Energetics and Calorimetry for the ESAT

Measuring Energy Changes

When a chemical reaction occurs, energy is either released into the surroundings (exothermic) or absorbed from the surroundings (endothermic). In a laboratory setting, this energy change is often measured using a technique called calorimetry. By performing a reaction in a container called a calorimeter, we can monitor the temperature change of a known mass of a substance, usually water or an aqueous solution, to calculate the quantity of heat energy transferred.

The Calorimetry Equation

To calculate the heat energy change, $q$, for a process, we use the following formula:

$q = mc\Delta T$

Where:

1. $q$ is the heat energy transferred, measured in Joules ($J$).
2. $m$ is the mass of the substance undergoing the temperature change, measured in grams ($g$). In solution calorimetry, this is the mass of the solution itself.
3. $c$ is the specific heat capacity of the substance. This is the amount of energy required to raise the temperature of 1 gram of a substance by 1 Kelvin ($K$) or 1 degree Celsius ($^{\circ}C$). For water and dilute aqueous solutions, $c$ is taken as $4.18\text{ J g}^{-1}\text{ K}^{-1}$.
4. $\Delta T$ is the change in temperature, calculated as $T_{\text{final}} - T_{\text{initial}}$.

Molar Enthalpy Changes

The heat energy $q$ refers to the energy change for the specific amounts used in the experiment. To find the molar enthalpy change, $\Delta H$, we must relate this energy to the amount of substance in moles, $n$. The enthalpy change is usually expressed in kilojoules per mole ($\text{kJ mol}^{-1}$).

The expression used is:

$\Delta H = -\frac{q}{1000n}$

The negative sign in the formula ensures the correct sign convention for $\Delta H$. If the temperature increases (exothermic), $q$ is positive, making $\Delta H$ negative. If the temperature decreases (endothermic), $q$ is negative, making $\Delta H$ positive.

Solution Calorimetry

In solution calorimetry, two reactants are mixed in an insulated container, such as a polystyrene cup. The cup acts as a calorimeter with low heat capacity and good insulation to minimise heat loss to the surroundings.

Worked Example: Neutralisation

A student mixes $50.0\text{ cm}^3$ of $1.00\text{ mol dm}^{-3}$ hydrochloric acid (HCl) with $50.0\text{ cm}^3$ of $1.00\text{ mol dm}^{-3}$ sodium hydroxide (NaOH). Both solutions were initially at $19.5^{\circ}C$. The maximum temperature reached was $26.0^{\circ}C$. Calculate the enthalpy change of neutralisation for this reaction.

Step 1: Calculate the mass of the solution. Assuming the density of the solution is $1.00\text{ g cm}^{-3}$, the total volume is $50.0 + 50.0 = 100.0\text{ cm}^3$. Therefore, $m = 100.0\text{ g}$.

Step 2: Calculate the temperature change. $\Delta T = 26.0 - 19.5 = 6.5\text{ K}$.

Step 3: Calculate the heat energy change. $q = mc\Delta T = 100.0 \times 4.18 \times 6.5 = 2717\text{ J} = 2.717\text{ kJ}$.

Step 4: Calculate the number of moles of water formed. $n(\text{HCl}) = \text{concentration} \times \text{volume} = 1.00 \times 0.0500 = 0.0500\text{ mol}$. Since the ratio in the equation $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ is $1:1$, the number of moles of water formed is $0.0500\text{ mol}$.

Step 5: Calculate the enthalpy change per mole. $\Delta H = -\frac{2.717}{0.0500} = -54.3\text{ kJ mol}^{-1}$.

Combustion Calorimetry

To measure the enthalpy change of combustion, a known mass of fuel is burned in a spirit burner to heat a known volume of water in a copper calorimeter or beaker.

Worked Example: Combustion of Ethanol

$0.46\text{ g}$ of ethanol ($M_r = 46.0$) was burned, and the heat produced was used to heat $150.0\text{ g}$ of water. The temperature of the water increased by $20.0^{\circ}C$. Calculate the enthalpy change of combustion of ethanol.

Step 1: Calculate the heat energy change of the water. $q = mc\Delta T = 150.0 \times 4.18 \times 20.0 = 12540\text{ J} = 12.54\text{ kJ}$.

Step 2: Calculate the moles of ethanol burned. $n = \text{mass} / M_r = 0.46 / 46.0 = 0.010\text{ mol}$.

Step 3: Calculate the molar enthalpy change. $\Delta H = -\frac{12.54}{0.010} = -1254\text{ kJ mol}^{-1}$.

Assumptions and Sources of Error

Calculated values for enthalpy changes in school laboratories are often lower than data book values due to several factors:

1. Heat loss: Energy is lost to the air and the container rather than being transferred entirely to the water or solution.
2. Incomplete combustion: When using spirit burners, the fuel may not burn completely, producing carbon or carbon monoxide instead of carbon dioxide, which releases less energy.
3. Specific heat capacity of the calorimeter: The container itself (the copper can or glass beaker) absorbs some heat energy, which is often ignored in simple calculations.
4. Density and heat capacity of solutions: We often assume solutions have the same density ($1.00\text{ g cm}^{-3}$) and specific heat capacity ($4.18\text{ J g}^{-1}\text{ K}^{-1}$) as pure water, which may not be strictly accurate for concentrated solutions.