Energy cycles in reactions

The energy needed to **break one mole** of a particular bond in the **gaseous** state (always a positive value).

**Endothermic** — breaking a bond always **costs** (absorbs) energy.

**Exothermic** — forming a bond always **releases** energy.

$\Delta H = \Sigma(\text{bonds broken}) - \Sigma(\text{bonds made})$.

The reaction is **exothermic** — more energy was released making bonds than was used breaking them.

The reaction is **endothermic** — breaking bonds cost more energy than was released making them.

A bond (e.g. C–H) exists in many molecules with slightly different strengths, so the booklet gives an **average**; ΔH is therefore an **estimate**.

Only when **all species are gaseous**, because bond enthalpy is defined for the gaseous state.

Only the bonds that **break or form** — unchanged bonds (spectator bonds) cancel out.

**Higher** — a larger bond enthalpy means a stronger bond that needs more energy to break.

Because the bond enthalpies are **averages**, so the calculated ΔH is only an **estimate**.

The total enthalpy change of a reaction is the **same** whatever route is taken, because ΔH depends only on the initial and final states.

A property that depends only on the **current state** of the system, not on the path taken to reach it (enthalpy is one).

Because enthalpy is a **state function**, so ΔH is **path-independent** — you can add up the steps of an alternative route.

Its **sign is reversed** (the magnitude stays the same).

ΔH is **doubled** — multiply ΔH by the same factor as the equation.

$\Delta H^{\ominus} = \Sigma\,\Delta H_{f}^{\ominus}(\text{products}) - \Sigma\,\Delta H_{f}^{\ominus}(\text{reactants})$.

**Zero** by definition (e.g. O_{2}(g), C(s) graphite).

**With** an arrow → **add** its ΔH; **against** it (reverse) → **subtract** its ΔH (flip the sign).

Forgetting to **multiply** a step by the number of moles in the target equation.

To find a ΔH that **cannot be measured directly** (e.g. the reaction is too slow or has side reactions).

The ΔHf equation **is** a Hess cycle — going down to the elements (reverse ΔHf of reactants) and up to the products (ΔHf of products).

The enthalpy change when **1 mol** of a compound forms from its **elements in their standard states** (100 kPa, stated T).

The enthalpy change when **1 mol** of a substance is **completely burned in oxygen** under standard conditions; always **negative**.

**Zero** — e.g. O_{2}(g), N_{2}(g), C(graphite); there is nothing to form.

$\Delta H^{\ominus} = \sum \Delta H_{f}^{\ominus}(\text{products}) - \sum \Delta H_{f}^{\ominus}(\text{reactants})$.

$\Delta H^{\ominus} = \sum \Delta H_{c}^{\ominus}(\text{reactants}) - \sum \Delta H_{c}^{\ominus}(\text{products})$.

Both reactants and products burn down to the **same products** (CO_{2} + H_{2}O), so the Hess cycle runs the other way → **reactants − products**.

A pressure of **100 kPa** and a stated temperature (usually **298 K**), with all substances in their standard states.

Enthalpy is a **state function** — ΔH depends only on the start and end states, so a 'paper' Hess route gives the same answer as experiment.

Forgetting to multiply each value by its **stoichiometric coefficient** (e.g. 2 H_{2}O) or forgetting an **element is zero**.

**Negative** (exothermic) — a quick check that you used the correct rule.

**kJ mol⁻¹** (kilojoules per mole).