Conservation of energy in a reacting system

The law of conservation of energy is a scientific law that states that the total energy of an isolated system remains constant, although energy may be transferred or converted from one form to another. In chemistry, this idea sits behind energy cycles: once we account for the energy absorbed and released in the separate steps, we can predict the overall energy change.

A chemical reaction rearranges atoms. Some bonds in the reactants have to be broken, and new bonds in the products have to form. Breaking bonds is endothermic because energy is needed to separate bonded atoms. Forming bonds is exothermic because energy is released when attractions between nuclei and shared electrons are established.

Bond enthalpy and average values

A bond enthalpy is an enthalpy change required to break one mole of a specified covalent bond in gaseous molecules by homolytic fission under standard conditions. Its unit is kJ mol⁻¹. Homolytic fission is a bond-breaking process in which the two bonding electrons are shared equally between the atoms. A radical is a species that contains an unpaired electron, usually shown with a dot, such as H•.

For example, the H–H bond enthalpy corresponds to:

[ ext{H}_2(g) \rightarrow 2 ext{H}\bullet(g) \qquad \Delta H = +436 ext{ kJ mol}^{-1} ]

where (\Delta H) is the molar enthalpy change of the process (kJ mol⁻¹). The positive sign matters: it tells you that energy has been absorbed.

An average bond enthalpy is a mean bond enthalpy obtained from the same type of bond measured in a range of different gaseous compounds. Data-booklet bond enthalpies are average values. A C–H bond in methane does not have exactly the same electronic environment as a C–H bond next to an oxygen atom, a double bond, or a halogen. Even in a single molecule, bonds written as the same type are not necessarily in identical surroundings.

That is why bond enthalpy calculations give estimates. They usually differ from experimental enthalpy changes because the data are averaged, and because bond enthalpy calculations treat the species as gaseous covalent molecules; they do not properly include intermolecular forces in liquids and solids.

Calculating enthalpy change from bond enthalpies

The calculation uses a simple bit of energy accounting: add the energy needed to break bonds, then subtract the energy released when bonds form.

[ \Delta H = \sum ext{BE}{ ext{broken}} - \sum ext{BE}{ ext{formed}} ]

where BE is the bond enthalpy of a bond counted in the balanced equation (kJ mol⁻¹). Average bond enthalpy values are supplied in the data booklet.

In practice, draw full structural formulae, count every bond broken in the reactants, count every bond formed in the products, multiply by the correct bond enthalpies, then subtract formed from broken. The balanced equation matters because the answer is for the reaction as written.

For example, in an addition reaction such as ethene reacting with hydrogen bromide, the structures reveal the hidden bookkeeping: a C=C bond and an H–Br bond are replaced by a C–C bond and a C–Br bond, with one extra C–H bond in the product. A common mistake is to count atoms instead of bonds.

Stability, bond length and polarity

Bond enthalpy is linked to bond strength. In general, shorter bonds are stronger because the bonded nuclei are closer to the shared electron density, so separating them takes more energy. Bond polarity can matter too: if a bond has greater ionic character, electrostatic attraction may increase bond strength, although size and overlap are also important. Don’t turn it into a single-rule topic; bond length, bond order, atom size and polarity all affect the value.

This links neatly to carbon–halogen bonds in nucleophilic substitution. The C–I bond is weaker and longer than the C–Cl bond, so it is easier to break; other factors being comparable, halogenoalkanes with weaker C–X bonds react faster in substitution. That is the chemical reason behind the usual reactivity trend from chloroalkanes to iodoalkanes.

Why the pathway does not matter

Hess’s law says that the enthalpy change for a reaction is the same whichever route is taken, as long as the initial and final states are identical. It follows directly from conservation of energy.

A state function is a property whose change depends only on the initial and final states, not on the route between them. Enthalpy is one of these. So if reactants A form products B directly, or go via an intermediate C, the overall enthalpy change stays the same.

[ \Delta H_x = \Delta H_y + \Delta H_z ]

where (\Delta H_x) is the molar enthalpy change for the direct route A → B (kJ mol⁻¹), (\Delta H_y) is the molar enthalpy change for A → C (kJ mol⁻¹), and (\Delta H_z) is the molar enthalpy change for C → B (kJ mol⁻¹).

Using Hess’s law in calculations

You’ll usually meet Hess’s law in two standard forms. They look different on the page, but both do the same energy accounting.

In the summation of equations method, known thermochemical equations are rearranged until they add up to the target equation. Reverse an equation, and you reverse the sign of its enthalpy change. Multiply an equation by a factor, and you multiply its enthalpy change by that same factor. Species on both sides cancel, just like terms in algebra.

In the enthalpy cycle method, the target reaction is drawn across the top, with both sides linked to a common set of substances underneath, often combustion products or elements in their standard states. Follow a route around the cycle. If you go against an arrow, change the sign of that enthalpy value.

It helps to write state symbols whenever the data depend on them. Forming H₂O(l), for example, is not the same enthalpy change as forming H₂O(g). Hess’s law only works cleanly when the substances and their states match exactly.

Experimental Hess cycles

Hess’s law helps when a reaction is hard to measure directly. A thermal decomposition, for example, may be too hot, too slow or too messy for simple calorimetry. Instead, two related reactions can be measured in a calorimeter, then their enthalpy changes can be combined in a cycle to work out the missing value.

When calorimetry data are used in a Hess calculation, the usual assumptions still matter. Heat loss to the surroundings, heat absorbed by the cup or thermometer, incomplete reaction, evaporation and uncertain extrapolation of temperature all affect the measured enthalpy changes. The cycle is exact in principle; the experimental numbers placed into it may not be.

Standard conditions and standard states

A standard state is the physical state and form in which a pure substance is most stable under standard conditions. In this syllabus, standard conditions are usually 298 K and 100 kPa. The standard-state detail matters: carbon as graphite and carbon as diamond count as different starting substances.

The symbol (\Delta H_c^\ominus) means standard enthalpy change of combustion, where (\Delta H_c^\ominus) is the molar enthalpy change when one mole of a substance burns completely in oxygen under standard conditions (kJ mol⁻¹). Combustion enthalpies are usually negative because complete combustion releases energy.

A standard enthalpy change of combustion is an enthalpy change when one mole of a substance in its standard state undergoes complete combustion in oxygen under standard conditions. For hydrocarbons and many oxygen-containing organic compounds, complete combustion gives CO₂(g) and H₂O(l), unless the question says otherwise.

For example, the standard combustion equation for butane is:

[ ext{C}4 ext{H}{10}(g) + \frac{13}{2} ext{O}_2(g) \rightarrow 4 ext{CO}_2(g) + 5 ext{H}_2 ext{O}(l) ]

Fractional coefficients are allowed in these definitions because the equation has to refer to exactly one mole of the substance being combusted.

Standard enthalpy of formation

The symbol (\Delta H_f^\ominus) means standard enthalpy change of formation, where (\Delta H_f^\ominus) is the molar enthalpy change when one mole of a compound forms from its elements in their standard states under standard conditions (kJ mol⁻¹).

A standard enthalpy change of formation is an enthalpy change when one mole of a substance is formed from its constituent elements in their standard states under standard conditions. For butane, the formation equation is:

[ 4 ext{C}(s, ext{ graphite}) + 5 ext{H}_2(g) \rightarrow ext{C}4 ext{H}{10}(g) ]

The standard enthalpy of formation of an element in its standard state is zero. Not because the element stores no energy, but because the enthalpy scale needs a reference point. Standard-state elements are chosen as that reference.

Key differences between standard enthalpy changes of combustion and formation.

An allotrope is a structural form of an element in the same physical state but with different bonding or arrangement of atoms. Diamond and graphite are allotropes of carbon, so their formation enthalpies are different. Graphite is the standard state of carbon, so (\Delta H_f^\ominus) for graphite is 0 kJ mol⁻¹; diamond is not the reference form, so its value is not zero.

Writing equations from definitions

For combustion, start with one mole of the substance plus oxygen, then balance to the complete-combustion products. For formation, start with the elements in their standard states and end with one mole of the compound. In both cases, include state symbols, since the data-booklet values depend on them.

Combustion and formation data are supplied in the data booklet. The skill isn't memorising the numbers. It's turning the definition into the correct balanced equation, then using the data consistently.

Formation data route

The symbol (\Delta H^\ominus) means the standard molar enthalpy change for the reaction as written, measured in kJ mol⁻¹. With formation data, Hess’s law compares the enthalpy needed to form the products from elements with the enthalpy needed to form the reactants from elements.

[ \Delta H^\ominus = \sum(\Delta H_f^\ominus ext{ products}) - \sum(\Delta H_f^\ominus ext{ reactants}) ]

Include the coefficients from the balanced equation in the sums. If the equation contains 3 mol of CO₂, multiply the CO₂ formation value by 3. Elements in their standard states contribute zero.

A good sign check is: products minus reactants for formation data. You’re comparing where the products sit on the enthalpy scale with where the reactants sit on that same scale.

Combustion data route

With combustion data, the common endpoint is complete combustion products. For many organic reactions, all carbon ends up as CO₂ and all hydrogen ends up as H₂O. Hess’s law then gives:

[ \Delta H^\ominus = \sum(\Delta H_c^\ominus ext{ reactants}) - \sum(\Delta H_c^\ominus ext{ products}) ]

Again, multiply each data-booklet value by its coefficient in the balanced equation. The order is now reactants minus products, the opposite of the formation expression. This is where careless errors tend to appear.

Choosing and checking the method

Use formation data when the data table gives (\Delta H_f^\ominus) values for the species in the reaction. Use combustion data when all the species can be linked to common combustion products. Both methods use Hess’s law; they just choose different common reference substances in the cycle.

The equations for formation and combustion data are in the data booklet, but the data-booklet equation won’t rescue poor chemistry. You still need the balanced equation, correct state symbols, and correct multiplication by stoichiometric coefficients.

Values calculated from tabulated standard enthalpies may still be approximate if the data are rounded or if the actual reaction conditions do not match the standard states. They are usually more reliable than average bond enthalpy estimates for reactions involving liquids and solids, because standard formation and combustion values refer to real substances in specified states rather than averaged gas-phase bonds.