Calorimetry

Introduction to Calorimetry

Calorimetry is the experimental technique used to measure the heat transferred during a physical or chemical process. The device used for this measurement is called a calorimeter. The principle behind calorimetry is the law of conservation of energy: the heat released by a system is equal to the heat absorbed by the surroundings (and vice versa).

Basic Principle

\[ q_{\text{system}} + q_{\text{surroundings}} = 0 \]

or simply:

\[ q_{\text{reaction}} = -q_{\text{calorimeter}} \]

The heat absorbed by the calorimeter and its contents is calculated using:

\[ q = mc\Delta T \]

Where:

• \( q \) = heat absorbed or released (J or kJ)
• \( m \) = mass of the substance (g)
• \( c \) = specific heat capacity (J g⁻¹ °C⁻¹)
• \( \Delta T \) = change in temperature (°C or K)

Simple Calorimeter (Polystyrene Cup Calorimeter)

Title: Simple Polystyrene Cup Calorimeter

Uses: Suitable for measuring enthalpy changes in aqueous solutions, such as neutralization reactions, dissolution processes, and displacement reactions.

Construction:

• A polystyrene cup (acts as an insulator)
• A lid with a hole for the thermometer
• A thermometer (0.1°C precision)
• A stirring rod

Advantages:

• Inexpensive and disposable
• Good insulation properties
• Quick to set up

Limitations:

• Not suitable for reactions involving gases
• Some heat loss to surroundings occurs despite insulation
• Cannot measure combustion reactions

Experimental Procedure (Neutralization):

1. Measure a known volume of acid into the polystyrene cup.
2. Record the initial temperature.
3. Measure a known volume of base.
4. Add the base to the acid, stir, and record the highest temperature reached.
5. Calculate the heat released and the enthalpy change per mole.

Worked Example:

Problem: 50 cm³ of 0.5 mol/dm³ HCl at 24.0°C was mixed with 50 cm³ of 0.5 mol/dm³ NaOH at 24.0°C in a polystyrene cup. The maximum temperature reached was 27.3°C. Calculate the enthalpy of neutralization. (Specific heat capacity of solution = 4.18 J g⁻¹ °C⁻¹, density = 1.0 g/cm³)

Solution:

\[ \Delta T = 27.3 – 24.0 = 3.3^\circ C \]

\[ m = 100 \, \text{cm}^3 \times 1.0 \, \text{g/cm}^3 = 100 \, \text{g} \]

\[ q = mc\Delta T = 100 \times 4.18 \times 3.3 = 1379.4 \, \text{J} = 1.3794 \, \text{kJ} \]

\[ n = 0.5 \, \text{mol/dm}^3 \times 0.050 \, \text{dm}^3 = 0.025 \, \text{mol} \]

\[ \Delta H_{\text{neut}} = -\frac{1.3794}{0.025} = -55.18 \, \text{kJ/mol} \]

 Bomb Calorimeter

Title: Bomb Calorimeter for Combustion Reactions

Uses: Specifically designed to measure the enthalpy of combustion of substances, particularly solids and liquids.

Construction:

• A strong stainless steel container (the “bomb”) that can withstand high pressure
• An ignition system with electrical wires
• A known mass of water surrounding the bomb
• A thermometer (often a digital thermistor with high precision)
• An insulated jacket to minimize heat loss

Procedure:

1. A precise mass of the sample is placed in a crucible inside the bomb.
2. The bomb is sealed and filled with pure oxygen at high pressure (about 25–30 atm).
3. The bomb is placed in the calorimeter containing a known mass of water.
4. The initial temperature of the water is recorded.
5. The sample is ignited electrically.
6. The temperature rise is recorded until a maximum is reached.
7. The heat released is calculated using the heat capacity of the calorimeter system.

Calibration:

Bomb calorimeters are calibrated using a substance with a known enthalpy of combustion, such as benzoic acid (ΔH_c° = -3227 kJ/mol). The heat capacity of the calorimeter (C_cal) is determined using:

\[ q_{\text{combustion}} = -C_{\text{cal}} \times \Delta T \]

Worked Example:

Problem: A bomb calorimeter has a heat capacity of 10.2 kJ/°C. When 0.50 g of ethanol (C₂H₅OH) was burned, the temperature of the calorimeter increased from 22.0°C to 24.5°C. Calculate the enthalpy of combustion of ethanol per mole. (Molar mass of ethanol = 46.0 g/mol)

Solution:

\[ \Delta T = 24.5 – 22.0 = 2.5^\circ C \]

\[ q_{\text{calorimeter}} = C_{\text{cal}} \times \Delta T = 10.2 \, \text{kJ/°C} \times 2.5^\circ C = 25.5 \, \text{kJ} \]

\[ q_{\text{combustion}} = -25.5 \, \text{kJ} \quad (\text{heat released by the reaction}) \]

\[ n_{\text{ethanol}} = \frac{0.50 \, \text{g}}{46.0 \, \text{g/mol}} = 0.01087 \, \text{mol} \]

\[ \Delta H_c^\circ = -\frac{25.5}{0.01087} = -2346 \, \text{kJ/mol} \]

Note: This value differs from the theoretical value (-1367 kJ/mol) because the example uses simplified numbers. In practice, calibration ensures accuracy.

Comparison of Calorimeters

Relationship between ΔH and ΔU:

\[ \Delta H = \Delta U + \Delta n_g RT \]

Where:

• \( \Delta n_g \) = change in moles of gas (moles of gaseous products – moles of gaseous reactants)
• \( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \)
• \( T \) = temperature in Kelvin

For reactions involving only solids and liquids,

\[ \Delta H \approx \Delta U \]