6.8 Enthalpy of Formation

1. A2 Ca(s) + O₂(g) → 2 CaO(s)
   A student who thinks every equation must use whole-number coefficients picks this. It forms 2 mol of CaO, so its ΔH° is twice ΔH°_f of CaO(s).
2. BCa(s) + ½ O₂(g) → CaO(s) Correct
   One mole of CaO(s) is formed from its elements in their standard states, calcium metal and oxygen gas, so ΔH° for this equation is ΔH°_f of CaO(s). The fraction ½ is needed to keep exactly one mole of product.
3. CCa²⁺(g) + O²⁻(g) → CaO(s)
   A student who thinks a formation reaction builds the compound from the particles it is made of picks this. The reactants must be the elements in their standard states, Ca(s) and O₂(g), not gaseous ions, which are far higher in energy.
4. DCa(OH)₂(s) → CaO(s) + H₂O(g)
   A student who thinks any reaction that produces one mole of a compound is its formation reaction picks this; heating calcium hydroxide does produce CaO(s). The reactant is a compound, not the elements, so ΔH° for this reaction is not ΔH°_f of CaO(s).

Working No calculation. A formation reaction makes exactly 1 mol of the compound from its elements in their standard states: Ca(s) and O₂(g). Balancing with 1 mol CaO requires ½ O₂(g): Ca(s) + ½ O₂(g) → CaO(s).

CED 6.8.A.1 · Read this in Fix

1. A−566.0 kJ/mol_rxn
   A student who subtracts the products' values from the reactants' picks this: 2(−393.5) − 2(−110.5) = −566.0. That is ΔH° for the reverse reaction, 2 CO(g) + O₂(g) → 2 CO₂(g). The reactant sum is subtracted from the product sum, giving +566.0 kJ/mol_rxn.
2. B+283.0 kJ/mol_rxn
   A student who uses each ΔH°_f once, ignoring the coefficients, picks this: (−110.5) − (−393.5) = +283.0. The equation shows 2 mol of CO₂ and 2 mol of CO, so each value is multiplied by 2.
3. C−221.0 kJ/mol_rxn
   A student who takes ΔH° of a reaction that produces CO(g) to be the ΔH°_f of the CO formed picks this: 2(−110.5) = −221.0. This reaction forms CO from CO₂, not from its elements, so the ΔH°_f of CO₂ must also be included.
4. D+566.0 kJ/mol_rxn Correct
   Products minus reactants, each multiplied by its coefficient: [2(−110.5) + 0] − [2(−393.5)] = +566.0 kJ/mol_rxn. O₂(g) is an element in its standard state, so its ΔH°_f is zero. The positive sign fits a decomposition that needs a large input of energy.

Working ΔH° = ΣΔH°_f,products − ΣΔH°_f,reactants = [2(−110.5) + 0] − [2(−393.5)] = −221.0 + 787.0 = +566.0 kJ/mol_rxn. O₂(g) is an element in its standard state (ΔH°_f = 0); the H₂O(g) value is not needed.

CED 6.8.A.1 · Read this in Fix

1. A223 kJ Correct
   ΔH° = [−393.5 + 2(−285.8)] − (−74.8) = −890.3 kJ/mol_rxn, using the value for liquid water. One mole of reaction uses 2 mol of O₂, so 0.500 mol of O₂ corresponds to 0.250 mol of reaction, and the heat released is 0.250 × 890.3 = 223 kJ.
2. B445 kJ
   A student who reads kJ/mol_rxn as 'per mole of O₂' picks this: 0.500 × 890.3 = 445 kJ. The equation consumes 2 mol of O₂ per mole of reaction, so 0.500 mol of O₂ is only 0.250 mol of reaction.
3. C890 kJ
   A student who takes ΔH° to be the heat released whatever amount reacts picks this. ΔH° = −890.3 kJ is for one mole of reaction, which consumes 2 mol of O₂; here only 0.500 mol of O₂ reacts.
4. D151 kJ
   A student who uses each ΔH°_f once, ignoring the coefficient 2 of H₂O, picks this: ΔH° = (−393.5 − 285.8) − (−74.8) = −604.5 kJ/mol_rxn, then 0.250 × 604.5 = 151 kJ. Two moles of H₂O(l) form per mole of reaction, so its ΔH°_f is multiplied by 2, giving −890.3 kJ/mol_rxn.

Working ΔH° = [−393.5 + 2(−285.8)] − [−74.8 + 2(0)] = −965.1 + 74.8 = −890.3 kJ/mol_rxn (H₂O(l), not H₂O(g), as the equation shows). One mole of reaction consumes 2 mol O₂, so 0.500 mol O₂ is 0.250 mol of reaction. Heat released = 0.250 mol_rxn × 890.3 kJ/mol_rxn = 223 kJ.

CED 6.8.A.1 · Read this in Fix

1. AEnergy is needed to break the O–H bonds inside the water molecules.
   A student who thinks vaporization breaks covalent bonds within molecules picks this. Water vapor still consists of H₂O molecules; only the attractions between molecules are overcome.
2. BEnergy is needed to heat the liquid water up to its boiling point first.
   A student who thinks water becomes a gas only at its boiling point picks this. Both ΔH°_f values refer to 25°C, and water evaporates at 25°C, so the +44.0 kJ/mol is for the change of state alone, with no heating involved.
3. CEnergy is needed to overcome the attractions between water molecules. Correct
   In the liquid, water molecules attract one another strongly. Turning the liquid into gas separates the molecules, which requires energy, so H₂O(g) is higher in enthalpy than H₂O(l) and ΔH° = −241.8 − (−285.8) = +44.0 kJ/mol.
4. DEnergy is needed to make every water molecule grow to a larger size.
   A student who thinks particles expand when a substance becomes a gas picks this. The molecules stay the same size; the gas occupies more volume because the molecules are farther apart.

Working No calculation needed beyond ΔH° = −241.8 − (−285.8) = +44.0 kJ/mol. The process separates whole H₂O molecules from one another; energy is absorbed to overcome the attractions between molecules, so the gas is higher in enthalpy than the liquid.

CED 6.8.A.1 · Read this in Fix

1. AThe flame also transferred heat to the aluminum and the air, not just the water. Correct
   Only the heat absorbed by the water was measured. Heat that warmed the aluminum can and the surrounding air was not counted, so the heat per mole of methane calculated from the water's temperature rise is too small and the measured ΔH is less negative than −802 kJ/mol_rxn.
2. BThe calculation gave O₂(g) a ΔH°_f of zero, but O₂(g) has its own value.
   A student who thinks every element has its own nonzero ΔH°_f picks this. O₂(g) is an element in its standard state, so its ΔH°_f is zero by definition, and the calculated value is not missing a term.
3. CThe student burned less than 1 mol of methane, so a smaller ΔH is expected.
   A student who thinks ΔH depends on the amount burned picks this. The student's result is expressed per mole of reaction, so a smaller sample gives a smaller q but the same ΔH.
4. DThe calculation should not have multiplied the ΔH°_f of H₂O(g) by 2.
   A student who thinks each ΔH°_f is used once regardless of coefficients picks this. Two moles of H₂O(g) form per mole of reaction, so the ΔH°_f of H₂O(g) must be multiplied by 2.

Working No calculation. The calculation from ΔH°_f values is correct: [−393.5 + 2(−241.8)] − [−74.8] = −802.3 kJ/mol_rxn, with O₂(g) at zero and H₂O(g) multiplied by 2. In the open-can experiment only the heat absorbed by the water is counted; heat passing to the can and the air is missed, so the measured value is less negative.

CED 6.8.A.1 · Read this in Fix

1. A−2088 kJ/mol_rxn
   A student who uses each ΔH°_f once, whatever the coefficient, picks this: −2044 + (−44.0). Four moles of water form per mole of reaction, so the change is 4 × (−44.0 kJ).
2. B−2220 kJ/mol_rxn Correct
   Only the water term in the products' sum changes. Each mole of H₂O(l) has a ΔH°_f that is 44.0 kJ more negative than that of H₂O(g), and 4 mol of water form, so ΔH° = −2044 kJ + 4(−44.0 kJ) = −2220 kJ/mol_rxn.
3. C−1868 kJ/mol_rxn
   A student who calculates ΔH° as reactants minus products picks this, since a more negative ΔH°_f for a product then raises ΔH° by 4 × 44.0 kJ. In products minus reactants, a more negative product value makes ΔH° more negative.
4. D−2044 kJ/mol_rxn
   A student who thinks ΔH° depends only on which substances react and form, not on their physical states, picks this. H₂O(l) has a different ΔH°_f from H₂O(g), so ΔH° for the reaction changes.

Working ΔH° = ΣΔH°_f(products) − ΣΔH°_f(reactants). Only the water term changes: 4 mol H₂O(l) in place of 4 mol H₂O(g) changes the products' sum by 4 × [−285.8 − (−241.8)] kJ = −176.0 kJ. ΔH° = −2044 kJ + (−176.0 kJ) = −2220 kJ/mol_rxn.

CED 6.8.A.1 · Read this in Fix