Calculate heat energy using calorimetry. Enter mass of water/solution, specific heat capacity (default 4.18 J/g/°C for water), temperature change, and moles of substance to find molar enthalpy change.
Calculate ΔH for a reaction using standard enthalpies of formation: ΔH°rxn = Σ ΔHf°(products) − Σ ΔHf°(reactants). Enter each species, its ΔHf° (kJ/mol), its stoichiometric coefficient, and whether it is a product or reactant.
ΔH = Σ(bonds broken in reactants) − Σ(bonds formed in products). Enter bond type, bond enthalpy (kJ/mol), number of bonds, and whether it is in reactants (broken) or products (formed).
Enthalpy (H) is a thermodynamic quantity representing the total heat content of a system at constant pressure. The enthalpy change (ΔH) is the difference in enthalpy between products and reactants:
Exothermic reactions: Products have lower enthalpy than reactants. ΔH is negative. Heat is released to surroundings. Temperature of surroundings rises.
Endothermic reactions: Products have higher enthalpy than reactants. ΔH is positive. Heat is absorbed from surroundings. Temperature of surroundings falls.
Standard enthalpy values (ΔH°) are measured at 298 K (25°C) and 100 kPa (1 bar), with all species in their standard states.
| Symbol | Quantity | Unit |
|---|---|---|
| q | Heat energy transferred | J |
| m | Mass of solution | g |
| c | Specific heat capacity | J/g/°C |
| ΔT | Temperature change | °C |
100 cm³ of water, temperature rises 15.0°C
q = 100 × 4.18 × 15.0 = 6270 J = 6.27 kJ
If 0.0500 mol reacted: ΔH = −6.27 / 0.0500 = −125.4 kJ/mol (negative because exothermic)
Hess's Law states that the total enthalpy change for a chemical reaction is independent of the route taken, provided initial and final conditions are the same. This is a consequence of the conservation of energy.
ΔHf° values (kJ/mol): CH₄(g) = −74.8, O₂(g) = 0, CO₂(g) = −393.5, H₂O(l) = −285.8
ΔH°rxn = [1×(−393.5) + 2×(−285.8)] − [1×(−74.8) + 2×0]
= [−393.5 − 571.6] − [−74.8]
= −965.1 − (−74.8) = −890.3 kJ/mol
Note that the ΔHf° of any element in its standard state (e.g. O₂(g), Na(s), C(graphite)) is zero by definition.
Bond breaking is endothermic (+). Bond forming is exothermic (−). Mean bond enthalpies are averages and apply to gaseous species only.
Bonds broken: 1×H−H (436) + 1×Cl−Cl (243) = +679 kJ
Bonds formed: 2×H−Cl (432) = −864 kJ
ΔH = 679 − 864 = −185 kJ/mol
| Bond | kJ/mol | Bond | kJ/mol |
|---|---|---|---|
| H−H | 436 | C−C | 347 |
| H−O | 463 | C=C | 612 |
| H−N | 391 | C≡C | 838 |
| H−Cl | 432 | C−H | 413 |
| Cl−Cl | 243 | C=O | 743 |
| O=O | 498 | C−O | 358 |
| N≡N | 945 | N−H | 391 |
The enthalpy change when 1 mole of a compound is formed from its elements in their standard states at 298 K and 100 kPa.
| Compound | ΔHf° (kJ/mol) |
|---|---|
| CO₂(g) | −393.5 |
| H₂O(l) | −285.8 |
| H₂O(g) | −241.8 |
| CH₄(g) | −74.8 |
| C₂H₅OH(l) | −277.7 |
| NH₃(g) | −46.1 |
| HCl(g) | −92.3 |
The enthalpy change when 1 mole of substance is completely burned in excess oxygen at standard conditions. Always negative (combustion is exothermic).
| Fuel | ΔHc° (kJ/mol) |
|---|---|
| CH₄(g) methane | −890 |
| C₂H₅OH(l) ethanol | −1371 |
| C(s) graphite | −393.5 |
| H₂(g) | −286 |
A Born-Haber cycle is a Hess's Law cycle used to calculate the lattice enthalpy of an ionic compound, which cannot be measured directly. The cycle breaks down the formation of an ionic compound from its elements into individual steps, each with a measurable enthalpy change.
Solving: ΔHlattice = −411 − 107 − 496 − 122 − (−349) = −787 kJ/mol
Enthalpy change (ΔH) is the heat energy transferred at constant pressure during a chemical reaction. For exothermic reactions, ΔH is negative and energy is released to the surroundings (temperature rises). For endothermic reactions, ΔH is positive and energy is absorbed from the surroundings (temperature falls). The magnitude indicates how much energy per mole of reaction.
The calorimetry formula q = mcΔT calculates heat energy transferred: q is heat in joules, m is the mass of solution in grams, c is specific heat capacity (4.18 J/g/°C for water), and ΔT is the temperature change in °C. For an exothermic reaction, the solution temperature rises (ΔT positive, q positive for solution), so ΔH of reaction is negative — always remember to apply the sign convention.
Hess's Law states that the total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same. This follows from the first law of thermodynamics (conservation of energy). It allows ΔH to be calculated for reactions that cannot be measured directly by constructing a thermochemical cycle using known enthalpy values.
The standard enthalpy of formation (ΔHf°) is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions (298 K, 100 kPa). By definition, the ΔHf° of any element in its standard state is zero (e.g. O₂(g), Na(s), C(graphite) all have ΔHf° = 0).
ΔH(reaction) = Σ(bond enthalpies of bonds broken in reactants) − Σ(bond enthalpies of bonds formed in products). Bond breaking is endothermic (+); bond forming is exothermic (−). This method uses mean bond enthalpies and is only an approximation because mean values do not account for the specific molecular environment. The method applies to reactions involving gaseous species.
A Born-Haber cycle is a Hess's Law cycle that allows the calculation of lattice enthalpy of an ionic compound, which cannot be measured directly. It links the standard enthalpy of formation to atomisation enthalpies, ionisation energies, and electron affinities. Using Hess's Law, the unknown lattice enthalpy can be calculated from the other known values in the cycle.
ΔHf° (standard enthalpy of formation) is the enthalpy change when exactly one mole of compound is formed from its elements in their standard states. ΔHc° (standard enthalpy of combustion) is the enthalpy change when exactly one mole of substance is completely burned in excess oxygen under standard conditions. Both are defined per mole and at standard conditions (298 K, 100 kPa), and combustion enthalpies are always negative.