AP Chemistry · Free-response skill

A calorimetry calculation, with sign and units

The move. Turn a coffee-cup calorimetry run into a molar enthalpy: compute q, find the moles of the substance, give ΔH with its sign and units, and explain how a heat leak biases the result.

How to do it

  1. Find the heat with q=mcΔTq = mc\Delta T for the solution, using the mass of the solution and ΔT as final minus initial. Write the unit, joules, and change to kilojoules when you need kJ/mol.
  2. The process you study and the solution trade heat, so their q values have opposite signs. Decide the sign of ΔH from which way the heat moved.
  3. Divide by the moles of the substance named in the question, found from its mass and molar mass.
  4. For an error, ask what it did to the measured temperature change, then follow that change through q=mcΔTq = mc\Delta T to the final answer.

Practice prompt Try it

A student dissolves 15.0 g of potassium nitrate, KNO₃ (101.11 g/mol), in 85.0 g of water in a coffee-cup calorimeter. The temperature of the solution falls from 25.0 °C to 12.6 °C. Assume the solution has a mass of 100.0 g and a specific heat of 4.18 J/(g·°C), and that the cup absorbs no heat.

Answer each part below. Show your work and include units.

  1. Find the amount of heat, in joules, that the solution lost.
  2. Find the number of moles of KNO₃ that dissolved.
  3. Find ΔH for dissolving KNO₃, in kJ per mole of KNO₃, with the correct sign.
  4. In fact the cup is not perfectly insulated, and heat from the warmer room leaks into the cold solution during the run. Say whether the ΔH the student calculates will be greater or less than the true value, and say why.

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Checklist

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0 of 5 points

Model answer, annotated

(a) ΔT = 12.6 − 25.0 = −12.4 °C. q = mcΔT = 100.0 g × 4.18 J/(g·°C) × (−12.4 °C) = −5.18 × 10³ J for the solution, so the solution lost 5.18 × 10³ J (5183 J).

  • Point 1 Uses final minus initial temperature and gives the heat with its unit, joules.

(b) Moles KNO₃ = 15.0 g ÷ 101.11 g/mol = 0.148 mol.

  • Point 2 Changes grams to moles with the molar mass given.

(c) The solution lost 5.18 kJ, so the dissolving absorbed 5.18 kJ: q for the process is +5.18 kJ. ΔH = +5.183 kJ ÷ 0.1484 mol = +34.9 kJ/mol of KNO₃. The positive sign shows that dissolving is endothermic.

  • Point 3 Flips the sign because the solution cooled, changes joules to kilojoules, and divides by the moles that dissolved.

(d) The calculated ΔH will be less than the true value. Heat leaking in from the room partly rewarms the solution, so its temperature falls by less than it would in a perfect calorimeter. A smaller ΔT gives a smaller q from q = mcΔT, so dividing by the same 0.148 mol gives a ΔH that is too small.

  • Point 4 States the direction of the error first.
  • Point 5 Traces the error through the measurement: heat leaks in, smaller ΔT, smaller q.