U6.6 · Introduction to Enthalpy of ReactionPractice · 反应焓入门
Lessons covered:U6.6 Introduction to Enthalpy of Reaction
- 1Multiple choice
Compare ΔH for two reactions: ① H₂(g) + ½O₂(g) → H₂O(l); ② H₂(g) + ½O₂(g) → H₂O(g). Which statement is correct?
ExplanationΔH depends on physical states. ① ΔH = −285.8 kJ/mol and ② ΔH = −241.8 kJ/mol. Think of ① as making H₂O(g) and then condensing it, which releases another 44.0 kJ. Condensing forms attractions between molecules; the O–H bonds inside them are unchanged.
- 2Numeric
Given C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l), ΔH = −2219.9 kJ/mol. A camping stove burns 11.0 g of propane (M = 44.09 g/mol) completely. How much heat is released?
kJExplanationn(C₃H₈) = 11.0 ÷ 44.09 = 0.2495 mol. C₃H₈ has a coefficient of 1, so each mole releases 2219.9 kJ: q = 0.2495 × 2219.9 = 554 kJ.
- 3Numeric
Given 2CO(g) + O₂(g) → 2CO₂(g), ΔH = −566.0 kJ/mol. A mixture of 0.300 mol CO and 0.100 mol O₂ is ignited and reacts completely. How much heat is released?
kJExplanationFind the limiting reactant first: CO 0.300 ÷ 2 = 0.150, O₂ 0.100 ÷ 1 = 0.100. O₂ gives the smaller ratio, so it runs out after 0.100 mol of O₂ reacts with 0.200 mol of CO. O₂ has a coefficient of 1, so each mole of O₂ consumed releases 566.0 kJ: q = 0.100 × 566.0 = 56.6 kJ. Basing it on CO gives 84.9 kJ, but there isn’t enough O₂; 0.100 mol of CO is left over.
- 4Multiple choice
Complete combustion of ethanol and propane: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH = −1366.7 kJ/mol; C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l), ΔH = −2219.9 kJ/mol. Comparing the heat released per mole and per gram, which statement is correct?
ExplanationPer mole: propane 2219.9 kJ > ethanol 1366.7 kJ. Per gram: propane 2219.9 ÷ 44.09 = 50.3 kJ/g, ethanol 1366.7 ÷ 46.07 = 29.7 kJ/g. The molar masses are similar, yet the heat per gram differs a lot: ethanol already contains an O atom, so it is in a sense partly oxidized already.
- 5Multiple choice
Red-hot coke reacts with steam: C(s) + H₂O(g) → CO(g) + H₂(g), ΔH = +131.3 kJ/mol. What happens to heat when this reaction produces 3.00 mol of H₂?
ExplanationΔH > 0, so the reaction is endothermic. Each mole of H₂ formed absorbs 131.3 kJ, so 3.00 mol absorbs 3.00 × 131.3 = 393.9 kJ. That is why heat must be supplied continuously (industry alternates by blowing air through to burn some of the coke).
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