MCAT content library / Chemistry & physics / Reaction enthalpy: combine Hess cycles and bond-energy estimates
Build the requested reaction from compatible data, keeping direction, stoichiometric amount, and physical state attached to each energy value.
About 14 minutes, including the worked examples and checks.
Useful foundations
On this page
- Concept: enthalpy describes a change of state
- Pattern: reverse, scale, and cancel before adding
- Application: combine a three-step isomer cycle
- Rule: formation data provide a shared reference
- Application: a bond-energy estimate uses a different data model
- Use the 5P routine on an enthalpy calculation
- Check your reasoning
- Continue learning
- Sources
Concept: enthalpy describes a change of state
Enthalpy is H = U + PV. For a process at constant pressure with only pressure–volume work, the heat absorbed by the system equals ΔH. An exothermic reaction has negative ΔH; an endothermic reaction has positive ΔH. Define the reaction and the material states before assigning either label. Water vapor and liquid water are different states, so changing the product phase changes the reaction enthalpy.2
Enthalpy is a state function: the change depends on the specified starting and ending states. The amount of substance matters. A thermochemical equation written with ΔH = −80 kJ for conversion of two moles is associated with that stoichiometric amount; conversion of one mole by the same net reaction corresponds to −40 kJ. Reporting an energy “per mole” requires naming the mole of reaction or species being used.4
Pattern: reverse, scale, and cancel before adding
Hess’s law combines compatible thermochemical equations to construct a target reaction. Reversing a step reverses its enthalpy sign. Multiplying every coefficient in a step multiplies its enthalpy by the same factor. After those operations, cancel species that appear in equal amounts on both sides. Add the adjusted enthalpies only after checking that the remaining equation is the requested target.4
This is algebra on state changes, not a claim that the reaction actually follows those steps as a mechanism. A calculation can use a convenient hypothetical path through intermediates that never accumulate in the experiment. All data must refer to compatible temperatures, pressures, compositions, and phases, or include the required corrections.2
Application: combine a three-step isomer cycle
Consider four hypothetical gaseous isomers A, B, C, and D with the same molecular formula. The following original illustrative values all apply at one specified temperature and the same reference conditions. The target is A(g) → D(g). Values in the table are for the full equations as written.
| Step | Equation | ΔH |
|---|---|---|
| 1 | 2 A(g) → 2 B(g) | +60 kJ |
| 2 | C(g) → B(g) | +10 kJ |
| 3 | 2 C(g) → 2 D(g) | −50 kJ |
Halve step 1 to obtain A → B with +30 kJ. Reverse step 2 to obtain B → C with −10 kJ. Halve step 3 to obtain C → D with −25 kJ. B and C cancel when the transformed equations are added, leaving exactly A → D. The target enthalpy is +30 − 10 − 25 = −5 kJ for one mole of A converted.
A state-level check catches sign errors. Set A’s relative molar enthalpy to zero. Then B lies at +30 kJ/mol, C at +20 kJ/mol, and D at −5 kJ/mol. C lies below B because the supplied C → B step is endothermic. These levels reproduce the same A-to-D difference. They do not specify activation barriers or reaction speeds.
If the target changes to 2 D → 2 A, reverse and double the complete result to get +10 kJ. This is a new target, so retaining −5 kJ would fail both direction and amount. If a source step instead used liquid D, it could not be substituted without also accounting for the phase change needed to reach gaseous D.
Rule: formation data provide a shared reference
A standard formation enthalpy describes forming one mole of a substance from its constituent elements in their reference forms under standard-state conditions. The formation enthalpy of an element in its reference form is zero by convention; this does not mean its total enthalpy is zero. Another allotrope or an isolated gaseous atom need not have zero formation enthalpy.5
Use ΔH°reaction = ΣνΔH°f,products − ΣνΔH°f,reactants, where ν is each stoichiometric coefficient. The superscript ° identifies standard states, not a mandatory temperature. The table’s temperature must still be specified; 298.15 K is common. Modern standard pressure is 1 bar, while older tables may use 1 atm, so keep the convention consistent.2
For an original symbolic example, let X2(g) and Y2(g) be the reference forms of hypothetical elements and let ΔH°f[XY(g)] = −45 kJ/mol. For 2 XY(g) → X2(g) + Y2(g), the result is [0 + 0] − [2(−45)] = +90 kJ. The coefficient two belongs in the arithmetic even though the formation value is tabulated per mole of XY.
Application: a bond-energy estimate uses a different data model
Breaking a stable covalent bond requires energy; forming the corresponding bond releases energy. A gas-phase estimate therefore uses the sum of bond energies for bonds broken minus the sum for bonds formed. Average bond energies pool different molecular environments, so the estimate can differ from an enthalpy calculated using compound-specific thermochemical data.3
For H2(g) + Br2(g) → 2 HBr(g), suppose a passage supplies rounded bond energies of 436, 193, and 366 kJ/mol for H–H, Br–Br, and H–Br. Breaking one mole of each reactant bond costs 629 kJ; forming two moles of H–Br bonds releases 732 kJ. The estimated reaction enthalpy is 629 − 732 = −103 kJ. The negative sign comes from greater energy release on bond formation, not from bond breaking releasing energy.
Before comparing this result with a measurement, inspect the model. A reaction starting with liquid bromine needs a vaporization contribution before gas-phase bond data can describe it. Averaged bond values, temperature differences, experimental uncertainty, and omitted phase terms can explain disagreement. A disagreement alone does not disprove energy conservation or establish a new reaction mechanism.3
Use the 5P routine on an enthalpy calculation
Prioritize the exact target equation and its phases. Paraphrase each supplied value as energy for a stated amount and direction. Prognose which steps need reversal or scaling. Pick the combination whose species cancel correctly. Post-mortem the sign, units, coefficients, and compatibility of the data. A negative reaction enthalpy does not by itself establish a fast rate or the favored direction under every mixture condition.
This lesson addresses the reaction enthalpy, formation data, Hess combinations, and bond-energy scope in AAMC category 5E.1
ACTIVE PRACTICE · EXPERIMENT AND TRANSFER
Reverse and Scale Thermochemical Steps Before Summing
The original Hess cycle supplies 2A(g) → 2B(g) with +60 kJ; C(g) → B(g) with +10 kJ; and 2C(g) → 2D(g) with −50 kJ. All are hypothetical gaseous isomers at matched reference conditions.
Transformed contributions: + () + () kJ.
: ΔH = .
The intermediate B and C cancel, leaving only target-state change.
Reversing the final reaction reverses every contribution; scaling multiplies both the coefficients and the heat. These algebraic steps do not assert the reaction actually follows that mechanism.
Check your reasoning
Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.
Check 1
For hypothetical gaseous isomers at compatible conditions, 2 P → 2 Q has ΔH = +48 kJ and R → Q has ΔH = +7 kJ. What is ΔH for P → R?
A. −31 kJ
Review this choice. The first step must be halved, not reversed. The target starts with P, so P → Q remains endothermic.
B. −17 kJ
Review this choice. This has the magnitude of the correct result but the opposite sign. It describes R → P.
C. +17 kJ
Correct. Half of the first step gives P → Q at +24 kJ. Reversing R → Q gives Q → R at −7 kJ. Their sum is +17 kJ.
D. +31 kJ
Review this choice. This adds +7 kJ without reversing the second step. To consume Q and produce R, that step must contribute −7 kJ.
The principle to carry forward
Cancel the intermediate in the desired direction before adding adjusted enthalpies.
Check 2
At specified conditions, 3 M(g) → 3 N(g) has ΔH = −81 kJ. What is the enthalpy change for conversion of 2 mol N(g) to 2 mol M(g)?
A. −54 kJ
Review this choice. Scaling by two-thirds gives the correct magnitude, but the requested reverse direction changes the sign to positive.
B. +27 kJ
Review this choice. This is the reverse change for one mole. The question asks for conversion of two moles.
C. +54 kJ
Correct. Reversing gives +81 kJ for three moles; multiplying by 2/3 gives +54 kJ for two moles.
D. +162 kJ
Review this choice. The original −81 kJ applies to three moles, not one. Normalize the amount before scaling to two moles.
The principle to carry forward
Direction and amount are separate operations; both act on ΔH.
Check 3
A gas-phase estimate using average bond energies differs from a constant-pressure measurement for a reaction with one liquid reactant. Which explanation should be investigated first?
A. Bond breaking releases the energy missing from the estimate.
Review this choice. Breaking stable bonds requires energy. Reversing that sign would create a bookkeeping error rather than explain the model mismatch.
B. Average gas-phase bond data and the omitted phase contribution may not match the measured process.
Correct. The estimate and experiment may describe different physical states, and average bond data need not equal compound-specific values. Check those differences before interpreting the discrepancy.
C. The measured enthalpy must depend on the catalyst because the reaction is faster.
Review this choice. A catalyst can change the reaction pathway and rate without changing the enthalpy difference between the same initial and final states.
D. A mismatch proves that energy is not conserved during the reaction.
Review this choice. A model-data discrepancy can arise from missing terms, approximations, or measurement error. It does not establish failure of conservation.
The principle to carry forward
Compare the physical states and the meaning of each dataset before comparing numerical answers.
Continue learning
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- Entropy and temperature: explain when a process is favored
- Free energy and equilibrium: distinguish standard from actual conditions
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Sources
1. AAMC: Chemical and Physical Foundations, Content Category 5E
2. OpenStax Chemistry 2e, 5.3 Enthalpy
3. OpenStax Chemistry 2e, 7.5 Strengths of Ionic and Covalent Bonds
4. Florida State University, CHM1045 Enthalpy Lecture
5. UCLA Illustrated Glossary of Organic Chemistry: Enthalpy of formation