LaboratoryCourse level: AP Chemistry

Lab: Coffee-Cup Calorimetry and Heat of Reaction

What you'll be able to do: Turn raw calorimetry measurements into an enthalpy of reaction in kJ/mol and explain the main sources of error.

Best after: Specific Heat and Calorimetry Calculations

Introduction

A polystyrene cup, a thermometer and some water are enough to measure an enthalpy change. The chemistry is in the last two steps, converting joules of water heating into kilojoules per mole of reaction, and being honest about what the cup lets escape.

These are recommended, not required. You can start this lesson at any time.

Learning objectives

  • Convert a measured temperature change into an enthalpy of reaction in kJ/mol
  • Identify the limiting reactant in a calorimetry experiment
  • Explain the effect of heat loss on the measured value
  • Justify the assumptions made about solution density and specific heat

Why constant pressure matters

A coffee cup is open to the atmosphere, so the reaction happens at constant pressure. Under that condition the heat exchanged is equal to the enthalpy change, which is exactly the quantity thermochemical equations report.

The connection qp = ΔH assumes the pressure is effectively constant and the only work is pressure-volume work. For an electrical or electrochemical apparatus doing additional work, that short statement needs modification. In ordinary cup experiments, measure the temperature change of the solution and cup, then infer the chemical heat through conservation; the thermometer does not directly read a reaction enthalpy.

q(p) = ΔH

The four-step procedure

Measure total solution mass and ΔT; compute solution heat (and calibrated cup heat when supplied); use an energy balance to obtain reaction heat; then divide by the amount corresponding to the requested molar basis. For ν mol of limiting reactant per mole of reaction as written, ξ = n(limiting)/ν and ΔrH = q(reaction)/ξ. For 2 H₂ + O₂ → 2 H₂O, reacting 0.20 mol H₂ corresponds to 0.10 mol of reaction, not 0.20 mol. Dividing by moles of H₂ instead reports heat per mole H₂.

Suppose 50.0 mL of 1.00 M HCl reacts with 40.0 mL of 1.00 M NaOH. The supplied amounts are 0.0500 mol H⁺ and 0.0400 mol OH⁻, so 0.0400 mol water forms and acid remains in excess. With additive volumes, density 1.00 g/mL, c = 4.18 J/(g·°C), ΔT = 6.00 °C and Ccal = 10.0 J/°C, qsurroundings = (90.0)(4.18)(6.00) + (10.0)(6.00) = 2317.2 J. Thus qreaction = −2.3172 kJ and ΔH = −2.3172/0.0400 = −57.9 kJ/mol water. Dividing by the excess 0.0500 mol acid would assign heat to acid that did not react.

Name the basis in kJ per mole of reaction as written, reactant consumed, or product formed. Those values differ when the relevant stoichiometric coefficient is not 1.

Standard assumptions

Dilute aqueous solutions are treated as having density 1.00 g/mL and specific heat 4.18 J/g C. The polystyrene cup is assumed to absorb negligible heat unless a calorimeter constant is supplied. Both assumptions are approximations and should be stated when reporting a result.

Both starting solutions should have known, preferably equal temperatures. If their temperatures differ, account for their thermal mixing before attributing the whole thermometer change to reaction. The water-like density and c approximations are most reasonable for dilute solutions; concentrated salts or acids can differ appreciably. Include all solution mass, but do not double-count the cup if the supplied calibration already represents the whole calorimeter assembly.

Sources of error

For an exothermic reaction, heat escaping to air or absorbed by an unaccounted cup can reduce the measured solution temperature rise, making the inferred molar enthalpy less negative. A concentration error has no universal direction: if reaction heat is held fixed, overestimating reacting moles reduces |q/n| and underestimating them increases it; in a real run the reaction extent and measured heat may also change. Trace each error through q and the chosen mole denominator separately.

Record temperature before, during and after mixing rather than one late reading. Extrapolating the post-reaction cooling trend back toward the mixing time estimates the rise before heat loss, if the lab method permits it. For an endothermic event the surroundings can supply heat from the room, reducing the observed cooling and biasing a positive ΔH toward zero. Incomplete reaction also reduces measured heat, but its effect on molar ΔH depends on whether the denominator uses the actually reacted amount or the planned amount. Trace both parts of q/n.

Plotting temperature against time and extrapolating the cooling line back to the moment of mixing recovers the peak temperature that heat loss hid.

Bomb calorimetry, briefly

A bomb calorimeter is sealed and rigid, so its volume is constant. It measures internal energy change rather than enthalpy, and it is calibrated with a known reaction to obtain a heat capacity in J/C for the whole assembly.

Include the cup's heat when it is calibrated

If the calorimeter has a measured heat capacity, both solution and cup absorb or release heat. Add qsolution = mcΔT and qcup = CcalΔT before changing sign for the reaction. Omitting the cup understates the magnitude of reaction heat when both warm during an exothermic event.

Coffee-cup reaction heat balance. Read this visual alongside the section; the text alternative describes its scientific content and assumptions.
q_reaction = -(mcΔT + CcalΔT) when the cup and solution are the surroundings.

Key ideas

Definition
Coffee-cup calorimeter

An insulated open container used for constant-pressure measurements in solution.

Rule
Enthalpy from heat

At constant pressure the measured heat equals the enthalpy change of the reaction.

Rule
Per-mole reporting

For reaction enthalpy per mole as written, divide q(reaction) in kJ by ξ = n(limiting)/ν(limiting). For a named species basis, divide by moles of that species consumed or formed.

Assumption
Solution properties

Dilute aqueous solutions are taken as 1.00 g/mL and 4.18 J/g C.

Key concept
Effect of heat loss

Lost heat lowers the measured DT and makes exothermic ΔH values less negative than the true value.

Equation
Molar enthalpy and reaction extent

ΔrH = q(reaction)/ξ; ξ = n(limiting)/ν(limiting)

  • q(reaction) = heat for the reacting amount, in kJ
  • ξ = reaction extent in mol of reaction as written
  • ν(limiting) = positive stoichiometric coefficient of the limiting reactant

Worked examples

Worked example 1

50.0 mL of 1.00 M HCl at 21.0 C is mixed with 50.0 mL of 1.00 M NaOH at 21.0 C. The temperature rises to 27.8 C. Find ΔH of neutralisation in kJ/mol.

Try it first: Work out the total solution mass and the moles of water formed before doing any energy arithmetic.

    0 of 5 steps revealed.

    Worked example 2

    In the same experiment the student reports -52 kJ/mol. Give the most likely reason and state whether the true value is more or less negative.

      0 of 3 steps revealed.

      Worked example 3

      2.00 g of NH₄NO₃ (M = 80.04 g/mol) dissolves in 100.0 g of water and the temperature falls from 22.4 C to 20.9 C. Find ΔH of solution in kJ/mol.

        0 of 5 steps revealed.

        Worked example 4

        A reaction warms 100.0 g of solution by 5.00 °C. Take c = 4.18 J/(g·°C), Ccal = 20.0 J/°C, and 0.0500 mol reacted. Find ΔH per mole.

        Try it first: Account for both solution and cup.

          0 of 3 steps revealed.

          Common mistakes

          Common mistakes and better checks
          MistakeWhy it's wrongCheck instead
          Reporting ΔH with the same sign as the solution q.The solution and the reaction exchange heat in opposite directions.Flip the sign once, deliberately, before dividing by moles.
          Dividing by the moles of the excess reactant.Only the limiting reactant sets how much reaction actually occurred.Compare moles of both reactants first.
          Using only the water mass when a solid dissolves in it.The dissolved solid is part of the solution being warmed or cooled.Add the solute mass to the solvent mass.
          Claiming heat loss makes an exothermic ΔH too negative.Lost heat lowers DT, which shrinks the magnitude.Heat loss always drags an exothermic result toward zero.

          Scroll horizontally if needed to read all columns.

          After the lesson · Question bank

          Independent practice

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          What you should now know

          At constant pressure, reaction heat equals the enthalpy change for the amount reacted, provided only pressure-volume work is involved. Find solution heat from q = mcΔT, include a calibrated cup heat if supplied, and reverse the surroundings’ heat sign. State the molar basis: per mole of reaction as written, divide by reaction extent ξ = n(limiting)/ν(limiting); per mole of a named reactant or product, divide by that species’ stoichiometric amount. Retain guard digits and round only the reported result. Heat loss can make an exothermic result less negative.

          • Constant pressure means the measured heat is ΔH
          • Use the total solution mass, solvent plus solute
          • Flip the sign from solution to reaction exactly once
          • Divide by reaction extent or the explicitly requested species amount to reach kJ/mol
          • Heat loss biases exothermic results toward zero

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          Sources and further reading

          • Chemistry 2e, Section 5.2: Calorimetry
            Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 5.2
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            Lesson-specific rights and provenance need review.

            The current online Chemistry 2e edition lists CC BY-NC-SA 4.0. The source version used for this lesson and its applicable rights remain unverified.

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            Recorded attribution / license claim (applicability unverified): Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0.

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