MCAT topics → Thermodynamics and Kinetics → Enthalpy and Hess's Law

Enthalpy and Hess's Law

MCAT trap: Forgets to flip the sign of ΔH when reversing a step reaction in Hess's law. Reversing a reaction flips the sign of ΔH; multiplying by a coefficient scales the magnitude proportionally.

Enthalpy is one of the most tested thermodynamic concepts on the MCAT, and it shows up in almost every test section that touches chemistry — general chem, organic, even biochemistry (think ATP hydrolysis or metabolic pathways). At its core, enthalpy (H = U + PV) is a state function that tracks heat flow at constant pressure, which is the condition for most biological and chemical reactions. When pressure is constant, ΔH = q, and that single equation is the bridge between the abstract definition and everything practical.

The MCAT tests enthalpy from multiple angles. It'll ask you to recall definitions and sign conventions (exothermic = negative ΔH, products lower in energy), apply Hess's law to string together reactions and calculate an unknown ΔH, interpret energy diagrams from passage data, and estimate ΔH from bond energies. The bond energy angle is particularly common in organic passages where you don't have formation data but you do have structures. Each of these angles requires a slightly different skill — and the exam will mix them in the same passage.

What makes this topic genuinely tricky isn't the math — it's the conceptual traps. Students consistently flip the bond energy formula, forget to flip the sign when reversing a Hess's law step, or incorrectly place products above reactants on an exothermic energy diagram. These aren't careless errors; they reflect a shallow mental model. If you build the right picture — energy is released when bonds form, energy is absorbed when bonds break, and reversing a reaction literally runs it backward — these mistakes become impossible.

Common misconceptions

Common mistake
Wrong: When a reaction is reversed in Hess's law, only the magnitude of ΔH changes, not the sign.
Right: Reversing a reaction flips the sign of ΔH; multiplying by a coefficient scales the magnitude proportionally.
When you reverse a reaction, you're literally running it in the opposite direction — so every joule that was released is now absorbed, and vice versa. The sign of ΔH must flip. Forgetting this is usually a procedural error: students focus on getting the target species to cancel and lose track of what they did to each equation. Always annotate each step with its modified ΔH immediately after you reverse or scale it, before moving on.
Common mistake
Wrong: ΔH from bond energies equals bonds formed minus bonds broken.
Right: ΔH from bond energies equals energy of bonds broken minus energy of bonds formed (breaking requires energy input, forming releases energy).
Breaking bonds always requires energy input (endothermic), and forming bonds always releases energy (exothermic). The formula ΔH ≈ Σ(bonds broken) − Σ(bonds formed) follows directly from this: you pay energy to break reactant bonds, then get energy back when product bonds form, and the net is the difference. Inverting this gives you the wrong sign every time — a reaction that should be exothermic will appear endothermic. Anchor the formula by remembering: 'I break first, then I build.'
Common mistake
Wrong: In an exothermic reaction, products have higher energy than reactants.
Right: In an exothermic reaction, products have lower energy than reactants; the energy difference is released as heat (ΔH < 0).
Energy diagrams plot potential energy on the y-axis. In an exothermic reaction, the system loses energy to the surroundings — meaning products end up at a lower energy state than reactants. The 'downhill' drop IS the energy released as heat. Students who confuse this often conflate activation energy (which is a hump above reactants) with the overall reaction energy. The activation energy hump can be large even when the overall ΔH is very negative — these are independent features of the diagram.
Common mistake
Wrong: Enthalpy change equals heat transfer under any conditions.
Right: ΔH equals heat transfer only at constant pressure; at constant volume, heat transfer equals ΔU.
ΔH = q is only valid at constant pressure — this is not a universal identity. At constant volume (like in a bomb calorimeter), no PV work is done, so all the heat exchanged equals ΔU, not ΔH. The MCAT will sometimes specify the conditions of a calorimetry experiment precisely to test whether you apply the right relationship. If the problem says 'constant pressure' or describes an open container, use ΔH = q. If it says 'bomb calorimeter' or 'constant volume,' use ΔU = q.
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What the exam tests

  1. Know the formal definition of enthalpy (H = U + PV) and understand why ΔH equals heat transferred only at constant pressure, not under all conditions.
  2. Use Hess's law to calculate an unknown ΔH by combining a series of reactions — this includes reversing reactions (and flipping the sign of ΔH) and scaling reactions by coefficients (and scaling ΔH proportionally).
  3. Estimate ΔH for a reaction using average bond energies: add up the energy required to break all bonds in reactants, subtract the energy released forming all bonds in products.
  4. Interpret energy diagrams and apply sign conventions correctly — exothermic reactions have ΔH < 0 and products sit lower than reactants on the diagram; endothermic reactions are the opposite.

Can you avoid these mistakes?

Given these two reactions: (1) A → B, ΔH = +50 kJ and (2) C → B, ΔH = −30 kJ, what is ΔH for the reaction A → C? Show your reasoning step by step.
A reaction breaks one C–C bond (347 kJ/mol) and two C–H bonds (413 kJ/mol each), and forms one C=O bond (745 kJ/mol) and one O–H bond (463 kJ/mol). Estimate ΔH and determine whether the reaction is exothermic or endothermic.
A student draws an energy diagram for an exothermic reaction and places the products higher than the reactants, arguing that 'the products gained the energy that was released.' What is wrong with this model, and how should the diagram look?
A bomb calorimeter measures 25 kJ of heat released during a combustion reaction. A student reports this as ΔH = −25 kJ. Is this correct? Under what conditions would ΔH equal −25 kJ for this reaction?

Related topics

First Law of Thermodynamics and Internal Energy Calorimetry and Heat Capacity Entropy and the Second Law Gibbs Free Energy and Spontaneity

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