Example 5.3. Calculating DH °r from standard enthalpies of formation

Problem:Carbon monoxide reducesFe₂O₃ to iron

Fe₂O_3(s) + 3 CO_(g) = 2 Fe_(s) + 3 CO_2(g)

Calculate DH °_r for the reaction from standard enthalpies of formation of the reactants and products. ΔH_f°(Fe₂O₃) = –822.2 kJ×mol^-1, ΔH_f°(CO) = –110.5 kJ×mol^-1 and ΔH_f°(CO₂) = –393.5 kJ×mol^-1.

Analysis: We will use the equation

DH °_reaction = ΣΔH_f °_products – ΣΔH_f °_reactants

The coefficients in the chemical equation specify the amounts of matter of each substance. So to get the total amount of energy contributed by each reactant or product we multiply its ΔH_f° by its coefficient in the equation. As iron is an elementary substance, stable under standard conditions, ΔH_f°(Fe) = 0.

Solution: for the reaction given

DH °_r = 3ΔH_f °(CO₂) – (ΔH_f°(Fe₂O₃) + 3 ΔH_f°(CO)) = 3×(–393.5) – (–822.2 + 3×(–110.5)) = – 26.8 kJ

Answer: The DH °_r for the reaction is – 26.8 kJ.

In addition to heats of formation there are other thermodynamical characteristics of compounds. The examples are heat of dissolution (DH for dissolution of 1 mol substance in sufficient amount of solvent) and heat of combustion (DH for reaction of 1 mol substance with excess of O₂). Heats of dissolution and heats of combustion are used in calculations.

The heat of reaction can be calculated, using sums of the heats of combustion of all products and reactants (each ΔH_comb° should be multiplied by the corresponding coefficient in the equation):

DH°_r = ΣΔH_comb°(reactants) – ΣΔH_comb°(products)

Example 5.4. Using the heats of combustion to calculate DH°_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:

These data should be used to calculate the heat effectfor the reaction

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(orenthalpy 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°)

Date: 2015-12-11; view: 721