Thermochemistry

Which reactions release energy, which absorb it, and — more importantly — which can actually happen. Two scalar quantities (enthalpy and entropy) combine into a third (Gibbs free energy) that answers "will this react?" for every reaction at every temperature.

1. Why thermochemistry matters

Chemistry has two big questions. "Will this reaction happen?" and "How fast?" Thermochemistry answers the first one. Kinetics answers the second. You need both, but thermochemistry comes first because it tells you whether the reaction is worth waiting for in the first place.

At a higher level, thermochemistry is the chemistry version of classical thermodynamics: energy, temperature, heat, and entropy. The physics track emphasizes engines and phase transitions. The chemistry track emphasizes reactions — energy changes when bonds break and form, enthalpies of solution, and the equilibrium constants that come out the other side.

Why should you care?

• Energy technology. Every battery, fuel cell, and combustion engine is a controlled thermochemical process. You can't optimize what you can't calculate, and cell voltages come from $\Delta G$.
• Materials and metallurgy. Smelting, alloying, oxide reduction, phase stability — all of it is free-energy bookkeeping. The Ellingham diagram that metallurgists use to decide whether aluminum can reduce iron oxide is just $\Delta G$ vs. temperature.
• Biology. ATP hydrolysis releases $\approx 30$ kJ/mol and powers the cell. Every enzymatic reaction is coupled to a favorable free-energy change. Metabolism is literally thermochemistry on a conveyor belt.
• Environmental and climate. The atmospheric CO$_2$ uptake by the ocean is a thermochemical equilibrium that shifts with temperature. Global warming models need $\Delta H$ and $\Delta S$ for every relevant reaction.

2. Vocabulary cheat sheet

3. Enthalpy and heat

When a reaction occurs, some energy is released or absorbed as heat. The amount of heat exchanged, at constant pressure, is called the enthalpy change:

Most combustion reactions are exothermic. Burning methane:

Heat capacity

If you put heat into a substance but no reaction occurs, the temperature rises. The constant of proportionality is the heat capacity:

4. Calorimetry

Calorimetry is the experimental technique for measuring $\Delta H$. You run a reaction in an insulated container and track the temperature change of the surroundings — usually a known mass of water. Conservation of energy does the rest:

Worked example: coffee-cup calorimetry

You dissolve 2.0 g of solid NaOH in 100.0 mL of water in a styrofoam cup. The temperature rises from 22.0°C to 26.6°C. Find $\Delta H$ per mole of NaOH.

1. Heat absorbed by water: $q = m c \Delta T = 100 \cdot 4.18 \cdot 4.6 = 1923$ J.
2. That's heat gained by the surroundings, so $q_\text{rxn} = -1923$ J.
3. Moles of NaOH: $2.0 / 40.0 = 0.050$ mol.
4. $\Delta H = q_\text{rxn} / n = -1923 / 0.050 = -38\,460$ J/mol $= -38.5$ kJ/mol.

The published value is $-44.5$ kJ/mol — off by ~15%, because a styrofoam cup is a lousy adiabatic boundary and loses heat. The method is right; the instrument is crude. That's why the bomb calorimeter exists.

5. Hess's law and standard enthalpies of formation

Enthalpy is a state function, meaning $\Delta H$ only depends on the initial and final states — not on the path between them. That simple fact has enormous practical consequences. It means you can compute $\Delta H$ for any reaction by adding and subtracting $\Delta H$ values from any other reactions that sum (algebraically) to the one you want. That's Hess's law.

In practice you tabulate $\Delta H_f^\circ$, the standard enthalpy of formation, for every compound, and compute any reaction's $\Delta H$ from the difference of sums:

Worked example: burning methane via formation enthalpies

Compute $\Delta H^\circ$ for CH$_4$(g) + 2 O$_2$(g) $\to$ CO$_2$(g) + 2 H$_2$O(l).

From tables (kJ/mol): $\Delta H^\circ_f[\text{CH}_4(g)] = -74.8$; $\Delta H^\circ_f[\text{O}_2(g)] = 0$ (element); $\Delta H^\circ_f[\text{CO}_2(g)] = -393.5$; $\Delta H^\circ_f[\text{H}_2\text{O}(l)] = -285.8$.

Which matches the experimental value. Hess's law lets you calculate combustion, oxidation, and fuel energies for thousands of reactions from a compact table of a few hundred formation enthalpies.

6. Bond enthalpies

If you don't have formation data but you know a reaction's bond changes, you can estimate $\Delta H$ from bond enthalpies — average energies for breaking specific kinds of bonds:

For the methane combustion above: break 4 C-H (413 kJ/mol each) and 2 O=O (498). Form 2 C=O (799 kJ/mol in CO$_2$) and 4 O-H (463). Estimate:

Compare with the tabulated value of $-890$ kJ/mol. An error of 88 kJ/mol is about 10% — reasonable for a method that requires no formation data. The systematic underestimate reflects the fact that bond enthalpies don't quite capture the stability of CO$_2$.

7. Entropy and the second law

Exothermic reactions are common but not universal. Ammonium nitrate dissolves in water endothermically and instantly — the water gets cold and yet the reaction happens. Why?

The answer is entropy. The universe prefers states with more accessible microstates. A jumbled solution of ions has vastly more microstates than a single crystal, so even though dissolving absorbs energy, the resulting increase in disorder makes it worth doing. The formal statement is the second law of thermodynamics:

Where entropy comes from, informally

Entropy counts the number of ways a system can arrange its microscopic degrees of freedom. More ways = more entropy. Some patterns:

• Phase. Gas $\gg$ liquid $>$ solid. Gases have far more accessible positions and velocities than liquids, which are in turn more flexible than solids.
• Temperature. Hotter = more entropy. Adding heat gives molecules more kinetic energy and more accessible states.
• Mixing. Two separate pure substances have lower entropy than the mixed version. Mixing adds $\Delta S_\text{mix} = -R(x_1\ln x_1 + x_2\ln x_2)$.
• Molecule count. A reaction that produces more molecules than it consumes generally increases entropy, because more independent particles = more microstates.
• Complexity. A flexible linear molecule has higher entropy than a rigid ring. The pieces of a protein have higher entropy than the folded state; that's why protein folding is entropically costly.

At absolute zero, a perfect crystal has exactly one microstate and $S = 0$. That's the third law of thermodynamics, and it lets you tabulate absolute entropies rather than just changes.

8. Gibbs free energy and spontaneity

Keeping track of the universe is inconvenient. Fortunately, at constant $T$ and $P$, you can fold everything into a single system-only quantity that you can compute from tables:

The four quadrants of spontaneity

Because $\Delta G = \Delta H - T\Delta S$ has two signed terms, there are four possibilities:

Two quadrants are temperature-sensitive, and the crossover temperature is exactly $T = \Delta H/\Delta S$. That's the boiling point of a liquid, the melting point of a solid, and the temperature at which a reaction switches from spontaneous to non-spontaneous.

Worked example: when does water boil?

For H$_2$O($l$) $\to$ H$_2$O($g$), $\Delta H = +40.7$ kJ/mol and $\Delta S = +109$ J/(mol·K).

1. At $T = 298$ K: $\Delta G = 40700 - 298 \times 109 = 40700 - 32482 = +8218$ J/mol. Positive — water doesn't spontaneously boil at room temperature, as expected.
2. Set $\Delta G = 0$ and solve for $T$: $T = 40700 / 109 = 373$ K $= 100°$C. The boiling point.
3. At $T = 400$ K: $\Delta G = 40700 - 400 \times 109 = -2900$ J/mol. Negative — water does boil, as you'd expect on a stove.

Notice the entropy change is what drives the transition at high temperature. The enthalpy says "don't boil"; the entropy says "boil if you can afford it." The temperature sets the exchange rate between these two votes, and 100°C is where they balance.

9. Interactive: spontaneity explorer

Pick $\Delta H$ and $\Delta S$ for a hypothetical reaction and slide the temperature to see $\Delta G$ change sign. The gray band marks the crossover temperature $T_\text{switch} = \Delta H/\Delta S$ where $\Delta G = 0$.

10. Thermochemistry in code

Three practical calculations. Hess's law from formation enthalpies, bond-enthalpy estimation, and Gibbs crossover.

import numpy as np

# ---------- 1. Hess's law from formation enthalpies ----------
# dHf in kJ/mol; None for elements in their standard state (zero).
dHf = {
    "CH4(g)":      -74.8,
    "O2(g)":        0.0,
    "CO2(g)":     -393.5,
    "H2O(l)":     -285.8,
    "H2O(g)":     -241.8,
}

def dH_rxn(reactants, products):
    # each argument is a list of (species, stoichiometric coefficient) tuples
    return (sum(n * dHf[sp] for sp, n in products)
            - sum(n * dHf[sp] for sp, n in reactants))

# Methane combustion: CH4 + 2 O2 -> CO2 + 2 H2O(l)
dH = dH_rxn([("CH4(g)", 1), ("O2(g)", 2)],
            [("CO2(g)", 1), ("H2O(l)", 2)])
print(f"methane combustion dH = {dH:.1f} kJ/mol")   # -890.3

# ---------- 2. Bond-enthalpy estimate ----------
# Average bond enthalpies in kJ/mol
BE = {
    "C-H": 413, "O=O": 498,
    "C=O": 799, "O-H": 463,
}

def dH_bonds(broken, formed):
    return sum(n * BE[b] for b, n in broken) - sum(n * BE[b] for b, n in formed)

# CH4 + 2 O2 -> CO2 + 2 H2O: break 4 C-H + 2 O=O, form 2 C=O + 4 O-H
dH_est = dH_bonds([("C-H", 4), ("O=O", 2)],
                   [("C=O", 2), ("O-H", 4)])
print(f"bond-enthalpy estimate dH = {dH_est} kJ/mol")   # -802

# ---------- 3. Gibbs free energy and crossover temperature ----------
def dG(dH_kJmol, dS_JmolK, T):
    return dH_kJmol - T * dS_JmolK / 1000.0

def T_crossover(dH_kJmol, dS_JmolK):
    return dH_kJmol * 1000.0 / dS_JmolK   # kelvin

# Water vaporization: dH = +40.7 kJ/mol, dS = +109 J/(mol K)
print(f"boiling point of water = {T_crossover(40.7, 109):.1f} K")  # ~373
print(f"dG at 298 K = {dG(40.7, 109, 298):.2f} kJ/mol")  # +8.22
print(f"dG at 400 K = {dG(40.7, 109, 400):.2f} kJ/mol")  # -2.90

import math

# Pure Python version.

R = 8.314  # J/(mol K)

def dG(dH, dS, T):
    # dH in kJ/mol, dS in J/(mol K), T in K
    return dH - T * dS / 1000.0

def K_from_dG(dG_kJmol, T):
    # dG0 = -RT ln K  =>  K = exp(-dG0 / RT)
    return math.exp(-dG_kJmol * 1000.0 / (R * T))

# Simple check: water's boiling point
T_boil = 40.7 * 1000.0 / 109.0
print(f"water bp = {T_boil:.1f} K")  # 373.4

# Haber process: N2 + 3 H2 -> 2 NH3
#   dH = -92 kJ/mol, dS = -199 J/(mol K)
print(f"Haber dG at 298 K = {dG(-92, -199, 298):.2f} kJ/mol")  # ~-33, favorable
print(f"Haber dG at 700 K = {dG(-92, -199, 700):.2f} kJ/mol")  # ~+47, unfavorable

# K at 298 K for Haber (dG0 = -33 kJ/mol)
print(f"K(298) = {K_from_dG(-33, 298):.2e}")  # ~6e5

# Calorimetry: dissolve NaOH and measure T rise
def dH_calorimetry(m_water_g, dT_C, n_moles):
    # assume c_water = 4.18 J/(g K), ignore calorimeter heat cap
    q_water = m_water_g * 4.18 * dT_C      # J absorbed by water
    q_rxn = -q_water                     # by conservation
    return q_rxn / n_moles / 1000.0        # kJ/mol

print(f"NaOH dH est = {dH_calorimetry(100, 4.6, 0.050):.1f} kJ/mol")  # ~-38.5

Three notes:

• Temperature in Gibbs formulas is always in kelvin. Using celsius is a common and painful mistake.
• Be careful with units when $\Delta H$ is in kJ/mol but $\Delta S$ is in J/(mol·K) — the factor of 1000 trips people up constantly.
• Both $\Delta H$ and $\Delta S$ vary weakly with temperature. For small temperature ranges (a few hundred K), treating them as constants is fine. For wider ranges, you need heat-capacity data and Kirchhoff's law.

11. Cheat sheet

See also