Enthalpy changes cannot be measured directly because we cannot measure the total heat content of a substance. Instead, we measure the temperature change of the surroundings during a chemical reaction. This experimental technique is called calorimetry.
🔑 Key Principle
In calorimetry, the chemical reaction is the system, and the liquid medium (usually water or an aqueous solution) is the surroundings. We assume that all the heat released or absorbed by the reaction is transferred to or from this liquid medium. By measuring the temperature change of the liquid, we can calculate the heat energy transferred.
The experimental process of measuring the heat energy change of a chemical reaction, typically by monitoring the temperature change of a known mass of water or solution.
The amount of heat energy required to raise the temperature of one gram of a substance by one Kelvin (or one degree Celsius). For water and dilute aqueous solutions, the value is taken to be \( 4.18\text{ J g}^{-1}\text{ K}^{-1} \).
The Calorimetry Equation
To calculate the heat energy transferred during a reaction, we use the following equation:
\[ q = mc\Delta T \]
Where:
- \( q \) is the heat energy transferred, in Joules (\( \text{J} \)).
- \( m \) is the mass of the surroundings being heated or cooled, in grams (\( \text{g} \)). In solution calorimetry, this is the mass of the solution (often assumed to have a density of \( 1.0\text{ g cm}^{-3} \), so \( 50\text{ cm}^3 = 50\text{ g} \)). In combustion calorimetry, this is the mass of the water in the copper cup, not the fuel.
- \( c \) is the specific heat capacity (\( 4.18\text{ J g}^{-1}\text{ K}^{-1} \) for water).
- \( \Delta T \) is the change in temperature (\( T_{\text{final}} - T_{\text{initial}} \)), in Kelvin (\( \text{K} \)) or degrees Celsius (\( ^\circ\text{C} \)).
A temperature change has the same numerical value in Kelvin as it does in degrees Celsius. For example, if a temperature rises from \( 20.0\text{ }^\circ\text{C} \) to \( 25.5\text{ }^\circ\text{C} \), the change is \( 5.5\text{ }^\circ\text{C} \) or \( 5.5\text{ K} \). You do not need to add 273.15 to the difference, only to absolute temperatures!
Calculating Enthalpy Change per Mole (\( \Delta H \))
Once \( q \) has been calculated, we must convert it into the molar enthalpy change, \( \Delta H \), which is measured in \( \text{kJ mol}^{-1} \):
\[ \Delta H = -\frac{q}{n \times 1000} \]
Where:
- \( n \) is the amount in moles of the limiting reactant (for solution reactions) or the fuel burned (for combustion reactions).
- We divide by 1000 to convert the heat energy \( q \) from Joules (\( \text{J} \)) to kilojoules (\( \text{kJ} \)).
- A negative sign is added if the reaction is exothermic (temperature increased), and a positive sign is used if the reaction is endothermic (temperature decreased).
Experimental Setups
There are two main experimental methods required for AQA A-Level Chemistry: solution calorimetry and combustion calorimetry.
1. Solution Calorimetry (Displacement, Neutralisation, or Dissolving)
This is carried out using a simple polystyrene cup calorimeter. Polystyrene is an excellent thermal insulator, which helps minimise heat loss to the surroundings. A lid is placed on top to reduce heat loss through convection, and the cup is placed inside a glass beaker for stability and additional insulation.
Assume the density of the solution is \( 1.00\text{ g cm}^{-3} \) and its specific heat capacity is \( 4.18\text{ J g}^{-1}\text{ K}^{-1} \).
Solution:
1. Identify the mass of the solution surroundings (\( m \)):
\[ m = 50.0\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 50.0\text{ g} \]
2. Calculate the temperature change (\( \Delta T \)):
\[ \Delta T = 31.2 - 19.5 = 11.7\text{ K} \]
3. Calculate the heat energy transferred (\( q \)):
\[ q = m \times c \times \Delta T = 50.0 \times 4.18 \times 11.7 = 2445.3\text{ J} = 2.4453\text{ kJ} \]
4. Calculate the moles of the limiting reactant copper(II) sulfate (\( n \)):
\[ n = c \times V = 1.00\text{ mol dm}^{-3} \times \frac{50.0}{1000}\text{ dm}^3 = 0.0500\text{ mol} \]
5. Calculate the enthalpy change (\( \Delta H \)):
\[ \Delta H = -\frac{q}{n} = -\frac{2.4453}{0.0500} = -48.9\text{ kJ mol}^{-1}\text{ (3 s.f.)} \]
The negative sign must be included because the temperature increased, which indicates an exothermic reaction.
2. Combustion Calorimetry
This is used to measure the standard enthalpy of combustion of a liquid fuel (e.g., alcohol). A known mass of fuel is burned in a spirit burner under a copper can (calorimeter) containing a measured mass of water. The heat from the flame heats the water. We weigh the spirit burner before and after the experiment to determine the mass of fuel burned.
Errors and Improvements in Calorimetry
Simple lab calorimetry experiments often yield enthalpy values that are significantly lower than data book values. You must be able to explain these errors and suggest improvements in exam questions:
| Experimental Source of Error | Effect on Results | Suggested Improvement |
|---|---|---|
| Heat loss to the surroundings (air, thermometer, cup). | Measured temperature change (\( \Delta T \)) is too small; calculated \( \Delta H \) is less exothermic. | Use a lid, wrap the cup in mineral wool insulation, or use draft shields around the flame. |
| Incomplete combustion of fuel (soot forms on copper can). | Less heat energy is released per gram of fuel; calculated \( \Delta H \) is less exothermic. | Ensure a plentiful supply of oxygen (e.g., burn in pure oxygen using a bomb calorimeter). |
| Evaporation of fuel from the wick of the spirit burner. | The measured mass of fuel burned is too high; calculated \( \Delta H \) is less exothermic. | Put the cap back on the spirit burner immediately after extinguishing the flame. |
| Heat capacity of calorimeter is ignored (heat goes into copper can/polystyrene cup). | The calculated heat energy (\( q \)) is too low; calculated \( \Delta H \) is less exothermic. | Include the heat capacity of the container in the calculations, or calibrate the system. |
Temperature Extrapolation (Cooling Curves)
For slow reactions, heat loss to the surroundings happens simultaneously with the reaction. This prevents the solution from reaching its theoretical maximum temperature. To correct for this heat loss, we can record the temperature at regular intervals (e.g., every minute) both before and after mixing the reactants, and plot a cooling curve.
💡 Correcting for Heat Loss Step-by-Step
- Record the temperature of the solution in the cup every minute for 3 minutes.
- At the 4th minute, add the second reactant but do not record the temperature (allow time for mixing).
- Continue recording the temperature every minute from the 5th minute to the 10th (or 15th) minute.
- Plot a graph of temperature against time.
- Draw a line of best fit through the points before mixing, and another line through the cooling points after mixing.
- Extrapolate the cooling line back to the 4th minute (the mixing time).
- The difference between this extrapolated value and the initial temperature line at the 4th minute gives the corrected, theoretical maximum temperature change (\( \Delta T \)).
Calculate the enthalpy change of the reaction, assuming the specific heat capacity of the solution is \( 4.18\text{ J g}^{-1}\text{ K}^{-1} \). (The magnesium was in excess).
Solution:
1. Calculate the corrected temperature change (\( \Delta T \)):
\[ \Delta T = 38.6 - 20.2 = 18.4\text{ K} \]
2. Calculate the heat energy transferred (\( q \)):
\[ q = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 18.4\text{ K} = 7691.2\text{ J} = 7.6912\text{ kJ} \]
3. Calculate the moles of hydrochloric acid that reacted:
\[ n(\text{HCl}) = c \times V = 0.500\text{ mol dm}^{-3} \times \frac{100.0}{1000}\text{ dm}^3 = 0.0500\text{ mol} \]
4. Write the balanced equation to find the limiting reactant stoichiometry:
\[ \text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)} \]
The standard enthalpy change is calculated per mole of the reaction as written, or per mole of Mg reacting. Since 2 moles of \( \text{HCl} \) react per mole of Mg, the moles of reaction is:
\[ n_{\text{reaction}} = \frac{0.0500\text{ mol}}{2} = 0.0250\text{ mol} \]
5. Calculate the enthalpy change (\( \Delta H \)):
\[ \Delta H = -\frac{q}{n_{\text{reaction}}} = -\frac{7.6912}{0.0250} = -308\text{ kJ mol}^{-1}\text{ (3 s.f.)} \]
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