Standard Enthalpy of Formation: Calculating Reaction Enthalpy

Question: Calculate the standard enthalpy of combustion of liquid ethanol, C2H5OH(l), to produce gaseous carbon dioxide and liquid water at 298 K. The standard enthalpies of formation are: Δ_fH°[C2H5OH(l)] = −277.0 kJ/mol, Δ_fH°[CO2(g)] = −393.5 kJ/mol, and Δ_fH°[H2O(l)] = −285.8 kJ/mol.

Step 1: Write the balanced chemical equation for the combustion of 1 mole of ethanol:

C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l)

Step 2: Identify the standard enthalpy of formation for each substance:

Δ_fH°[C₂H₅OH(l)] = −277.0 kJ/mol

Δ_fH°[O₂(g)] = 0.0 kJ/mol (element in reference standard state)

Δ_fH°[CO₂(g)] = −393.5 kJ/mol

Δ_fH°[H₂O(l)] = −285.8 kJ/mol

Step 3: Calculate the total enthalpy of formation for all products:

Σ ν_p Δ_fH°(products) = [2 × (−393.5 kJ/mol)] + [3 × (−285.8 kJ/mol)]

2 × (−393.5) = −787.0 kJ

3 × (−285.8) = −857.4 kJ

Sum of products = −787.0 + (−857.4) = −1644.4 kJ

Step 4: Calculate the total enthalpy of formation for all reactants:

Σ ν_r Δ_fH°(reactants) = [1 × (−277.0 kJ/mol)] + [3 × 0.0 kJ/mol] = −277.0 kJ

Step 5: Compute the standard enthalpy of reaction:

Δ_rH° = Σ ν_p Δ_fH°(products) − Σ ν_r Δ_fH°(reactants)

Δ_rH° = (−1644.4 kJ) − (−277.0 kJ)

Δ_rH° = −1644.4 + 277.0 = −1367.4 kJ/mol

Verification by second route (summing constituent step reactions):

Formation of 2 mol CO₂ gives −787.0 kJ. Formation of 3 mol H₂O gives −857.4 kJ. Decomposition of 1 mol C₂H₅OH into elements gives −(−277.0) = +277.0 kJ. Adding all three: −787.0 − 857.4 + 277.0 = −1367.4 kJ/mol. Both methods yield identical results.

Answer: −1367.4 kJ/mol

Question: The thermite reaction between solid aluminium and solid iron(III) oxide produces molten iron and solid aluminium oxide according to the equation: 2 Al(s) + Fe2O3(s) → Al2O3(s) + 2 Fe(s). Calculate the standard enthalpy change of this reaction using the following data: Δ_fH°[Fe2O3(s)] = −824.2 kJ/mol and Δ_fH°[Al2O3(s)] = −1675.7 kJ/mol.

Step 1: Check the balanced reaction equation and stoichiometric coefficients:

2 Al(s) + 1 Fe₂O₃(s) → 1 Al₂O₃(s) + 2 Fe(s)

Reactants: 2 moles of Al(s), 1 mole of Fe₂O₃(s)

Products: 1 mole of Al₂O₃(s), 2 moles of Fe(s)

Step 2: Assign formation enthalpies to elemental reference states:

Δ_fH°[Al(s)] = 0.0 kJ/mol

Δ_fH°[Fe(s)] = 0.0 kJ/mol

Step 3: Calculate the product term:

Σ ν_p Δ_fH°(products) = [1 × (−1675.7 kJ/mol)] + [2 × 0.0 kJ/mol] = −1675.7 kJ

Step 4: Calculate the reactant term:

Σ ν_r Δ_fH°(reactants) = [2 × 0.0 kJ/mol] + [1 × (−824.2 kJ/mol)] = −824.2 kJ

Step 5: Subtract reactants from products:

Δ_rH° = (−1675.7 kJ) − (−824.2 kJ)

Δ_rH° = −1675.7 + 824.2 = −851.5 kJ

Verification by second route:

Difference in oxide stability = (−1675.7) − (−824.2). Computing step by step: −1675.7 + 800 = −875.7; −875.7 + 24.2 = −851.5 kJ. The arithmetic matches exactly.

Answer: −851.5 kJ

Question: The standard enthalpy of reaction for the combustion of gaseous methane is −890.3 kJ/mol: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l). Given that Δ_fH°[CO2(g)] = −393.5 kJ/mol and Δ_fH°[H2O(l)] = −285.8 kJ/mol, calculate the standard enthalpy of formation of methane, Δ_fH°[CH4(g)].

Step 1: Set up the reaction enthalpy expression from the balanced chemical equation:

Δ_rH° = { Δ_fH°[CO₂(g)] + 2 × Δ_fH°[H₂O(l)] } − { Δ_fH°[CH₄(g)] + 2 × Δ_fH°[O₂(g)] }

Step 2: Substitute the known numerical values into the product summation:

Σ products = (−393.5 kJ) + [2 × (−285.8 kJ)]

2 × (−285.8) = −571.6 kJ

Σ products = −393.5 + (−571.6) = −965.1 kJ

Step 3: Express the reactant summation with unknown x = Δ_fH°[CH₄(g)]:

Since Δ_fH°[O₂(g)] = 0 kJ/mol, Σ reactants = x + 0 = x

Step 4: Formulate the linear equation:

−890.3 kJ = −965.1 kJ − x

Rearranging to solve for x:

x = −965.1 kJ − (−890.3 kJ)

x = −965.1 + 890.3 = −74.8 kJ/mol

Verification by second route:

Forward calculation check: Products − Reactants = −965.1 − (−74.8) = −965.1 + 74.8 = −890.3 kJ/mol. This re-derives the given reaction enthalpy accurately.

Answer: −74.8 kJ/mol