Calorimetry with a Limiting Reagent and an Overstated Uncertainty
A general chemistry student is assigned to determine the enthalpy of neutralization of HCl(aq) with NaOH(aq). She mixes 50.0 mL of 1.00 M HCl with 50.0 mL of 1.00 M NaOH in a coffee-cup calorimeter. The initial temperature of both solutions is 22.4 °C; after mixing, the highest temperature is 29.2 °C. She calculates q = 2.84 kJ and ΔH = -56.8 kJ/mol, close to the accepted -57.3 kJ/mol. Her partner points out that the calorimeter was not calibrated, the solutions were not exactly at the same temperature, and the final temperature was read 30 s after mixing, when it had already begun to fall. The student also has a second task: predict the temperature change if she repeats with 50.0 mL of 1.00 M HCl and 75.0 mL of 0.800 M NaOH. She has only one calorimeter and a thermometer with 0.2 °C graduations.
1. Evaluate the student's calculated ΔH. For each identified problem (uncalibrated calorimeter, unequal initial temperatures, late final-temperature reading), state whether it would make the measured ΔH more negative, less negative, or leave it roughly unchanged, and why.
Read the worked answer
The arithmetic is q = (100.0 g)(4.18 J/g°C)(6.8°C) = 2.84 kJ for 0.0500 mol, so ΔH = -56.8 kJ/mol. An uncalibrated calorimeter that absorbs heat makes the measured ΔT too small, so q and |ΔH| are too small (less negative). If the two solutions were not at the same initial temperature, the reported ΔT is ambiguous: if the acid was cooler than the base, the final temperature underestimates the true mixing rise and |ΔH| is too small; if warmer, the opposite. Reading the final temperature 30 s after mixing, after the peak has passed, makes ΔT too small and |ΔH| less negative.
All three tend to bias toward a less negative value, yet the result is already close to -57.3, suggesting either compensating errors or that the effects are small relative to the thermometer resolution. The student cannot claim better than about ±2–3 kJ/mol.
2. For the second experiment, identify the limiting reactant, calculate the expected temperature change, and explain how the different total mass affects the result.
Read the worked answer
Moles HCl = 0.0500 L × 1.00 M = 0.0500 mol. Moles NaOH = 0.0750 L × 0.800 M = 0.0600 mol. HCl is limiting because 1 mol HCl reacts with 1 mol NaOH and 0.0500 < 0.0600.
Heat released = 0.0500 mol × 57.3 kJ/mol = 2.865 kJ, assuming the same ΔH. Total solution mass ≈ 50.0 g + 75.0 g = 125.0 g (assuming density 1.00 g/mL and negligible heat capacity of the calorimeter). ΔT = q/(mc) = 2865 J/(125.0 g × 4.18 J/g°C) = 5.48 °C. The larger total mass means the same heat produces a smaller temperature rise than in the first experiment (6.8 °C), even though more NaOH is present; excess NaOH does not react. If the calorimeter heat capacity is significant, the actual ΔT will be smaller.
3. The student wants to report ΔH = -56.8 ± 0.1 kJ/mol. Critique this uncertainty and propose a calibration/analysis plan that would justify a defensible uncertainty.
Read the worked answer
±0.1 kJ/mol is not defensible. The thermometer reads to 0.2 °C, so a single ΔT of 6.8 °C has an uncertainty of at least about 0.2–0.3 °C, which is 3–4% of ΔT; propagated, that alone is about ±2 kJ/mol. The uncalibrated calorimeter, heat loss, and incomplete mixing add more.
A defensible plan: calibrate the calorimeter electrically or with a known reaction (for example, dissolving a known amount of KCl or neutralizing a standard), determine its heat capacity, and correct q. Record temperature every 5 s before and after mixing, plot T vs time, and extrapolate the post-mixing line back to the mixing time to estimate the instantaneous ΔT; run at least three trials and report mean ± standard deviation. Also use the same initial temperature for both solutions and a thermometer with 0.1 °C or better resolution. A realistic report might be -57 ± 2 kJ/mol, not ±0.1.