This practical guide outlines how to determine the enthalpy change of a reaction in aqueous solution (such as a neutralisation or displacement reaction) using a simple coffee-cup calorimeter. Accurate temperature readings over time and graphical extrapolation are used to correct for heat loss to the surroundings.
Core Specification Link
This practical is directly linked to topic 3.1.4 Energetics, focusing on the relation \( q = mc\Delta T \), standard enthalpy definitions, and Hess's law calculations.
Aim
To measure the temperature change over time for the reaction between hydrochloric acid and sodium hydroxide solution, and to calculate the enthalpy change of neutralisation.
Equipment List
- Expanded polystyrene cup (acts as the calorimeter)
- Plastic lid with a hole for the thermometer
- Glass beaker (250 cm³, to support the polystyrene cup)
- Thermometer (graduated to 0.1 °C or 0.2 °C) or temperature probe
- Measuring cylinders (50 cm³)
- Stirring rod (glass)
- Stopwatch or timer
Experimental Method
- Place the polystyrene cup inside the 250 cm³ glass beaker to provide stability and additional insulation.
- Using a measuring cylinder, transfer exactly 25.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid (HCl) into the polystyrene cup.
- Place the thermometer into the acid . Start the stopwatch and record the temperature of the acid every 30 seconds for 2 minutes to establish a baseline.
- Measure 25.0 cm³ of 1.00 mol dm⁻³ sodium hydroxide solution (NaOH) in a second measuring cylinder.
- At exactly 2 minutes and 30 seconds, do not record the temperature. Instead, quickly add the sodium hydroxide solution to the polystyrene cup.
- Immediately place the lid on the cup, insert the thermometer, and stir the reaction mixture gently.
- Record the temperature of the solution every 30 seconds starting from 3 minutes, up to 7 minutes.
- Plot a graph of temperature (y-axis) against time (x-axis) . Draw two lines of best fit: one through the temperature readings taken prior to mixing, and one through the readings taken during cooling after mixing.
- Extrapolate both lines to the time of mixing (2.5 minutes) to determine the theoretical maximum temperature change (ΔT) if no heat loss had occurred.
Safety & Risk Assessment
| Hazard | Risk | Precaution |
|---|---|---|
| 1.00 mol dm⁻³ Hydrochloric acid | Skin and eye irritation. | Wear safety goggles and a lab coat. Wash any skin splashes immediately. |
| 1.00 mol dm⁻³ Sodium hydroxide | Skin irritation and potential eye damage. | Wear safety goggles. Wash skin spills immediately with water. |
| Glassware and thermometer | Cuts from broken glass. | Handle with care. Avoid using the thermometer as a stirring rod. Place in secure positions. |
Results & Calculations
Temperature-Time Extrapolation Graph
Because heat loss starts as soon as the reaction begins, the temperature change cannot be read simply as the difference between initial and maximum temperatures. Instead, we plot the temperature against time and extrapolate back to the time of mixing (2.5 minutes) as shown below:
Assume specific heat capacity \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\) and density of solution \(\rho = 1.00\text{ g cm}^{-3}\).
Step 1: Calculate the total mass of the solution (\(m\))
\[ \text{Total volume} = 25.0 + 25.0 = 50.0\text{ cm}^3 \] \[ \text{Mass } m = \text{volume} \times \text{density} = 50.0\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 50.0\text{ g} \]Step 2: Calculate the heat energy transferred (\(q\))
\[ q = m \times c \times \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.80\text{ K} = 1421.2\text{ J} = 1.4212\text{ kJ} \]Step 3: Calculate the moles of limiting reagent used (\(n\))
\[ n(\text{HCl}) = \text{concentration} \times \text{volume (dm}^3) = 1.00\text{ mol dm}^{-3} \times 0.0250\text{ dm}^3 = 0.0250\text{ mol} \] \[ n(\text{NaOH}) = 1.00\text{ mol dm}^{-3} \times 0.0250\text{ dm}^3 = 0.0250\text{ mol} \]Since the reaction ratio is 1:1, 0.0250 moles of water are formed.
Step 4: Calculate the standard molar enthalpy change (ΔH)
\[ \Delta H_{\text{neut}} = -\frac{q}{n} = -\frac{1.4212\text{ kJ}}{0.0250\text{ mol}} = -56.8\text{ kJ mol}^{-1} \]The enthalpy change of neutralisation is -56.8 kJ mol⁻¹ (3 significant figures).
Sources of Error & Improvements
| Error Source | Classification | Consequence & Mitigation |
|---|---|---|
| Heat loss to the surroundings | Systematic | Heat escapes through the cup walls and lid, reducing the maximum recorded temperature and making ΔH less exothermic. Mitigation: Use a double-insulated polystyrene cup, add a lid, or use a vacuum flask. |
| Heat capacity of the cup and thermometer | Systematic | A small amount of heat is absorbed by the cup and the thermometer rather than the water. Mitigation: Account for the calorimeter constant (heat capacity of calorimeter) in calculations. |
| Uncertainty in thermometer reading | Random | Reading errors on analogue thermometers introduce uncertainty. Mitigation: Use a digital temperature probe connected to a datalogger. |
| Assuming density and specific heat capacity of solution are equal to water | Systematic | Solutes modify specific heat capacity and density. Mitigation: Use precise literature specific heat capacities for NaCl solutions. |
Exam questions, marked instantly
Write your answer and the AI examiner checks it against the mark scheme, showing exactly which marks you earned and what to add.
Exam-style question
3 marksExplain why extrapolating the cooling curve gives a more accurate temperature change than using the highest temperature recorded.
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- Heat is lost to the surroundings throughout the reaction [1]
- So the highest recorded temperature is lower than the true maximum temperature [1]
- Extrapolating back to the time of mixing estimates the temperature rise if no heat had been lost [1]
Exam-style question
2 marksA student dissolves anhydrous copper(II) sulfate in water in a polystyrene cup. State two assumptions made when calculating the enthalpy change.
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Any two from:
- The density of the solution is 1.00 g cm⁻³, so the mass of solution equals the volume of water [1]
- The specific heat capacity of the solution is the same as water, 4.18 J g⁻¹ K⁻¹ [1]
- All the heat released is transferred to the solution, with none lost to the surroundings or the cup [1]
CPAC Skills Assessed
- CPAC 2: Design and apply investigative approaches to measure thermal energy changes.
- CPAC 3: Safely use chemical substances and basic calorimetry apparatus.
- CPAC 4: Make and record temperature measurements over time with appropriate precision.
When calculating \(q = mc\Delta T\), ensure that \(m\) is the mass of the solution being heated (in this neutralisation reaction, the sum of both solution volumes: 50.0 g), not the moles or the mass of any dry reagent. Do not forget to include the negative sign in the final answer if the temperature increased (exothermic reaction).