Example 5.4. Using the heats of combustion to calculate h°f

Problem: The heats of combustion of H₂, graphite and acetylene are equal to –286 kJmol^1, –393.5 kJ mol^1 and –1301kJmol^1, respectively (the products are returned to 25 and 1 atm). Calculate the standard heat of formation of acetylene.

Solution: The equation of combustion of 1 mol of of H₂, graphite and acetylene are:

(1)	C(graphite) + O2(g) → CO2(g)	ΔH1° = –393.51 kJ
(2)	H2(g) + ½ O2(g) → H2O(l)	ΔH2° = –286 kJ
(3)	C2H2(g) + 5/2 O2(g) → 2 CO2(g)  + H2O(l)	ΔH3° = –1301kJ

These data should be used to calculate the heat effect for the reaction

(4)

2 C(graphite) + H2(g) → C2H2(g)

ΔH4° = ?

The heat of reaction can be calculated, using sums of the heats of combustion of reactants and the heat of combustion of acetylene

ΔH₄° = ΣΔH_comb°(reactants) – ΣΔH_comb°(products) = 2 ΔH₁° + ΔH₂°  ΔH₃°

(The same result can be obtained applying Hess’s law.)

Substituting heats of combustion we obtain

ΔH_f°(C₂H_2(g)) = 2(–393.5) + (–286) – (–1301) = 228 kJ

Answer: The standard heat of formation of C₂H_2(g) is 228 kJmol^1.

5.1.2 Bond Energy and Heat Effect of Reaction

Bond energy is the amount of energy, required to break 1 mol of chemical bonds. Thus, for H₂, the energy adsorbed in the following reaction is the bond energy.

H_2(g) → 2 H_(g)

The energy needed to break all the bonds in 1 mol of a chemical substance (either an elementary substance or a chemical compound) to form atoms is called the atomization energy (or enthalpy of atomization), denoted as ΔH_at. Its value is equal the sum of all the bond energies in a molecule. Thus, for water, the atomization energy adsorbed in a process

H₂O_(g) → O_(g) +2 H_(g)

Standard enthalpies of atomization (ΔH_at) have been determined for many elementary substances (some examples are given in table 5.1).

Table. 5.1. Standard enthalpies of atomization (ΔH_at)

Elementary substance	ΔHat, kJ/mol
H2(g)	436.0
Li	161
Be	327
B	555
C(graphite)	715
Na	108
Si	454

Example 5.5. Calculating bond energies from thermodynamic data

Problem: The standard heat of formation of methane, CH₄ is –74.9 kJmol^1. The H-–H bond energy is 436 kJmol^1; standard enthalpy of atomization for carbon is given in the table 5.1. Calculate the C–H bond energy.

Solution: The equation of methane formation is

C(graphite) + 2 H2(g) → CH4(g)

ΔH1° = –74.9 kJ

We can envision an alternative path from elementary substances to the compound that follows a series of successive reactions:

(1)	C(graphite) → C(g)	ΔH1° = 715 kJ
(2)	2 H2(g) → 4 H(g)	ΔH2° = 872 kJ
(3)	C(g) + 4 H(g) → CH4(g)	ΔH3° = –ΔHat° for CH4

The heat of methane formation is equal to the sum of the heats reactions (1), (2) and (3).

ΔH_f° = ΔH₁° + ΔH₂° + ΔH₃°

The only unknown quantity is ΔH₃°, which is negative of the atomization energy (negative, because step 3 involves the formation of chemical bonds). Substituting and solving for the atomization energy we obtain

ΔH_at° = ΔH₁° + ΔH₂° – ΔH_f° = 715 + 872 – (–74.9) = 1662 kJ

This quantity is the total amount of energy needed to break all four C–H bonds in 1 mol of CH₄. Division by 4 gives average bond energy of 415 kJmol^1.

Answer: The average C–H bond energy is 415 kJmol^1.

Enthalpy of ionization. To calculate energy of crystalline lattice for ionic solid, enthalpies of ionization and electron affinities are used. The enthalpy of ionization is the enthalpy change that accompanies formation of 1 mole of cations from isolated atoms or other cations in a ground state. The electron affinity is the energy that is realized or adsorbed when electron(s) are added to 1 mole of isolated atoms or anions in a ground state.