5.7: Enthalpy Calculations

Hess's Law And Enthalpy of Reaction

Exercise \(\PageIndex{1}\)

What is the overall chemical equation that results from the sum of the given steps?

2 C(s) + 2 H₂O(g) → 2 CO(g) + 2 H₂(g)

CO(g) + H₂O(g) → CO₂(g) + H₂(g)

CO(g) + 3 H₂(g) → CH₄(g) + H₂O(g)

Answer

2 C(s) + 2 H₂O(g) → CO₂(g) + CH₄(g)

Exercise \(\PageIndex{2}\)

Determine the heat of evaporation of carbon disulfide,
CS₂(l) → CS₂(g)
given the enthalpies of reaction below.

C(s) + 2 S(s) → CS₂(l)                         Δ_rH° = +89.4 kJ/mol-rxn

C(s) + 2 S(s) → CS₂(g)                        Δ_rH° = +116.7 kJ/mol-rxn

Answer

+27.3 kJ

Reverse (1) and add to (2)                    ΔH_rxn = -ΔH₁ + ΔH₂

(1) CS₂(l) → C(s) + 2 S(s)                               = -89.4 kJ/mol + 116.7 kJ/mol = 27.3 kJ/mol

(2) C(s) + 2 S(s) → CS₂(g) 

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CS₂ (l) → CS₂ (g)

 

Enthalpy of Formation

Exercise \(\PageIndex{3}\)

Determine the standard enthalpy of formation of Fe₂O₃(s) given the thermochemical equations below.

Fe(s) + 3 H₂O(l) → Fe(OH)₃(s) + 3/2 H₂(g)                    Δ_rH° = +160.9 kJ/mol-rxn

H₂(g) + 1/2 O₂(g) → H₂O(l)                                          Δ_rH° = –285.8 kJ/mol-rxn

Fe₂O₃(s) + 3 H₂O(l) → 2 Fe(OH)₃(s)                              Δ_rH° = +288.6 kJ/mol-rxn

Answer

–824.2 kJ/mol-rxn

1. (Fe(s) + 3 H₂O(l) → Fe(OH)₃(s) + 3/2 H₂(g)                    Δ_rH° = +160.9 kJ/mol-rxn) * 2

    2 Fe(s) + 6 H₂O(l) → 2 Fe(OH)₃(s) + 3 H₂(g)                    Δ_rH° = +321.8 kJ/mol-rxn

2. (H₂(g) + 1/2 O₂(g) → H₂O(l)                                          Δ_rH° = –285.8 kJ/mol-rxn) * 3

     3 H₂(g) + 3/2 O₂(g) → 3 H₂O(l)                                     Δ_rH° = –857.4kJ/mol-rxn

3. (Fe₂O₃(s) + 3 H₂O(l) → 2 Fe(OH)₃(s)                              Δ_rH° = +288.6 kJ/mol-rxn) Reverse Equation or ( * -1)

    2 Fe(OH)₃(s) → Fe₂O₃(s) + 3 H₂O(l)                               Δ_rH° = -288.6 kJ/mol-rxn

4. Add the equations and Δ_rH°.

2 Fe(s) + 6 H₂O(l) → 2 Fe(OH)₃(s) + 3 H₂(g)                    Δ_rH° = +321.8 kJ/mol-rxn

3 H₂(g) + 3/2 O₂(g) → 3 H₂O(l)                                       Δ_rH° = –857.4kJ/mol-rxn

2 Fe(OH)₃(s) → Fe₂O₃(s) + 3 H₂O(l)                                 Δ_rH° = -288.6 kJ/mol-rxn

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2 Fe(s) + 3/2 O₂(g) → Fe₂O₃(s)                                       ΔH_f = -824.2 kJ/mol

Exercise \(\PageIndex{4}\)

Determine the standard enthalpy of formation of calcium carbonate from the thermochemical equations given below.

Ca(OH)2(s) → CaO(s) + H2O(l)                                Δ_rH° = 65.2 kJ/mol-rxn

Ca(OH)2(s) + CO2(g) → CaCO3(s) + H2O(l)              Δ_rH° = −113.8 kJ/mol-rxn

C(s) + O2(g) → CO2(g)                                           Δ_rH° = −393.5 kJ/mol-rxn

2 Ca(s) + O2(g) → 2 CaO(s)                                    Δ_rH° = −1270.2 kJ/mol-rxn

Answer

−1207.6 kJ/mol-rxn

 

Enthalpy of Reaction from Standard Enthalpies of Formation

Exercise \(\PageIndex{5}\)

Calculate ΔrH° for the combustion of ammonia, using standard molar enthalpies of formation:

4 NH₃(g) + 7 O₂(g) → 4 NO₂(g) + 6 H₂O(l)

Molecule             ΔfH° (kJ/mol-rxn)

NH₃(g)                      –45.9

NO₂(g)                      +33.1

H₂O(l)                       –285.8

Answer

–1398.8 kJ/mol-rxn

ΔH°rxn = Σ ΔH°_f(products) - Σ ΔH°_f(reactants)

           = [(4 * ΔH°_f(NO₂)) + (6 * ΔH°_f(H₂O))] - [(4 * ΔH°_f(NH₃)) + (7 * ΔH°_f(O₂))]

           = [(4 * 33.1) + (6 * -285.8)] - [(4 * -45.9) + (7 * 0)]

           = -1398.8 kJ/mol-rxn

Exercise \(\PageIndex{6}\)

What is ΔrH° for the following phase change?

LiF(s) → LiF(l)

Substance           ΔH°f (kJ/mol-rxn)

LiF(s)                     –616.93

LiF(l)                      –598.65

Answer

18.28 kJ/mol-rxn