Chemical Thermodynamics · TOPIC LESSON
Keep the equation, amount, physical state and energy sign attached to every enthalpy value.
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BEFORE YOU START
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01 / UNDERSTAND & APPLY
Enthalpy H=U+PV is a state function. Under constant pressure with only pressure–volume work, the heat absorbed by a closed reacting system equals its enthalpy change. A negative reaction ΔH means the system releases heat in that constant-pressure setting; a positive value means it absorbs heat. Neither sign alone establishes spontaneity or speed.
A thermochemical equation specifies reactants, products, coefficients, physical states and conditions. The associated energy refers to the reaction extent represented by those coefficients. If A+2B→C has Δ_rH=−120 kJ mol⁻¹ of reaction, consuming 0.500 mol A with enough B corresponds to half a mole of reaction and an enthalpy change of −60.0 kJ.
Reversing a reaction reverses the enthalpy sign. Multiplying all coefficients by a factor multiplies the reaction enthalpy for that newly written equation by the same factor. Fractional coefficients are acceptable in thermochemical equations, especially when defining formation of exactly one mole of a substance. They are mole ratios, not a claim that an isolated half molecule reacts.
Physical states matter because liquid and gaseous water, or different carbon allotropes, have different enthalpies. A combustion equation producing H₂O(l) has a different enthalpy from one producing H₂O(g). Before combining data, align the states or include a phase-change correction. Omitting state labels can conceal a difference of many kilojoules.
ORIGINAL WORKED EXAMPLE
For 2A+B→3C, Δ_rH=−240 kJ per mole of the reaction as written. If 0.300 mol A reacts with excess B, find heat at constant pressure with only PV work.
Answer q_P=−36.0 kJ.
02 / UNDERSTAND & APPLY
If a reaction can be expressed as the sum of other reactions, its enthalpy change is the same sum of their enthalpy changes. The intermediate route can be hypothetical; it does not need to be a practical laboratory mechanism. What matters is that the combined initial and final states match the target exactly.
Treat the reaction equations like algebra. Reverse an equation if a target species is on the wrong side, scaling the enthalpy sign accordingly. Multiply an equation to obtain the required coefficient. Then add and cancel matching species only when their chemical identity and physical state agree. CO₂(g) does not cancel against CO₂(aq) without an additional dissolution step.
A useful strategy is to solve for species that appear in only one supplied equation. Their coefficients often determine the required multiplier immediately. After combining, rewrite the net equation and check each atom. Only then add the transformed energy values; a correct arithmetic sum attached to the wrong net reaction is not a solution.
Hess’s law does not say that heat is path independent in every process. Enthalpy is path independent. The equality between measured heat and ΔH requires the stated constant-pressure and work assumptions. Confusing these two statements can lead to using a heat measured under inappropriate conditions without conversion.
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ORIGINAL WORKED EXAMPLE
At the same temperature and states, A→B has ΔH=+35.0 kJ mol⁻¹ and B→C has ΔH=−90.0 kJ mol⁻¹. Find ΔH for 2C→2A.
Answer ΔH=+110 kJ for the equation 2C→2A.
03 / UNDERSTAND & APPLY
The standard molar enthalpy of formation Δ_fH° is the enthalpy change for forming one mole of a specified substance from its elements in their reference states at the stated temperature and standard pressure. For example, a formation equation for a compound must produce exactly one mole of that compound, even if fractional coefficients are needed for a diatomic element.
The standard formation enthalpy of an element in its reference form is defined as zero. This does not mean the element has zero absolute internal energy or zero entropy. A different allotrope or a gas-phase atom can have a nonzero formation enthalpy relative to the chosen elemental reference. At ordinary reference conditions graphite, rather than diamond, is the carbon reference form.
Because formation equations start from a common set of elemental references, subtracting reactant formation enthalpies from product formation enthalpies gives Δ_rH°=ΣνΔ_fH°(products)−ΣνΔ_fH°(reactants). Include every coefficient and physical state. A zero formation value makes a term vanish only for the correct reference-state element.
The degree symbol specifies standard states; it does not by itself set temperature to 298 K. Tables often use 298.15 K, and values at other temperatures require corresponding data or a correction. For solutions, the standard-state convention must also match. Do not silently combine unrelated temperatures or different aqueous conventions.
Δ_rH° = ΣνΔ_fH°(products) − ΣνΔ_fH°(reactants)
All values must refer to the same temperature and specified physical states.
ORIGINAL WORKED EXAMPLE
For CH₄(g)+2O₂(g)→CO₂(g)+2H₂O(l), use Δ_fH° values −75, 0, −394 and −286 kJ mol⁻¹ respectively. Calculate Δ_rH°.
Answer Δ_rH°=−891 kJ mol⁻¹ for the stated combustion equation and supplied rounded data.
04 / UNDERSTAND & APPLY
Standard combustion enthalpy describes complete combustion of one mole of a specified substance to stated products under the reference conditions. “Complete” and the product states matter: carbon normally ends as CO₂ in such an equation, while the water phase must be specified. Actual incomplete burning that produces CO or soot is a different reaction.
If two compounds with the same elemental composition burn to the same final products, the difference between their combustion enthalpies gives their interconversion enthalpy. This is another Hess cycle: both routes end at a common reference mixture. Check the exact formula and product amounts before subtracting.
A phase correction follows the same algebra. If a reaction produces ν moles of liquid water and the corresponding alternative produces vapor, add ν times the enthalpy of vaporization at the appropriate temperature. Vaporization absorbs energy, so a combustion ending in vapor is less exothermic than one ending in liquid when the other states match.
Heat values for fuels may use different water-product conventions, sometimes expressed through higher and lower heating values. Those names reflect whether condensation energy is included in the specified accounting. Never compare two quoted energy values without checking their basis, moisture content and reference conditions.
ORIGINAL WORKED EXAMPLE
A reaction producing 2 mol H₂O(l) has ΔH=−500 kJ for the equation. At the same temperature, ΔH_vap(H₂O)=+44.0 kJ mol⁻¹. What is ΔH for the same reaction producing 2 mol H₂O(g)?
Answer ΔH=−412 kJ for the vapor-product equation.
05 / UNDERSTAND & APPLY
A bond dissociation enthalpy refers to breaking a specified gas-phase bond, usually homolytically, into the specified gas-phase fragments. Breaking a stable bond requires energy. Forming the same bond from the corresponding separated fragments releases that energy. A reaction can release heat because formation of product bonds releases more than is required to break reactant bonds.
The elementary estimate Δ_rH≈ΣD(bonds broken)−ΣD(bonds formed) imagines gas-phase reactants converted to atoms or appropriate fragments, then assembled into products. Count all changed bonds with their multiplicities. Bonds unchanged on both sides cancel in a carefully constructed comparison, but do not cancel a bond merely because it has the same element pair if its environment changes and a precise value is required.
Average bond enthalpies are averages across molecular environments, so the result is approximate. A C–H bond in one molecule is not guaranteed to have exactly the average C–H value. Successive dissociation energies of a polyatomic molecule also differ because the remaining fragment changes after each bond is removed.
For condensed reactants or products, gas-phase bond estimates need phase corrections to compare with an experimental enthalpy involving liquids or solids. Ionic crystal reactions are not accurately represented by simply assigning one isolated “ionic bond energy” per formula unit; lattice and solvation terms describe collective and environmental effects.
Δ_rH ≈ ΣD_broken − ΣD_formed
Approximate gas-phase bond-enthalpy cycle; apply needed phase corrections separately.
ORIGINAL WORKED EXAMPLE
For H₂(g)+Cl₂(g)→2HCl(g), use D(H–H)=436, D(Cl–Cl)=243 and D(H–Cl)=431 kJ mol⁻¹. Estimate ΔH.
Answer Estimated ΔH≈−183 kJ mol⁻¹.
06 / UNDERSTAND & APPLY
An ionic-solid cycle may involve atomization of elements, ionization of gaseous atoms, electron attachment, and lattice formation from gaseous ions. Hess’s law connects those steps to the formation enthalpy of the solid. The numerical signs depend on how each step is written, especially lattice enthalpy.
Some sources define lattice enthalpy as the positive energy required to separate one mole of an ionic solid into gaseous ions; others report the negative enthalpy for lattice formation from those ions. These are opposite directions of the same idealized step. Write the actual equation instead of trying to infer the sign from the word “lattice.”
Dissolving an ionic solid can be represented by lattice separation followed by hydration of gaseous ions. Lattice separation is endothermic; hydration is usually exothermic. Their sum gives the solution enthalpy for the stated final concentration or limiting-dilution convention. A positive solution enthalpy does not by itself prove insolubility, because entropy also affects dissolution free energy.
Ionization enthalpies are positive for removing electrons from isolated gaseous species. Electron-gain enthalpy may be negative for a favorable first attachment, but adding an electron to an already negative ion can require energy. Do not assume every “electron affinity” number is reported with the same sign convention: some tables report the positive energy released rather than the signed attachment enthalpy.
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ORIGINAL WORKED EXAMPLE
A hypothetical salt requires +760 kJ mol⁻¹ to separate its lattice into gaseous ions. Hydration of those ions releases 810 kJ mol⁻¹. Estimate its solution enthalpy for the specified final solution state.
Answer ΔH_solution=−50 kJ mol⁻¹.
07 / UNDERSTAND & APPLY
From H=U+PV, a reaction between ideal-gas states at the same temperature gives Δ_rH≈Δ_rU+Δn_gasRT when condensed-phase PV differences are negligible. Here Δn_gas is gaseous product coefficients minus gaseous reactant coefficients for the equation as written. Solids, liquids and dissolved species do not enter this ideal-gas count.
A bomb calorimeter often supplies constant-volume reaction heat related to ΔU under its work assumptions. To obtain ΔH, use the gas-count correction with the reaction’s specified product phases. If combustion produces liquid water, that water is not counted as gas. Changing the phase changes both thermochemical data and Δn_gas.
The temperature dependence of reaction enthalpy follows d(Δ_rH)/dT=Δ_rC_P, with product heat capacities minus reactant heat capacities including coefficients. If Δ_rC_P is approximately constant, Δ_rH(T₂)=Δ_rH(T₁)+Δ_rC_P(T₂−T₁). This is the elementary Kirchhoff correction.
Use consistent joule or kilojoule units. If a phase transition lies between T₁ and T₂, split the path and include the transition enthalpy rather than applying one constant heat-capacity difference across it. Very broad temperature extrapolations can fail because heat capacities and even stable chemical species change.
Δ_rH≈Δ_rU+Δn_gasRT
Ideal gaseous components at the same T; condensed-phase PV contributions neglected.
Δ_rH(T₂)=Δ_rH(T₁)+Δ_rC_P(T₂−T₁)
Constant Δ_rC_P approximation with no unaccounted phase transition.
ORIGINAL WORKED EXAMPLE
For N₂(g)+3H₂(g)→2NH₃(g) at 300 K, a hypothetical measurement gives Δ_rU=−90.0 kJ mol⁻¹. Estimate Δ_rH using ideal gases.
Answer Δ_rH≈−95.0 kJ mol⁻¹.
08 / UNDERSTAND & APPLY
A calorimetric temperature rise measures heat gained by the solution and apparatus. In an effectively insulated experiment, q_reaction=−(mc+C_cal)ΔT if one common temperature change and constant heat capacities are appropriate. The reacting system’s heat has the opposite sign to the measured surroundings heat.
To obtain molar reaction enthalpy, divide by the actual reaction extent, determined from the limiting reagent and the balanced equation. Dividing by total moles mixed, or by the moles of an excess reagent, can give the wrong answer even when the heat calculation is correct. For neutralization, specify whether the result is per mole of acid, base, reaction or water formed.
The approximate constancy of strong-acid/strong-base neutralization enthalpy in dilute water reflects a common net ionic process, H⁺+OH⁻→H₂O. Weak-acid or weak-base neutralization includes additional ionization equilibria, so its observed heat need not match the strong/strong value. Concentration, dilution and other mixing effects can also matter.
Practical assumptions should be visible: no substantial heat leak, negligible evaporation, known heat capacity and appropriate reaction completion. If the apparatus heat capacity is ignored, state that it is assumed negligible. A temperature-rise measurement alone does not prove the intended reaction was the only process producing heat.
ORIGINAL WORKED EXAMPLE
Mix 25.0 mL of 1.00 M monoprotic strong acid with 40.0 mL of 1.00 M NaOH. An ideal insulated measurement gives q_reaction=−1.40 kJ after correcting apparatus and dilution effects. Find enthalpy per mole of water formed.
Answer −56.0 kJ per mole of water formed.
09 / UNDERSTAND & APPLY
These are original exercises with supplied data. Keep the complete reaction beside each number and state whether the answer is an energy for a sample or a molar reaction quantity. All hypothetical numbers serve the calculation and should not be mistaken for a source of tabulated chemical data.
Check signs by tracing the physical direction of each step. Check coefficients by canceling the actual equations. Check states before comparing or combining values. These three audits prevent most enthalpy-cycle mistakes before any calculator is needed.
ORIGINAL PRACTICE · 1
+240 kJ.
Reverse the reaction to change −80 to +80, then multiply both equation and enthalpy by three.
ORIGINAL PRACTICE · 2
The elemental reference is molecular O₂ gas, not isolated O atoms.
Formation of one mole O(g) from half a mole O₂ requires bond dissociation. The zero convention applies only to the specified reference form.
ORIGINAL PRACTICE · 3
−40 kJ mol⁻¹.
Products minus reactants gives −90−[(−20)+(−30)]=−90+50=−40.
ORIGINAL PRACTICE · 4
It is 25 kJ mol⁻¹ more positive for that equation.
Add the endothermic liquid-to-gas step to the original reaction. The vapor retains the extra enthalpy.
ORIGINAL PRACTICE · 5
−150 kJ.
Broken minus formed is 500−650=−150. Bond breaking costs energy; the larger formation release makes the net process exothermic.
ORIGINAL PRACTICE · 6
RT≈2.49 kJ mol⁻¹ of reaction.
ΔH−ΔU=ΔnRT=1×8.314×300 J mol⁻¹. The positive gas-count change gives a positive correction.
ORIGINAL PRACTICE · 7
−48.0 kJ mol⁻¹.
The correction is 20.0×100=2000 J mol⁻¹=+2.00 kJ mol⁻¹, giving −50.0+2.00=−48.0.
ORIGINAL PRACTICE · 8
−50.0 kJ mol⁻¹.
Reaction heat is the negative of the surroundings gain: −2.00 kJ. Divide by 0.0400 mol reaction to obtain −50.0 kJ mol⁻¹.
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