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
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.
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.
q(p) = ΔH
Measure the total solution mass, measure DT, compute q(solution) with q = mcDT, flip the sign for q(reaction), then divide by the moles of limiting reactant. Skipping the sign flip or the mole step is what turns an otherwise correct experiment into a wrong answer.
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.
Heat escapes through the lid gap and into the cup itself, so the measured temperature rise for an exothermic reaction is smaller than the true value and the magnitude of ΔH comes out too low. Slow mixing, a thermometer reading taken too late, and an inaccurate concentration all push in the same direction.
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.
An insulated open container used for constant-pressure measurements in solution.
At constant pressure the measured heat equals the enthalpy change of the reaction.
Divide q(reaction) in kJ by the moles of limiting reactant to obtain ΔH in kJ/mol.
Dilute aqueous solutions are taken as 1.00 g/mL and 4.18 J/g C.
Lost heat lowers the measured DT and makes exothermic ΔH values less negative than the true value.
ΔH = q(reaction) / n(limiting)
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.
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.
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.
Why it's wrong: The solution and the reaction exchange heat in opposite directions.
Check instead: Flip the sign once, deliberately, before dividing by moles.
Why it's wrong: Only the limiting reactant sets how much reaction actually occurred.
Check instead: Compare moles of both reactants first.
Why it's wrong: The dissolved solid is part of the solution being warmed or cooled.
Check instead: Add the solute mass to the solvent mass.
Why it's wrong: Lost heat lowers DT, which shrinks the magnitude.
Check instead: Heat loss always drags an exothermic result toward zero.
No practice questions are available for this topic yet. You can still practice the whole unit.
A coffee-cup calorimeter runs at constant pressure, so the heat measured is the enthalpy change directly. Measure the temperature change of the solution, use q = mcDT for the solution, flip the sign to get q(reaction), then divide by the moles of the limiting reactant to get ΔH in kJ/mol. Heat lost to the room and to the cup makes exothermic results look too small in magnitude, which is why extrapolating the cooling curve back to mixing time gives a better DT.
Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0. License