ΔH < 0 (heat released to the surroundings).ΔH > 0 (heat absorbed from the surroundings).| Symbol | Name | Definition (standard state) | Typical example |
|---|---|---|---|
| ΔH°f | Standard enthalpy of formation | Enthalpy change when 1 mol of a compound is formed from its constituent elements in their standard states. | H₂(g) + ½ O₂(g) → H₂O(l) |
| ΔH°c | Standard enthalpy of combustion | Enthalpy change when 1 mol of a substance burns completely in excess O₂ under standard conditions. | CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l) |
| ΔH°neut | Standard enthalpy of neutralisation | Enthalpy change when 1 mol of H⁺ reacts with 1 mol of OH⁻ to give liquid water (standard conditions). | HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) |
| ΔH°r | Standard enthalpy of reaction | Overall enthalpy change for any reaction, calculated from ΔH°f values of reactants and products. | Any balanced equation, e.g. the combustion of propane. |
For the reaction HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) the measured heat released is –57 kJ per mole of water formed. Hence, ΔH°neut = –57 kJ mol⁻¹.
ΔHrxn ≈ ∑ Ebroken − ∑ Eformed
| Bond | Bond energy (kJ mol⁻¹) | Bond | Bond energy (kJ mol⁻¹) |
|---|---|---|---|
| C–C | 347 | C=O | 799 |
| C–H | 413 | O–H | 463 |
| H–H | 436 | N≡N | 945 |
| Cl–Cl | 242 | O=O | 498 |
Because enthalpy is a state function, the total ΔH for a reaction is independent of the pathway taken. This allows us to combine known enthalpy changes to obtain the ΔH for a reaction that is difficult to measure directly.
Mathematical form:
ΔHoverall = ∑ ΔHi
Steps to apply Hess’s Law
At constant pressure the heat exchanged equals the enthalpy change of the reaction:
qrxn = −qsolution = −m c ΔT
To obtain the molar enthalpy change:
ΔH = qrxn / n (kJ mol⁻¹)
Ccal) by a calibration experiment. Then include it:
qtotal = CcalΔT + m c ΔT
At constant volume the heat exchanged equals the change in internal energy (ΔU):
qrxn = −CcalΔT = ΔU
For reactions involving gases, ΔH can be obtained from ΔU using:
ΔH = ΔU + ΔngRT
whereΔng is the change in the number of moles of gas.
ΔH°f is the enthalpy change for forming 1 mol of a compound from its elements in their standard states. By definition, ΔH°f = 0 for any element in its standard state.
| Compound | ΔH°f (kJ mol⁻¹) | Compound | ΔH°f (kJ mol⁻¹) |
|---|---|---|---|
| H₂O(l) | -285.8 | CO₂(g) | -393.5 |
| CH₄(g) | -74.8 | NH₃(g) | -46.1 |
| C₂H₆(g) | -84.0 | SO₂(g) | -296.8 |
| H₂(g) | 0 | O₂(g) | 0 |
Using ΔH°f values, the standard enthalpy change for any reaction is:
ΔH°rxn = ∑ ΔH°f, products − ∑ ΔH°f, reactants
Reaction (standard conditions)
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l)
ΔH°f values (kJ mol⁻¹)
Calculation
\[ \Delta H^\circ_{\text{comb}} = [3(-393.5) + 4(-285.8)] - [(-104.7) + 5(0)] \] \[ \Delta H^\circ_{\text{comb}} = (-1180.5 - 1143.2) - (-104.7) = -2323.7 + 104.7 = -2219.0\ \text{kJ mol}^{-1} \]The combustion of one mole of propane releases 2219 kJ** under standard conditions.
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