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Thermochemistry and Energetics

Thermochemistry is the study of heat changes that accompany physical and chemical processes. For the MCAT you should know the vocabulary (system, surroundings, state functions), the first and second laws of thermodynamics, the difference between ΔU and ΔH, calorimetry (q = mcΔT), Hess's law and bond enthalpies, and—most heavily tested of all—entropy (S) and Gibbs free energy (ΔG = ΔH − TΔS) as the predictor of spontaneity.

MCAT high-yield topics. This chapter maps to the AAMC Chemical & Physical Foundations outline: the first and second laws, state functions, enthalpy and Hess's law, bond enthalpies, calorimetry (q = mcΔT), entropy, and Gibbs free energy / spontaneity. The Gibbs and entropy sections are the most frequently tested—prioritize them.

Thermodynamics

Thermodynamics is the branch of science that deals with energy changes accompanying physical and chemical processes, especially the inter-conversion of heat and work. Chemical thermodynamics applies these ideas to chemical reactions; the predictions are independent of the path or the rate.

System / Surrounding / State Function Terms

System
The portion of the universe under study (e.g. the chemicals in a beaker).
Surroundings
Everything outside the system that can interact with it.
Boundary
The real or imaginary surface separating system from surroundings.
Open system
Exchanges both matter and energy with surroundings (e.g. boiling water in an open pan).
Closed system
Exchanges only energy, not matter (e.g. a stoppered flask).
Isolated system
Exchanges neither matter nor energy (e.g. a perfect thermos flask — an idealization).
State function
Property whose value depends only on the present state of the system, not on how it got there. Examples: U, H, S, G, T, P, V.
Path function
Property whose value depends on the path taken between states. Examples: q (heat) and w (work).
Standard state
Pure substance at 1 atm (101.3 kPa) and a specified temperature, conventionally 298 K. Quantities measured under these conditions are denoted by a degree sign, e.g. ΔH°.

Internal Energies

The internal energy U of a system is the total energy stored in it — the sum of the kinetic energies of all particles plus the potential energies of all interactions (intermolecular forces, bonds). U is a state function. Its absolute value cannot be measured, but changes ΔU can.

For a chemical reaction at constant volume in a closed system, the heat absorbed equals ΔU:

ΔU = qv (no expansion work)

This is what a bomb calorimeter measures.

First Law of Thermodynamics

The first law is the principle of conservation of energy applied to thermodynamic systems: energy can be transformed from one form to another but cannot be created or destroyed. Mathematically:

ΔU = q + w

Sign convention (IUPAC):

For pressure-volume (expansion) work at constant external pressure: w = −Pext ΔV.

Enthalpy

Enthalpy H is defined as H = U + PV. Like U, H is a state function. The change in enthalpy at constant pressure equals the heat absorbed:

ΔH = qp

This is why most laboratory reactions in open vessels are described by ΔH, not ΔU.

Relation between ΔH and ΔU

For reactions involving gases: ΔH = ΔU + ΔngasRT, where Δngas = (moles of gaseous products) − (moles of gaseous reactants). For reactions with no change in moles of gas, ΔH ≈ ΔU.

Standard enthalpy changes

Types of standard enthalpy change (per mole)
TypeSymbolDefinitionSign
FormationΔH°fForming 1 mol of compound from elements in standard statesUsually −ve (some +ve, e.g. NO)
CombustionΔH°cComplete combustion of 1 mol in excess O2Always −ve
NeutralizationΔH°neut1 mol H2O formed from H+ + OH (dilute)~ −57 kJ/mol (strong acid + strong base)
SolutionΔH°sol1 mol solute dissolved in excess solventCan be + or −
AtomizationΔH°at1 mol of gaseous atoms formed from elementAlways +ve
Bond enthalpyΔH°BEEnergy to break 1 mol of a specific bond in gas phaseAlways +ve
Lattice enthalpyΔH°L1 mol of ionic solid formed from gaseous ionsAlways −ve (release of energy)
HydrationΔH°hyd1 mol of gaseous ions dissolved in waterAlways −ve
Fusion / vaporizationΔH°fus / ΔH°vap1 mol melted / vaporized+ve (endothermic phase changes)

Calorimetry

Calorimetry measures heat flow from the temperature change of a known mass of water (or other substance). The core MCAT equation is:

q = m c ΔT

where q is heat (J), m is mass (g), c is the specific heat capacity (J g−1 °C−1; for water c = 4.18 J g−1 °C−1), and ΔT = Tfinal − Tinitial. If molar heat capacity C is used, q = n C ΔT.

Exothermic and Endothermic Reactions

Hess's Law

Hess's law of constant heat summation (Germain Hess, 1840): the total enthalpy change for a reaction is independent of the route taken from reactants to products, provided the initial and final states are the same.

This is a direct consequence of enthalpy being a state function. It allows enthalpy changes that cannot be measured directly to be calculated from those that can.

General form

ΔHreaction = ΣΔH°f(products) − ΣΔH°f(reactants)

Example — formation of CO

The direct combustion C(s) + ½O2(g) → CO(g) cannot be measured cleanly because some CO2 always forms. Hess's law lets us combine:

(i) − (ii): C(s) + ½O2(g) → CO(g); ΔHf(CO) = ΔH1 − ΔH2 = −110.5 kJ.

Applications of Hess's law

Bond Enthalpies

A bond enthalpy is the energy required to break one mole of a given bond in the gas phase; breaking bonds is always endothermic (+) and forming bonds is always exothermic (−). For a gas-phase reaction the MCAT lets you estimate ΔH from average bond enthalpies:

ΔHrxn = Σ(bonds broken) − Σ(bonds formed)

Note the sign carefully: this is "bonds broken minus bonds formed" (reactant bonds minus product bonds), the opposite ordering of the ΔH°f formula. Values are averages, so bond-enthalpy estimates are approximate.

Second Law and Entropy

The second law of thermodynamics states that the total entropy of the universe (system + surroundings) increases in any spontaneous process: ΔSuniverse = ΔSsystem + ΔSsurroundings > 0 for a spontaneous change, and = 0 at equilibrium.

Entropy S is a state function that measures the dispersal of energy and matter—often described as "disorder" or the number of accessible microstates. Units: J K−1 mol−1. The third law sets S = 0 for a perfect crystal at 0 K, so unlike H, absolute entropies (S°) can be tabulated.

Gibbs Free Energy and Spontaneity

Combining the first and second laws gives the single most tested equation in MCAT thermodynamics—the Gibbs free energy, which folds both enthalpy and entropy into one spontaneity criterion at constant T and P:

Gibbs free energy

ΔG = ΔH − TΔS (T in kelvin).
· ΔG < 0 → reaction is spontaneous (exergonic) in the forward direction.
· ΔG > 0 → non-spontaneous (endergonic); spontaneous in reverse.
· ΔG = 0 → system is at equilibrium.
Also: ΔG° = −RT ln K and ΔG = ΔG° + RT ln Q.

How the signs of ΔH and ΔS set spontaneity

Sign of ΔG from ΔH and ΔS
ΔHΔSSpontaneity
− (exothermic)+ (more disorder)Spontaneous at all temperatures (ΔG always < 0)
+ (endothermic)− (less disorder)Non-spontaneous at all temperatures (ΔG always > 0)
− (exothermic)− (less disorder)Spontaneous only at low T (enthalpy-driven)
+ (endothermic)+ (more disorder)Spontaneous only at high T (entropy-driven)

The crossover temperature at which a reaction switches between spontaneous and non-spontaneous is found by setting ΔG = 0, giving T = ΔH / ΔS.

Common trap. Heat (q) and work (w) are not state functions. Two reactions that go from the same start to the same end can absorb very different amounts of heat depending on whether they are run at constant pressure (q = ΔH) or constant volume (q = ΔU). Only U, H, S, G are path-independent.
Memory aid. "EXO leaves the system." Negative ΔH = exothermic = energy exits. Positive ΔH = endothermic = energy enters. Pair with: products lower than reactants ↔ exothermic.

Worked MCQs

Seven MCQs that capture the high-yield testing patterns for this chapter. Read the explanation even when you get the answer right — it's where the deeper concept lives.

Q1. Which of the following is NOT a state function?

  • Internal energy (U)
  • Enthalpy (H)
  • Work (w)
  • Pressure (P)

U, H, S, G, T, P, V are state functions — they depend only on the current state. Heat (q) and work (w) are path functions; the same change in state can involve different q and w depending on how the change is carried out.

Q2. The first law of thermodynamics is mathematically expressed as:

  • ΔG = ΔH − TΔS
  • ΔU = q + w
  • PV = nRT
  • ΔH = ΔU + PΔV

The first law is the conservation of energy: ΔU = q + w (heat added to the system + work done on the system). The other equations are real, but describe Gibbs free energy, the ideal gas law, and the H−U relation respectively.

Q3. An exothermic reaction is one in which:

  • ΔH is positive
  • ΔH is negative
  • ΔS is always negative
  • The system absorbs heat from surroundings

In an exothermic reaction the system loses heat to the surroundings, so ΔH < 0. Combustion, neutralization and respiration are textbook examples.

Q4. Hess's law is a direct consequence of:

  • Conservation of mass
  • Heat being a path function
  • Enthalpy being a state function
  • The ideal gas law

Because H depends only on the initial and final states, the total ΔH for any path from reactants to products is the same. Hess's law allows you to add or subtract reaction enthalpies as if you were doing algebra.

Q5. For the reaction N2(g) + 3H2(g) → 2NH3(g) at 298 K, Δngas equals:

  • +1
  • 0
  • +2
  • −2

Δngas = (moles of gaseous products) − (moles of gaseous reactants) = 2 − (1 + 3) = −2. This is what enters ΔH = ΔU + ΔngasRT.

Q6. A reaction has ΔH = +50 kJ/mol and ΔS = +100 J K−1 mol−1. Above what temperature does it become spontaneous?

  • 50 K
  • 200 K
  • 500 K
  • It is spontaneous at all temperatures

Set ΔG = ΔH − TΔS = 0 → T = ΔH / ΔS. Convert units: 50 000 J / 100 J K−1 = 500 K. With both ΔH and ΔS positive, the reaction is entropy-driven—ΔG becomes negative (spontaneous) only above 500 K.

Q7. For which process is ΔSsystem most clearly positive?

  • 2 NO2(g) → N2O4(g)
  • H2O(l) → H2O(s)
  • CaCO3(s) → CaO(s) + CO2(g)
  • N2(g) + 3 H2(g) → 2 NH3(g)

Entropy rises most when a gas is generated from a solid. Decomposition of CaCO3 creates a mole of CO2 gas from a solid, a large increase in disorder. The other three all decrease the moles of gas (or freeze a liquid), so ΔSsystem is negative.

Quick Recap

Test yourself. Take a timed Thermochemistry quiz or browse all Chemistry MCQs to lock these concepts in.