Energy cycles (Hess's law)

The average bond enthalpy is the energy needed to break one mole of a particular covalent bond in the gaseous state, averaged over many molecules. The values are listed in Section 12 of the data booklet (measured at 298.15 K).

Two facts drive every calculation:

• Breaking bonds is endothermic ($+$): energy must be put in.
• Forming bonds is exothermic ($-$): energy is released.

To estimate a reaction enthalpy, break every bond in the reactants and make every bond in the products:

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

This equation is not printed in the booklet - you must remember it. Work through the equation systematically: draw the structures, count each bond type, multiply by its bond enthalpy, then subtract.

Because the values are averages taken across different molecules, answers are only approximate. They also assume all species are gases; if a reactant or product is a liquid or solid, the bond-enthalpy method ignores the extra enthalpy involved in changing state, so it is less accurate than a formation-enthalpy calculation.

Hess's law states that the enthalpy change of a reaction is the same whatever route is taken from reactants to products, as long as the start and end conditions match. This works because enthalpy is a state function - it depends only on the current state, not on how you got there.

The practical use is to find a $\Delta H$ you cannot measure directly by going via an alternative route whose steps are known:

$$\Delta H_{\text{reaction}} = \Delta H_1 + \Delta H_2 + \dots$$

Drawing an energy cycle helps:

• Put the target reaction as the top (or direct) arrow.
• Put the known route as a series of arrows reaching the same products.
• The two routes must give the same total $\Delta H$.

Key rule: if you reverse a step, you reverse the sign of its $\Delta H$; if you multiply a reaction, you multiply its $\Delta H$ by the same factor.

A classic application is finding the enthalpy of formation of a compound (e.g. converting between C, $\text{CO}$ and $\text{CO}_2$) when one of the steps cannot be carried out cleanly on its own.

Two standard enthalpies are tabulated for use with Hess's law.

Standard enthalpy of formation, $\Delta H_f^{\ominus}$ is the enthalpy change when one mole of a substance forms from its elements in their standard states under standard conditions. By definition, $\Delta H_f^{\ominus}$ of an element in its standard state is zero (e.g. $\text{O}_2(g)$, $\text{C(graphite)}$). Values are in Section 13.

Standard enthalpy of combustion, $\Delta H_c^{\ominus}$ is the enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions. Values are in Section 14. Combustion is exothermic, so these are negative.

Standard conditions in the booklet use the standard state symbol $^{\ominus}$ and a stated temperature (usually 298 K) and pressure of 100 kPa.

Why two methods? Some substances are hard to form directly from elements but easy to burn, and vice versa. Choosing the table that matches your available data lets you reach the same answer by Hess's law - which is the point of the next subtopic.

The data booklet gives two ready-made Hess's-law equations.

From formation enthalpies (booklet):

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

From combustion enthalpies (booklet):

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

Notice the order is reversed between the two: formation is products minus reactants, combustion is reactants minus products. This is the single most common source of sign errors.

Method:

• Write and balance the equation first.
• Multiply each tabulated value by its stoichiometric coefficient.
• Apply the correct $\Sigma$ pattern for the data you are given.
• Set $\Delta H_f^{\ominus} = 0$ for any element in its standard state.

Think of the formation version as a triangle: elements at the bottom, reactants and products at the top. The direct route (reactants $\rightarrow$ products) equals "down to elements, then up to products", which is why reactant formation values are subtracted.

A Born-Haber cycle is an energy cycle for an ionic compound that lets you find the lattice enthalpy, which cannot be measured directly. It applies Hess's law around a loop with these steps:

• Atomisation of each element (e.g. $\text{Na(s)} \rightarrow \text{Na(g)}$, and $\tfrac{1}{2}\text{Cl}_2(g) \rightarrow \text{Cl(g)}$) - all endothermic.
• Ionisation energy to form cations (endothermic).
• Electron affinity to form anions (usually exothermic for the first electron).
• Lattice enthalpy - gaseous ions combining into the solid lattice (the reverse, lattice formation, is exothermic).

These must equal the enthalpy of formation of the ionic solid:

$$\Delta H_f^{\ominus} = \Delta H_{\text{atom}} + \Delta H_{\text{IE}} + \Delta H_{\text{EA}} - \Delta H_{\text{lattice}}$$

Tips:

• Track signs and directions of arrows carefully - reverse an arrow, reverse the sign.
• Multiply atomisation/ionisation by the number of atoms or charges (e.g. $\text{Mg}^{2+}$ needs two ionisation energies).
• Lattice enthalpy increases with higher ionic charge and smaller ionic radius (stronger electrostatic attraction).