Enthalpy and Calorimetry: Measuring the Heat of a Reaction

Drawing the boundary between system and surroundings

Every energy calculation starts by choosing a system, usually the reacting chemicals. Everything else is the surroundings, which in a calorimeter means mainly the water or solution and the vessel. Energy can cross the boundary as heat, q, or as work, w, and the change in the system's internal energy is ΔU = q + w.

Enthalpy, H = U + PV, is defined so that its change equals the heat exchanged at constant pressure when the only work is expansion work: ΔH = q_p. Because most bench chemistry happens at roughly constant atmospheric pressure, chemists tabulate enthalpy rather than internal energy.

Sign conventions for heat and enthalpy

Signs are always written from the system's point of view. Heat flowing into the system is positive; heat flowing out is negative. An exothermic reaction releases heat to the surroundings, so its ΔH is negative and the surroundings warm up. An endothermic reaction absorbs heat, so its ΔH is positive and the surroundings cool.

In a calorimeter you measure the surroundings, not the reaction. If the water warms, the water gained heat, and the reaction lost exactly that much. This gives the central relationship of calorimetry: q_reaction = −q_surroundings.

Heat capacity and the equation q = mcΔT

The heat needed to change the temperature of a substance is q = mcΔT, where m is mass, c is specific heat (the heat needed to warm one gram by one degree), and ΔT = T_final − T_initial. A change of one degree Celsius equals a change of one kelvin, so ΔT has the same value on either scale. Specific heats used below come from OpenStax Chemistry 2e Table 5.1, measured near 25 °C; for liquid water, c = 4.184 J/(g·°C).

Worked example 1. How much heat raises a 310.0 g block of silicon from 24.5 °C to 160.0 °C? The specific heat of silicon is 0.712 J/(g·°C). ΔT = 160.0 − 24.5 = 135.5 °C. q = (310.0 g)(0.712 J/(g·°C))(135.5 °C) = 2.99 × 10⁴ J, or 29.9 kJ. The specific heat has three significant figures, so the answer does too.

Worked example 2. A 450.0 g lead block at 88.0 °C is placed in 60.0 g of water at 20.5 °C in a foam cup. Lead's specific heat is 0.130 J/(g·°C), and water's is the value given above. The heat lost by the lead equals the heat gained by the water, so q_lead + q_water = 0: (58.5 J/°C)(T_f − 88.0) + (251.0 J/°C)(T_f − 20.5) = 0, where each coefficient is mass times specific heat. Solving, T_f = (58.5 × 88.0 + 251.0 × 20.5)/(58.5 + 251.0) = 33.3 °C. Check: the water gains about 251.0 × 12.8 ≈ 3.2 kJ, and the lead loses about 58.5 × 54.7 ≈ 3.2 kJ. The water's temperature moves far less than the lead's because its heat capacity is more than four times larger.

Coffee-cup calorimetry at constant pressure

A coffee-cup calorimeter is an insulated cup open to the air, so the process happens at constant pressure and the measured heat gives ΔH directly. For dilute aqueous solutions, the usual assumptions are that the solution has the density and specific heat of water and that the cup absorbs negligible heat.

Worked example 3. 60.0 mL of 0.800 mol/L nitric acid at 21.42 °C is mixed with 60.0 mL of 0.800 mol/L potassium hydroxide at the same temperature. The mixture reaches 26.75 °C. Find ΔH per mole of water formed. Mass of solution: 120.0 mL × 1.00 g/mL = 120. g. ΔT = 26.75 − 21.42 = 5.33 °C. q_solution = (120. g)(4.184 J/(g·°C))(5.33 °C) = 2.68 × 10³ J. So q_reaction = −2.68 × 10³ J.

Moles of water formed: each solution supplies 0.0600 L × 0.800 mol/L = 0.0480 mol, and the reaction H⁺(aq) + OH⁻(aq) → H₂O(l) is 1:1, so 0.0480 mol of water forms. ΔH = −2.68 kJ / 0.0480 mol = −55.8 kJ/mol. As a check, standard enthalpies of formation from OpenStax Appendix G give −285.83 − (−230.0) = −55.8 kJ/mol for the same net reaction, taking ΔH°f of H⁺(aq) as zero by convention.

Bomb calorimetry and the gap between ΔU and ΔH

Combustion is measured in a bomb calorimeter, a sealed steel vessel in a water bath. Because the volume is fixed, no expansion work is done and the measured heat is ΔU, not ΔH. The whole apparatus is calibrated as a unit, giving a calorimeter constant C_cal in kJ/°C, so q = C_cal × ΔT with no separate mass or specific heat.

Worked example 4. Burning 0.5000 g of liquid benzene, C₆H₆, in a calorimeter with C_cal = 10.25 kJ/°C raises the temperature by 2.038 °C. With IUPAC standard atomic weights rounded to two decimals (C 12.01, H 1.01), M = 78.12 g/mol, so n = 0.5000 g / 78.12 g/mol = 0.006400 mol. q_cal = (10.25 kJ/°C)(2.038 °C) = 20.89 kJ, so ΔU = −20.89 kJ / 0.006400 mol = −3264 kJ/mol, with four significant figures because every measured value has four.

To convert to enthalpy, use ΔH = ΔU + Δn_gas RT. In C₆H₆(l) + 15/2 O₂(g) → 6 CO₂(g) + 3 H₂O(l), gas moles fall from 7.5 to 6, so Δn_gas = −1.5 and the correction is (−1.5)(8.314 J/(mol·K))(298.15 K) = −3.72 kJ/mol. ΔH = −3264 − 3.72 = −3268 kJ/mol, consistent with the −3267.7 kJ/mol obtained from Appendix G formation enthalpies. The correction grows when a reaction makes or consumes many moles of gas.

Reading and scaling thermochemical equations

A thermochemical equation attaches ΔH to the reaction exactly as written. Double the coefficients and ΔH doubles; reverse the reaction and the sign flips. Physical states matter: forming liquid water releases more heat than forming water vapor, because condensation itself is exothermic.

To find the heat for a given amount of substance, convert to moles and use the ΔH per mole of reaction as a conversion factor. Burning 12.5 g of benzene, with ΔH = −3267.7 kJ per mole of benzene, releases (12.5 g / 78.12 g/mol) × 3267.7 kJ/mol = 523 kJ. Write it as q = −523 kJ for the reaction, or as 523 kJ of heat released.

Calorimetry mistakes that flip signs or lose heat

Calorimetry arithmetic is short, so most errors are conceptual.

A reliable final check is physical: if the water warmed, the reaction must have a negative ΔH.

• Missing minus sign: the temperature of the surroundings rose, so q_reaction is negative. Reporting +55.8 kJ/mol for a neutralization that warmed the water is a sign error.
• Wrong mass: in solution calorimetry, use the mass of the whole solution, not just the solute or one of the two mixed volumes.
• Per gram versus per mole: divide by moles of the specified species to get kJ/mol, and say which species.
• Mixing joules and kilojoules: q from mcΔT is in joules; convert before dividing into a kJ/mol answer.
• Treating bomb data as ΔH: a sealed bomb gives ΔU; apply the Δn_gas RT correction when ΔH is required.