Standard enthalpies of formation, Hess's law and an introduction to chemical bonding

Last week we measured $\Delta H$ with a calorimeter. Measuring every reaction is not practical; this week you will learn to calculate it from tables. There are two ways: a one-line formula with enthalpies of formation and adding equations with Hess’s law. Both are guaranteed questions on the midterm and the final. The second half of the page is the introduction to a new section: why and how atoms form bonds.

Golden rules:

1. ΔH°f = 0 for elements in their standard states. $\mathrm{O_2}$, $\mathrm{N_2}$, graphite are zero.
2. Products minus reactants, each multiplied by its coefficient.
3. Hess: if you reverse an equation the sign changes, if you multiply it the value changes.
4. A Lewis structure starts with counting the total valence electrons.

Standard state and the standard enthalpy of formation

The absolute value of enthalpy cannot be measured, only its change. That is why a reference point is chosen: the standard state. It is the most stable state of the pure substance at 1 bar pressure and the specified temperature (25 °C in tables).

Standard enthalpy of formation ($\Delta H^\circ_f$): the enthalpy change of the reaction in which 1 mol of a compound forms from its elements in their standard states.

$\mathrm{C(graphite) + 2H_2(g) + \tfrac12 O_2(g) \rightarrow CH_3OH(l)} \qquad \Delta H^\circ_f = -238.7\ \text{kJ/mol}$

A checklist when writing a formation equation:

• Exactly 1 mol of product on the right (there can be fractions on the left).
• Only elements on the left, in their standard states: $\mathrm{H_2(g)}$, $\mathrm{O_2(g)}$, $\mathrm{Br_2(l)}$, C(graphite), Na(s).
• The enthalpy of formation of an element itself is 0, because “forming from itself” is not a change.

A negative $\Delta H^\circ_f$ means the compound is more stable (lower in energy) than its elements. The 44.0 kJ/mol difference between $\mathrm{H_2O(l)}$ and $\mathrm{H_2O(g)}$ is the enthalpy of vaporization.

The enthalpy of reaction from enthalpies of formation

Do every reaction as a thought experiment in two steps: break the reactants down into their elements, and build the products from the elements. The result:

$\Delta H^\circ_{\text{rxn}} = \sum n_p\,\Delta H^\circ_f(\text{products}) - \sum n_r\,\Delta H^\circ_f(\text{reactants})$

Example 1. The oxidation of ammonia (the first step of nitric acid production):

$\mathrm{4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g)}$

products:   4(+90.25) + 6(−241.8) = 361.0 − 1450.8 = −1089.8 kJ
reactants:  4(−46.11) + 5(0)      = −184.4 kJ
ΔH° = −1089.8 − (−184.4) = −905.4 kJ

Example 2 (food energy). Glucose combustion $\mathrm{C_6H_{12}O_6(s) + 6O_2 \rightarrow 6CO_2 + 6H_2O(l)}$, $\Delta H^\circ_f(\text{glucose}) = -1273.3$ kJ/mol:

$\Delta H^\circ = [6(-393.5) + 6(-285.8)] - (-1273.3) = -2803\ \text{kJ/mol}$

Per gram, $-2803/180.16 = -15.6$ kJ/g. It agrees with the value in last week’s bomb calorimeter question.

PITFALL · the state of water. If the question says $\mathrm{H_2O(g)}$ use −241.8, if it says $\mathrm{H_2O(l)}$ use −285.8. In tables of enthalpies of combustion, liquid water is generally assumed.

PITFALL · forgetting to multiply by the coefficient. Table values are per mole. If there are 4 NO in the equation, you multiply by 4. This is the most common way to lose points.

The formula also works in reverse: if the ΔH of the reaction and the other ΔH°f values are known, a single unknown ΔH°f can be solved (finding the enthalpy of formation from the enthalpy of combustion).

Hess’s law

Since enthalpy is a state function, the total $\Delta H$ is the same whether the reaction takes place in one step or several. Hess’s law: if a reaction can be written as the sum of other reactions, its $\Delta H$ is the sum of their $\Delta H$ values.

The strategy:

1. Write the target equation.
2. For each species in the target, if it appears in only one of the given equations, catch it from there. Reverse or multiply that equation according to the side of the species in the target and its coefficient.
3. Apply the operation done to the ΔH as well.
4. Add them, and cancel the intermediate species that appear on both sides. Check that you have obtained the target.

Example 3. The target: $\mathrm{S(s) + \tfrac32 O_2(g) \rightarrow SO_3(g)}$. Given:

$\text{(a) } \mathrm{S + O_2 \rightarrow SO_2} \quad \Delta H = -296.8\ \text{kJ} \qquad \text{(b) } \mathrm{2SO_2 + O_2 \rightarrow 2SO_3} \quad \Delta H = -197.8\ \text{kJ}$

S is on the left in the target, 1 of it → (a) as is:   S + O2 → SO2              −296.8
SO3 is on the right in the target, 1 of it → (b) × 1/2:  SO2 + 1/2 O2 → SO3       −98.9
add (SO2 cancels):                                    S + 3/2 O2 → SO3          −395.7 kJ

The result is the same as $\Delta H^\circ_f(\mathrm{SO_3})$ in the table; as it should be.

TIP · start with the species that appears only once. Species that appear in many equations, like O₂, sort themselves out; leave them for the end and just check.

Introduction to chemical bonding

The topic of the next section is molecular shapes and polarity. It all rests on one question: how do atoms share or transfer their electrons?

Valence electrons and Lewis symbols

The electrons that take part in bonding are the valence electrons in the outermost shell. For main group elements their number is read from the group number (groups 1-2: 1-2, groups 13-18: 3-8). In a Lewis symbol, each valence electron is placed as a dot around the element symbol.

Octet rule: atoms tend to gather 8 valence electrons (a noble gas configuration) around them by bonding. For H this number is 2.

Ionic bonds

A metal (low ionization energy) gives up electrons, and a nonmetal (high electron affinity) takes them. The oppositely charged ions that form are held together by electrostatic attraction: an ionic bond. In the formula the charges must add up to zero:

Na· + ·Cl: (7 dots)  →  Na+ [Cl]−          NaCl
Ca (gives 2 e−) + 2 F (each takes one)  →  Ca2+ 2F−   CaF2
Al3+ and O2−  →  common multiple 6  →  Al2O3

Ionic solids have no molecules; each ion sits in a crystal lattice surrounded by oppositely charged neighbors. The energy that holds the lattice together is called the lattice energy: it increases as the ion charges get larger and the ions get smaller.

Covalent bonds

Between two nonmetals, electron transfer is not energetically favorable; instead they share electrons. Each shared pair is a covalent bond and is shown with a line.

• Bonding pair: a pair shared between two atoms.
• Lone pair: a pair that belongs to a single atom and does not take part in bonding.
• A single bond contains 2, a double bond 4, a triple bond 6 electrons. As the number of bonds increases, the bond gets shorter and stronger.

Simple examples: in $\mathrm{H_2O}$, O is bonded to two H with single bonds and has 2 lone pairs. In $\mathrm{NH_3}$, N carries three bonds and 1 lone pair. In $\mathrm{N_2}$ there is a triple bond: N≡N.

The first step in writing a Lewis structure is always to count the total valence electrons; for anions add as many electrons as the charge, for cations subtract. The detailed recipe, formal charge and resonance are next week’s topic.

Common mistakes