3.1.4 Energetics

Keep the sign tied to the system: exothermic reactions transfer heat to the surroundings and have ΔH < 0, while endothermic reactions take in heat and have ΔH > 0. Don’t decide the sign just from whether the surroundings feel hot or cold.

On an enthalpy profile, products below reactants means exothermic; products above reactants means endothermic. The vertical difference is ΔH.

With ΔH, remember that it is the heat-energy change for the system at constant pressure. Decide whether the reaction released or absorbed heat, then give the enthalpy change the correct sign.

Keep ΔH attached to the chemical equation as written. Reversing the reaction reverses the sign, and multiplying the whole equation multiplies the enthalpy change by the same factor.

A standard enthalpy value refers to substances in their standard states at 100 kPa and the stated temperature, often 298 K. Do not invent 298 K if a different temperature is given.

State symbols matter because “standard state” means the stable physical form of each substance under the stated standard conditions. A value for H₂O(l), for example, is not automatically interchangeable with one for H₂O(g).

For standard enthalpy of formation, form one mole of the compound from its elements in their standard states. For standard enthalpy of combustion, burn one mole of the substance completely in oxygen under standard conditions.

I’d learn the “one mole” and “standard states” wording exactly. The chemical equation must also represent that definition, which may require fractional coefficients for a formation equation.

For calorimetry, q = mcΔT gives the heat transferred to or from the material whose mass and specific heat capacity you are using. Keep m in the units expected by c, and take ΔT as final temperature minus initial temperature before dealing with the sign of the enthalpy change.

If water warms by 6.5 K, ΔT is +6.5 K for the water and q for the water is positive. For an exothermic reaction, the reaction enthalpy has the opposite sign because the reaction lost that heat. A value of c will be supplied when you need one; the method is what you need to know.

When you turn calorimetry data into a molar enthalpy change, work out q for the amount that reacted first. Convert q to kJ if needed, divide by the appropriate amount in moles and then give ΔH the correct sign.

If the temperature of the solution rises, the reaction released heat, so the reaction ΔH is negative even though the calculated q for the solution is positive.

In multi-step calorimetry, use q = mcΔT to find the heat first, then connect that heat to moles of reaction. A common route is q in J → convert to kJ → divide by reacting moles to obtain an enthalpy change in kJ mol⁻¹.

Keep the stages separate on paper. Find q for the measured temperature change, convert J to kJ if needed, then divide by the moles that actually underwent the reaction and apply the reaction sign.

Hess’s law says the enthalpy change for a reaction is independent of the route taken, provided the start and end states are the same. That lets you replace an awkward route with steps whose enthalpies are known.

When you reverse an equation, reverse the sign of ΔH. When you multiply an equation, multiply ΔH by the same factor. Keep those operations attached to the chemical equations.

For formation data, a very efficient route is ΔHᵣ = ΣΔHf(products) − ΣΔHf(reactants), including coefficients. Combustion cycles are often arranged the other way round, so follow the cycle arrows and assign each sign from the direction you travel. A memorised sign pattern is much less reliable.

If you draw a cycle, write the target reaction first and make sure every route connects exactly the same start and end states. The algebra should then agree with the arrows.

Treat a mean bond enthalpy as an averaged gas-phase bond-breaking value. The exact enthalpy of the same nominal bond changes with its molecular environment, so the mean value is necessarily an approximation.

That averaging matters when you use bond enthalpies to estimate a reaction enthalpy. The calculated value can differ from an experimental ΔH because the tabulated bond enthalpies are mean values taken from different gaseous compounds.

For the definition, I’d learn this precisely: mean bond enthalpy is the mean enthalpy change required to break one mole of a specified covalent bond in gaseous molecules by homolytic fission.

The word mean matters because the same bond type has slightly different enthalpies in different compounds, so the tabulated value is an average. The definition also specifies gaseous molecules and homolytic bond breaking.

Estimate reaction enthalpy with Σ(bond enthalpies broken) − Σ(bond enthalpies formed). Breaking bonds costs energy; forming bonds releases energy.

Count every bond that changes, including coefficients. Mean bond enthalpies are gas-phase averages, so the result is an estimate for the particular molecules. It is not an exact bond-energy calculation for their specific environments.

A calculation from mean bond enthalpies uses averaged gas-phase values, whereas formation or combustion data refer to specific substances and states. The results therefore need not agree exactly.

State differences are one reason for the mismatch. Mean bond enthalpies describe bonds in gaseous molecules, while formation/combustion data can include liquids or solids and the energetic effects associated with those actual states.