Enthalpy of combustion of ethanol: interpreting a value that is far too small
A student determines the enthalpy change of combustion of ethanol using a spirit burner beneath a copper can holding 100.0 g of water. The water temperature rises from 21.5 °C to 47.9 °C. The burner and its ethanol weigh 142.36 g before and 141.12 g after. She uses c = 4.18 J g⁻¹ K⁻¹ for water and M(C₂H₅OH) = 46.0 g mol⁻¹, and her thermometer reads to 0.5 °C. Her calculated value is far less exothermic than the data book value of −1367 kJ mol⁻¹. The teacher notes that the can was open to the air, the flame was about 3 cm below the base of the can, soot was deposited on the can, and the outside of the can became too hot to hold. A classmate suggests simply repeating the experiment with a larger mass of water, saying that this will 'get the right answer'.
1. Calculate the student's value for the enthalpy change of combustion of ethanol, showing each step and giving the sign and units.
Read the worked answer
Temperature rise ΔT = 47.9 − 21.5 = 26.4 K
Heat gained by water: q = mcΔT = 100.0 × 4.18 × 26.4 = 11 035 J = 11.04 kJ
Ethanol burned = 142.36 − 141.12 = 1.24 g; n = 1.24/46.0 = 2.70 × 10⁻² mol
ΔHc = −q/n = −11.04/0.0270 = −409 kJ mol⁻¹ (about −410 kJ mol⁻¹)
The negative sign is essential: combustion is exothermic, so energy is released. The calculation assumes that all the heat released went into the water and none into anything else - an assumption the next task attacks.
2. Explain, using the observations, why the value is so much less exothermic than −1367 kJ mol⁻¹. Would using a larger mass of water fix the problem?
Read the worked answer
The method only measures heat that reaches the water. Losses are large:
- hot gases and the flame heat the surrounding air, and heat is conducted/convected from the bare copper can;
- the flame sits 3 cm below the can, so much of its heat never reaches it;
- soot shows incomplete combustion, releasing less energy per mole (some carbon and CO are formed);
- water vapour in the products escapes, carrying its latent heat;
- the can and thermometer also absorb some heat, but this is small compared with the losses.
Using more water does not fix it. A larger mass gives a smaller ΔT (and therefore a larger percentage uncertainty from a thermometer reading to 0.5 °C) while the proportion of heat lost to the surroundings is unchanged. The measurement is not wrong because the water is too little; it is wrong because the calorimeter is open and unlagged.
Real improvements: a lid, a draught shield, a calorimeter with the burner enclosed, a spiral tube so the flame plays on the vessel, and calibrating the apparatus with a known electrical heater to find its heat capacity.
3. A second student, using the same apparatus, obtains a value of −1520 kJ mol⁻¹. Is this a better result? Explain what could produce it.
Read the worked answer
No - a value more exothermic than the accepted value is not 'better', it is impossible for this compound at these conditions, so it points to a systematic error that has made the mass of ethanol apparently burned too large (or the heat measured too large).
Since ΔHc = −mcΔT/(mass burned), the value is too negative if the calculated mass of ethanol lost is bigger than the mass actually combusted, or if ΔT or the water mass is overestimated.
Realistic causes:
- ethanol evaporates from the wick after the flame is extinguished, and the burner is reweighed only later, so the mass loss includes ethanol that never burned;
- the flame is allowed to burn briefly after the can is removed, or the burner is weighed before it has cooled, giving a false final mass;
- the water mass or ΔT is misread (e.g. reading 100.0 g as the can plus water, or reading the thermometer to 0.1 °C when it reads to 0.5 °C).
Better practice: extinguish the burner by replacing its cap, allow it to cool in a closed position, and reweigh promptly.