5.7: Enthalpy Calculations

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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)

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

(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

(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

(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

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

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