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Specified Practical Work

Determining enthalpy changes by calorimetry

  • 3 things to know
  • Component 3

Overview

Recall questions on the apparatus and procedure, the excess-reagent and matched-solution requirements of an indirect determination, applying q = mcΔT, extrapolating the temperature back to mixing, the main sources of error, and worked determinations checked against the accepted value.

What you need to know

Answer each one before you reveal it.

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Key term

Describe how a temperature against time graph is used to correct a calorimetry result for heat loss.

Key term

Explain why the enthalpy change for magnesium oxide reacting with carbon dioxide is found indirectly rather than by direct measurement.

Key term

Why is an experimental enthalpy change of combustion always less exothermic than the accepted value?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Which two quantities must be measured in a simple calorimetry experiment before an enthalpy change can be calculated?

    Model answer

    The mass of the solution being heated or cooled and the change in its temperature must both be measured, along with the amount in moles of the limiting reagent.

  2. 2

    Why is an expanded polystyrene cup with a lid used rather than a glass beaker?

    Model answer

    Polystyrene is a poor conductor so it loses little heat to the surroundings, it has a very small heat capacity of its own, and the lid cuts losses by evaporation and convection.

  3. 3

    Describe how a temperature against time graph is used to correct a calorimetry result for heat loss.

    Model answer

    The temperature is recorded at fixed intervals before the reagents are mixed and for several minutes afterwards, and the straight cooling section after the peak is extended backwards to the moment of mixing. The temperature read at that point gives the rise that would have been obtained with no heat loss.

  4. 4

    Why is the solution in a calorimetry experiment assumed to have the density and specific heat capacity of water?

    Model answer

    The solutions used are dilute and mostly water, so taking the density as 1.00 g cm⁻³ and the specific heat capacity as 4.18 J g⁻¹ K⁻¹ is a close enough approximation for the volume to be converted to a mass and the heat calculated.

  5. 5

    Why is the acid used in excess when the enthalpy change for a solid reacting with acid is measured?

    Model answer

    An excess ensures that all of the weighed solid dissolves and reacts, so the heat measured corresponds exactly to the known amount of solid rather than to an unknown fraction of it.

  6. 6

    When 0.500 g of magnesium oxide is added to 50.0 cm³ of excess hydrochloric acid the temperature rises by 8.9 °C. Calculate the enthalpy change per mole of magnesium oxide.

    Model answer

    The heat released is 50.0 × 4.18 × 8.9 = 1860 J, and the amount of MgO is 0.500 ÷ 40.3 = 0.0124 mol. Dividing gives 1.50 × 10⁵ J mol⁻¹, so the enthalpy change is −150 kJ mol⁻¹.

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