Calorimetry: When a Good Agreement Is Misleading
A student mixes 50.0 mL of 1.00 M HCl with 50.0 mL of 1.00 M NaOH in a coffee-cup calorimeter. Both solutions are initially at 22.0 °C. The highest temperature reached is 28.6 °C. The student calculates: q = (100.0 g)(4.18 J/g°C)(6.6 °C) = 2.76 kJ; moles = (0.0500 L)(1.00 M) = 0.0500 mol; ΔH = -2.76/0.0500 = -55.2 kJ/mol. The accepted value is -57.1 kJ/mol. The student writes: 'The reaction is less exothermic than the accepted value because the solutions were not at standard conditions (1 M and 25 °C).' The calorimeter has a known heat capacity of 15 J/°C, which the student ignored. The teacher must decide what feedback to give and whether the student's explanation is chemically valid.
1. Recalculate ΔH including the calorimeter heat capacity and compare with the student's value and the accepted value. Show all steps.
Read the worked answer
q_solution = m c ΔT = (100.0 g)(4.18 J/g°C)(6.6 °C) = 2759 J. q_calorimeter = C ΔT = (15 J/°C)(6.6 °C) = 99 J. Total heat released = 2759 + 99 = 2858 J = 2.858 kJ. mol H2O formed = 0.0500 L × 1.00 mol/L = 0.0500 mol (HCl and NaOH are equimolar, so neither is in excess). ΔH = −2.858 kJ / 0.0500 mol = −57.2 kJ/mol. This is close to the accepted −57.1 kJ/mol. The student's −55.2 kJ/mol was too small in magnitude because it assigned all heat to the solution and ignored the heat absorbed by the calorimeter walls.
2. Evaluate the student's explanation that 'nonstandard conditions' caused the difference. Is this chemically valid? What is the more likely explanation?
Read the worked answer
The student's 'nonstandard conditions' explanation is not chemically sound here. The reaction is H+(aq) + OH−(aq) → H2O(l); at 1.00 M the activities are not exactly 1, but the enthalpy of neutralization of strong acids and bases changes only slightly with concentration over this range, not by the 2 kJ/mol discrepancy. Standard state is 1 M, so the solutions are already near standard. The more likely explanation is the unaccounted calorimeter heat capacity, which is a real systematic error and accounts for nearly all of the difference. A smaller contribution could come from heat loss or from using 4.18 J/g°C and 1.00 g/mL for 1 M solutions rather than their actual density and specific heat.
3. The student repeats the experiment with 0.50 M HCl and 0.50 M NaOH and obtains ΔH = -49 kJ/mol. Explain why the apparent ΔH is different and whether this supports the student's original claim.
Read the worked answer
At 0.50 M, the same total volume contains half as many moles, so the temperature rise is roughly half (about 3.3 °C). The heat absorbed by the calorimeter and heat lost to the room are a larger fraction of the smaller total heat, so the apparent ΔH is less exothermic. The value −49 kJ/mol is therefore an artifact of the calorimeter's heat capacity and heat loss, not evidence that standard conditions matter. To compare concentrations fairly, calibrate the calorimeter with a known reaction (e.g., TRIS + HCl) or use a bomb calorimeter and apply the same heat-capacity correction. The result actually supports the need for calorimeter calibration, not the student's original claim.