Heat of neutralization benchmark: The average heat of neutralization for dilute strong acid–strong base reactions is approximately how many kcal per kmol of water formed?
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A6,680 kcal/kmol
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B13,360 kcal/kmol
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C2,481 kcal/kmol
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D97,302 kcal/kmol
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E1,360 kcal/kmol
Answer
Correct Answer: 13,360 kcal/kmol
Explanation
Introduction / Context:For dilute aqueous solutions of strong acids and strong bases, the heat of neutralization is nearly constant because the net ionic reaction is effectively H+ + OH− → H2O. Memorizing this benchmark helps in preliminary energy balances and calorimetry checks.
Given Data / Assumptions:
- Strong acid and strong base completely dissociated.
- Dilute solutions; negligible heats of dilution beyond the neutralization step.
- Standard reference temperature near ambient.
Concept / Approach:The molar heat of neutralization for H+ and OH− is about −13.6 kcal per mole of water formed (≈ −57 kJ/mol). Per kilomole of water, multiply by 1000 to get approximately −13,600 kcal/kmol. Rounded tabulated values often quote 13,360 kcal/kmol for convenience.
Step-by-Step Solution:Per mole: ΔH_neut ≈ −13.6 kcal/mol.Per kmol: −13.6 × 1000 ≈ −13,600 kcal/kmol.Common tabulation: ~ −13,360 kcal/kmol.
Verification / Alternative check:Converting to SI: 13,360 kcal/kmol × 4.1868 kJ/kcal ≈ 55,980 kJ/kmol ≈ 56 kJ/mol, consistent with standard values.
Why Other Options Are Wrong:6,680 and 2,481 kcal/kmol are far too small; 97,302 kcal/kmol is unrealistically large for this reaction.
Common Pitfalls:Confusing per mole with per kilomole; mixing sign convention (exothermic negative). This question asks magnitude.
Final Answer:13,360 kcal/kmol