Correcting a Coffee-Cup Calorimetry Calculation for Heat Capacity and Solution Properties
A student mixes 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH in a coffee-cup calorimeter. Both solutions are initially at 23.8 °C. After mixing, the highest temperature is 30.2 °C. The student uses the total volume as 100.0 mL, assumes a density of 1.00 g/mL and a specific heat of 4.184 J/(g·°C), and calculates ΔHneutralization = −53.6 kJ/mol. The teacher provides more accurate constants: solution density 1.02 g/mL, specific heat 3.98 J/(g·°C), and calorimeter heat capacity 45 J/°C. The student’s thermometer has ±0.2 °C precision, and the balance used to prepare solutions has ±0.5% accuracy. The teacher asks whether the student’s value can be used to identify the reaction as exothermic, and how to report a corrected value.
1. Calculate the corrected heat released and ΔH per mole of water formed. Show the sign convention and compare with the student’s −53.6 kJ/mol.
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Limiting reactant: 0.0500 L × 1.00 M = 0.0500 mol HCl and 0.0500 mol NaOH; the reaction is 1:1, so 0.0500 mol water forms. Corrected solution mass = 100.0 mL × 1.02 g/mL = 102 g. q_solution = m c ΔT = 102 g × 3.98 J/(g·°C) × (30.2 − 23.8) °C = 102 × 3.98 × 6.4 = 2.60 × 10^3 J. q_calorimeter = C ΔT = 45 J/°C × 6.4 °C = 2.9 × 10^2 J. Total heat released by reaction = 2.60 × 10^3 + 2.9 × 10^2 = 2.89 × 10^3 J, so q_rxn = −2.89 kJ. ΔH = q_rxn / 0.0500 mol = −57.7 kJ/mol. The student’s −53.6 kJ/mol used too little mass, a larger specific heat, and omitted the calorimeter, so it underestimates the exothermicity. The sign is negative, as expected for an exothermic neutralization.
2. Discuss whether the corrected value is precise enough to distinguish this reaction from a similar neutralization with ΔH = −55.0 kJ/mol. Include at least two sources of uncertainty.
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The corrected value is about −57.7 kJ/mol, but it is not precise enough to distinguish confidently from −55.0 kJ/mol. A temperature uncertainty of ±0.2 °C on each reading gives roughly ±0.3 °C on ΔT, or about ±4–5% on q; that is about ±2.5 kJ/mol on ΔH before considering heat loss. Heat loss through the coffee cup and imperfect mixing usually make the measured temperature rise too small, so the true exothermicity may be slightly larger in magnitude. Differences in solution density and specific heat with temperature also add uncertainty. To improve precision, insulate the calorimeter, calibrate its heat capacity electrically, use larger volumes or higher concentrations to get a larger ΔT, and repeat trials. Without those improvements, the two values overlap within reasonable uncertainty.