U6.8 · Enthalpy of FormationPractice · 生成焓
Lessons covered:U6.8 Enthalpy of Formation
- 1Multiple choice
For which equation does ΔH equal the standard enthalpy of formation of NaCl(s)?
ExplanationA formation equation makes 1 mol of the substance from elements in their standard states. At 25 °C sodium’s standard state is the solid and chlorine’s is Cl₂(g). The equation making 2 mol has ΔH = 2 × (−411.2); Na(g) isn’t sodium’s standard state; and forming the crystal from gaseous ions is the lattice-energy process, with no elements as reactants.
Review: U6.8 Enthalpy of Formation
- 2Numeric
Methane reacts with steam at high temperature to make hydrogen: CH₄(g) + H₂O(g) → CO(g) + 3H₂(g). ΔH°f (kJ/mol): CH₄(g) −74.8, H₂O(g) −241.8, CO(g) −110.5. Calculate ΔH.
kJ/molExplanationΔH = [−110.5 + 3 × 0] − [−74.8 + (−241.8)] = −110.5 − (−316.6) = +206.1 kJ/mol. H₂ is an element with zero enthalpy of formation. ΔH > 0, so the reaction is endothermic and needs a steady supply of heat.
Review: U6.8 Enthalpy of Formation
- 3Numeric
Ethene reacts with water to make ethanol: C₂H₄(g) + H₂O(l) → C₂H₅OH(l), ΔH = −44.2 kJ/mol. ΔH°f (kJ/mol): H₂O(l) −285.8, C₂H₅OH(l) −277.7. Find the enthalpy of formation of C₂H₄(g).
kJ/molExplanationLet x be the enthalpy of formation of C₂H₄(g): −44.2 = −277.7 − [x + (−285.8)], so x = −277.7 + 285.8 + 44.2 = +52.3 kJ/mol. It is positive, so making ethene from graphite and H₂ is endothermic.
Review: U6.8 Enthalpy of Formation
- 4Select all
At 25 °C, which of these have a standard enthalpy of formation of zero? Select all that apply.
More than one option may be correct. Choose all that apply, then press Submit.
ExplanationElements in their standard states have zero enthalpy of formation. At 25 °C mercury is a liquid, nitrogen is N₂(g), and iodine is the solid I₂(s), so these three are zero. Hydrogen’s standard state is H₂(g), not H atoms; I₂(g) isn’t iodine’s stable form at 25 °C; and chlorine is a gas at 25 °C, so Cl₂(l) isn’t its standard state.
Review: U6.8 Enthalpy of Formation
- 5Multiple choice
The standard enthalpy of formation of NO(g) is +90.3 kJ/mol. Which statement is correct?
ExplanationThe enthalpy of formation is ΔH for ½N₂(g) + ½O₂(g) → NO(g) = +90.3 kJ/mol > 0, endothermic, so NO sits above its elements. Decomposition is the reverse and releases 90.3 kJ. A bond enthalpy is the energy to break a bond in a gaseous molecule into atoms, which is a different quantity.
Review: U6.8 Enthalpy of Formation
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