Example 5.2. Applying Hess’s law by combining thermochemical equations

Problem: Given the following thermochemical equations

(1)	3 Mg(s) + 2 NH3(g) = Mg3N2(s) + 3 H2(g)	ΔH1° = – 371 kJ
(2)	NH3(g) = ½ N2(g) + 3/2 H2(g)	ΔH2° = 46 kJ

Calculate the heat effect for the reaction

(3)

3 Mg(s) + N2(g) = Mg3N2(s)

ΔH3° = ?

Analysis: To solve this problem, we have to combine equations (1) and (2) in such a way, that when they are added, everything cancels except the formulas in equation (3). (Let’s call this the target equation). Getting started it is better to look for formulas that appear only once in given equations (for example, Mg_(s) only appears in equation (1)). When a chemical formula appears in several equations, usually it isn’t obvious how the equations should be arranged to get this formula cancelled.

Solution: In the target equation 3Mg_(s) is on the left, so we use equations (1) without changes. In the target equation there are 1 mol N_2(g) on the left, but in equation (2) it is on the right with a coefficient ½. Therefore we have to reverse equation (2) and multiply its coefficients by 2, which gives equation (4) below. Notice, that we also multiply the ΔH° for equations (2) by 2 and also change its algebraic sign to get ΔH° for equation (4)

(4)

N2(g) + 3 H2(g) = 2 NH3(g)

ΔH° = – 92 kJ

Adding equations (1) and (4) gives

3 Mg_(s) + 2 NH_3(g) + N_2(g) + 3 H_2(g) = Mg₃N_2(s) + 3 H_2(g) + 2 NH_3(g)

Cancelling substances, that are the same on both sides, we get

3 Mg_(s) + N_2(g) = Mg₃N_2(s)

which is the target equation. Since it is obtained by adding of equations (1) and (4), its H °_r

(ΔH₃°) is obtained by adding of ΔH₁° and ΔH₄°.

ΔH₃° = ΔH₁° + ΔH₄° (or ΔH₃° = ΔH₁° + (- 2 ΔH₂°))

ΔH₃° = – 371 kJ + (– 92 kJ) = – 463 kJ

Answer: The enthalpy change for reaction (3) is therefore – 463 kJ

The standard enthalpy change for a reaction can be calculated, using sums of the standard enthalpies of formations of all products and reactants. Each ΔH_f ° should be multiplied by the corresponding coefficient in the equation:

H°_reaction = ΣΔH_f °_products – ΣΔH_f °_reactants

The above definition allows one to predict the enthalpy change of any reaction without knowing any more than the standard enthalpies of formation of the products and reactants.

Example 5.3. Calculating h °r from standard enthalpies of formation

Problem: Carbon monoxide reduces Fe₂O₃ to iron

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

Calculate H °_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

H °_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

H °_r = 3ΔH_f °(CO₂) – (ΔH_f°(Fe₂O₃) + 3 ΔH_f°(CO)) = 3(–393.5) – (–822.2 + 3(–110.5)) = – 26.8 kJ

Answer: The H °_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 (H for dissolution of 1 mol substance in sufficient amount of solvent) and heat of combustion (H 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):

H°_r = ΣΔH_comb°(reactants) – ΣΔH_comb°(products)