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Measuring enthalpy changes: R1.1.5 Practical considerations IB DP Chemistry Study Notes

Measuring enthalpy changes: R1.1.5 Practical considerations IB DP Chemistry Study Notes - New Syllabus 2025

Measuring enthalpy changes  – IB DP Chemistry- Study Notes

IITian Academy excellent Introduction to the Particulate Nature of Matter – Study Notes and effective strategies will help you prepare for your IB DP Chemistry 2025 exam.

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Reactivity 1.1.5 — Calorimetry Experiments

 Measuring Enthalpy Changes

Calorimetry is the experimental method used to measure the heat released or absorbed in a chemical or physical process. The basic setup involves a calorimeter (e.g., a polystyrene cup), thermometer, and a known mass of water or solution.

Formula Used:

\( q = mc\Delta T \)

Where:

  •  \( q \) = heat energy (J)
  •  \( m \) = mass of water or solution (g)
  •  \( c \) = specific heat capacity of water = \( 4.18\ \text{J g}^{-1}\text{K}^{-1} \)
  •  \( \Delta T \) = temperature change (K or °C)

Assumptions:

  •  No heat loss to the surroundings
  •  All the heat is absorbed by the water/solution
  •  The density of solution ≈ water (1 g cm−3)

Example 

25.0 cm3 of 2.00 mol dm−3 HCl is mixed with 25.0 cm3 of 2.00 mol dm−3 NaOH in a polystyrene cup. The temperature rises by 6.6 °C. Calculate the enthalpy change per mole of water formed.

▶️Answer/Explanation

Total volume = 50.0 cm3 → mass ≈ 50.0 g
\( q = mc\Delta T = 50.0 \times 4.18 \times 6.6 = 1379.4\ \text{J} = 1.379\ \text{kJ} \)
Moles of water formed = 0.025 mol
\( \Delta H = \frac{-1.379}{0.025} = -55.2\ \text{kJ mol}^{-1} \)

 Enthalpy Changes for Reactions in Solution

 Enthalpy changes for reactions in aqueous solution (e.g., neutralization, precipitation, dissolution) can be measured by calorimetry using a polystyrene cup to minimize heat loss. These reactions typically involve mixing two solutions and measuring the temperature change.

Formula Used: Again, we use

  • \( q = mc\Delta T \)

Then calculate molar enthalpy change:

  • \( \Delta H = \frac{-q}{n} \)

Where:

  •  \( n \) = moles of the limiting reagent

 The negative sign is included because exothermic reactions release heat

Important Considerations:

  •  Total mass = total volume of solutions (assuming 1.00 g cm−3 density)
  •  Use limiting reagent to calculate enthalpy change per mole

Example

50.0 cm3 of 1.00 mol dm−3 CuSO₄ is mixed with 50.0 cm3 of 1.00 mol dm−3 NaOH. A blue precipitate forms and the temperature rises from 21.0 °C to 27.4 °C. Calculate the enthalpy change per mole of limiting reagent.

▶️Answer/Explanation

Total mass = 100.0 g
\( \Delta T = 6.4\ ^\circ \text{C} \)
\( q = mc\Delta T = 100.0 \times 4.18 \times 6.4 = 2675.2\ \text{J} = 2.675\ \text{kJ} \)
Moles of CuSO₄ = 0.050 mol (same for NaOH, 1:2 ratio so CuSO₄ is limiting)
\( \Delta H = \frac{-2.675}{0.050} = -53.5\ \text{kJ mol}^{-1} \)

 Temperature Correction Graphs

In some calorimetry experiments, especially those involving slow heat exchange or delayed reaction (e.g., combustion or solid dissolving), we use a temperature-time graph to estimate the maximum temperature change more accurately.

Why needed?

 Heat loss to the surroundings before and after the reaction skews direct temperature readings.
 By extrapolating lines before and after the reaction, we determine the theoretical highest or lowest temperature the system would have reached.

How to use:

  • Record temperature at regular time intervals before, during, and after mixing or ignition.
  • Plot temperature vs time on a graph.
  • Draw best-fit lines for the “before” and “after” data.
  • Their intersection point is the corrected maximum temperature (used for \( \Delta T \)).

Important Notes:

  • This method assumes linear heat loss/gain rate.
  • Extrapolated value is more accurate than raw thermometer reading.

Example

In a neutralization experiment, a student adds HCl to NaOH and records temperature every 30 seconds. The graph of temperature vs time shows a gradual increase followed by a slow cooling. Extrapolation gives a maximum temperature of 29.8 °C, while the initial temperature was 22.0 °C. Calculate the heat released if the total mass of the solution was 110.0 g.

▶️Answer/Explanation

\( \Delta T = 29.8 – 22.0 = 7.8^\circ \text{C} \)
\( q = mc\Delta T = 110.0 \times 4.18 \times 7.8 = 3582.84\ \text{J} = 3.58\ \text{kJ} \)
If 0.050 mol of HCl was used, then:
\( \Delta H = \frac{-3.58}{0.050} = -71.6\ \text{kJ mol}^{-1} \)

 Enthalpy of Combustion Experiments

 The enthalpy of combustion is the heat released when 1 mole of a substance is burned completely in oxygen. It can be measured using a simple calorimeter where the substance is burned under a beaker of water.

Setup:

  • A known mass of a fuel (e.g., ethanol) is burned in a spirit burner.
  • The heat released is used to raise the temperature of a known mass of water in a metal calorimeter or beaker.
  • Measure the initial and final temperature of the water.

Equation: \( q = mc\Delta T \)

  • m = mass of water (g)
  • c = specific heat capacity of water = 4.18 J g⁻¹ K⁻¹
  • \( \Delta T \) = temperature change of the water

Enthalpy of combustion:

\( \Delta H_c = \frac{-q}{\text{mol of fuel burned}} \)

Common Errors:

  • Heat loss to the surroundings → actual \( \Delta H_c \) is more exothermic than measured.
  • Incomplete combustion leads to soot and less heat transfer.
  • Use of a draught shield and insulating the beaker improves accuracy.

Example

0.92 g of ethanol is burned and heats 100.0 g of water from 20.0 °C to 42.5 °C. Calculate the enthalpy of combustion of ethanol. Molar mass of ethanol = 46.0 g mol⁻¹.

▶️Answer/Explanation

\( \Delta T = 22.5^\circ \text{C} \)
\( q = 100.0 \times 4.18 \times 22.5 = 9405\ \text{J} = 9.405\ \text{kJ} \)
Moles of ethanol = \( \frac{0.92}{46.0} = 0.02\ \text{mol} \)
\( \Delta H_c = \frac{-9.405}{0.02} = -470.25\ \text{kJ mol}^{-1} \)

Note: Literature value for ethanol is around −1370 kJ mol⁻¹, so this experiment has significant heat loss.

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