U6.2 · Energy DiagramsPractice · 能量图
Lessons covered:U6.2 Energy DiagramsU6.5 Energy of Phase Changes
- 1Multiple choice
C₂H₂(g) + H₂(g) → C₂H₄(g), ΔH = −174.4 kJ/mol. Which statement about the energy diagram for this reaction is correct?
ExplanationΔH = H(products) − H(reactants) < 0, so the products sit 174.4 kJ below the reactants, for the amounts in the equation (1 mol each). The energy barrier (activation energy) is the climb from reactants to the transition state, and ΔH says nothing about its size.
Review: U6.2 Energy Diagrams
- 2Numeric
Given C₂H₂(g) + H₂(g) → C₂H₄(g), ΔH = −174.4 kJ/mol. Find ΔH for 2C₂H₄(g) → 2C₂H₂(g) + 2H₂(g).
kJ/molExplanationThe new equation is the original reversed and doubled. Reversing swaps the two levels on the diagram, so ΔH changes sign; doubling the amounts doubles the enthalpy difference. ΔH = −(−174.4) × 2 = +348.8 kJ/mol.
Review: U6.2 Energy Diagrams
- 3Multiple choice
When dissolves in water, the solution warms. Split dissolving into three steps: ① separating the and ions in the crystal (ΔH₁); ② separating some of the water molecules from each other (ΔH₂); ③ surrounding the ions with water molecules (hydration, ΔH₃). Which statement is correct?
ExplanationSeparating the ions overcomes the attraction between them and separating water molecules overcomes intermolecular attractions, so ΔH₁ and ΔH₂ are both positive. Ions attracting water molecules releases energy, so ΔH₃ < 0. The overall ΔH = ΔH₁ + ΔH₂ + ΔH₃ is negative, so hydration releases more. The and ions are still there after dissolving; no new substance forms.
Review: U6.2 Energy Diagrams
- 4Multiple choice
Which statement must be true for an endothermic process?
ExplanationEndothermic means ΔH = H(products) − H(reactants) > 0: the products sit higher on the energy diagram, so they are higher in energy and less stable. An endothermic process doesn’t need a burner: ammonium nitrate dissolves at room temperature, taking the heat it needs from the surrounding water, which is why the solution cools.
Review: U6.2 Energy Diagrams
- 5Numeric
Taking the elements in their standard states (H₂ and O₂) as zero, at 25 °C 1 mol of H₂O(g) has an enthalpy of −241.8 kJ and 1 mol of H₂O(l) has −285.8 kJ. Find ΔH for H₂O(l) → H₂O(g).
kJ/molExplanationΔH = H(final) − H(initial) = (−241.8) − (−285.8) = +44.0 kJ/mol. On the diagram H₂O(g) sits 44.0 kJ above H₂O(l), so vaporizing absorbs heat. This is the enthalpy of vaporization at 25 °C, a little larger than the 40.7 kJ/mol at 100 °C.
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