Calorimetry of Ethanol Combustion: Incomplete or Just Lossy?
Theoretical classroom model only. Use the supplied constants and idealizations; this is not authorization to handle unknown samples, run a combustion experiment or alter an operating chemical plant. A student determines the enthalpy of combustion of ethanol using a spirit burner under a copper can containing 150.0 g of water. The burner loses 2.50 g of ethanol. The water temperature rises from 21.0 °C to 41.0 °C. The copper can has a mass of 85.0 g and a specific heat capacity of 0.385 J g−1 K−1; the water is assumed to have c = 4.18 J g−1 K−1 and density 1.00 g cm−3. The student calculates ΔHc = −231 kJ mol−1 and compares this with a data-book value of −1367 kJ mol−1. The teacher comments that the result is 'not wrong, just incomplete'. A second student repeats the experiment behind a draught shield and obtains −780 kJ mol−1. The thermometer has an uncertainty of ±0.5 °C, and the balance reads to ±0.01 g.
1. Recalculate ΔHc, including the heat absorbed by the copper can, and comment on the sign and the units.
Read the worked answer
q(water) = 150.0 × 4.18 × 20.0 = 12 540 J. q(can) = 85.0 × 0.385 × 20.0 = 654.5 J. q(total) = 13 194.5 J = 13.19 kJ. n(ethanol) = 2.50 / 46.07 = 0.05426 mol. ΔHc = −13.19 / 0.05426 = −243 kJ mol−1. The negative sign means heat is released (exothermic) and the value is per mole of ethanol burned. The can correction is small (about 5%) and does not explain the large gap from −1367 kJ mol−1.
2. Identify and rank at least three reasons for the difference between the calculated value and −1367 kJ mol−1. Which reason is likely to be largest, and what evidence supports your ranking?
Read the worked answer
Reasons include: (i) heat loss to the surroundings by convection and radiation; (ii) incomplete combustion, producing soot or CO rather than CO2; (iii) evaporation of unburned ethanol from the wick; (iv) heating of the burner and the surrounding air; (v) non-standard conditions and heat capacity of products. A more exothermic repeated result behind a draught shield is consistent with improved heat capture, but a controlled comparison would be needed to isolate and rank the causes confidently. Incomplete combustion would also be visible as soot and would make the result less exothermic, but it is unlikely to account for a factor of six on its own. If each of the two temperature readings has uncertainty ±0.5 °C, the worst-case uncertainty in their difference is ±1.0 °C, or 5% of the20 °C rise. An independent-error root-sum-square model would give a different, explicitly stated estimate. Either is much smaller than the observed discrepancy.
3. A third student says: 'A bomb calorimeter would give exactly −1367 kJ mol−1.' Evaluate this claim.
Read the worked answer
The claim is too strong. A bomb calorimeter reduces heat loss and ensures complete combustion at constant volume, so it measures ΔU rather than ΔH. To compare with the data-book ΔH, you must convert using ΔH = ΔU + ΔngRT, where Δng is the change in moles of gas for the combustion equation. Calibration, impurities in the sample, and the choice of reference temperature also introduce uncertainty. So a bomb calorimeter would give a value much closer to −1367 kJ mol−1, but not 'exactly' it.