesel store about 46 MJ/kg and about 34 MJ per liter; lithium-ion batteries store roughly 0.5 to 0.9 MJ/kg (lesson 11 explains why). Food energy is the same chemistry run slowly: oxidizing glucose releases 2,803 kJ/mol, or about 15.6 kJ/g, and fats release more than twice as much per gram as carbohydrates because they are less oxidized. Nutrition labels use rounded values of 4, 4 and 9 Calories per gram for carbohydrate, protein and fat.