Enthalpy (HSC SSCE Chemistry): Revision Notes

Enthalpy

What is enthalpy?

Enthalpy is a fundamental concept in chemistry that helps us understand energy changes during chemical reactions. Think of enthalpy as a measure of the total energy stored within a substance or group of substances. This energy is primarily the chemical energy held in the bonds between atoms.

Change in enthalpy ($\Delta H$)

The change in enthalpy for a chemical reaction, represented by the symbol $\Delta H$, is defined as the heat absorbed per mole of specified reactant or product when the reaction occurs at constant pressure.

Most laboratory experiments occur at constant atmospheric pressure (in containers open to the air). This means that the heat absorbed or released during a reaction is a direct measure of $\Delta H$.

Understanding $\Delta H$

The change in enthalpy represents the difference in energy between products and reactants:

$\Delta H = \text{enthalpy of products} - \text{enthalpy of reactants}$

Endothermic and exothermic reactions

Chemical reactions can either absorb or release energy, which determines the sign of $\Delta H$:

Endothermic reactions absorb heat from their surroundings. In these reactions, the products have more energy than the reactants because energy has been absorbed during the reaction. For endothermic reactions, $\Delta H$ is positive ($\Delta H > 0$).

Exothermic reactions release heat to their surroundings. In these reactions, the products have less energy than the reactants because energy has been released. For exothermic reactions, $\Delta H$ is negative ($\Delta H < 0$).

Graphical representation

The relationship between enthalpy changes and the energy levels of reactants and products can be visualised in energy diagrams:

Writing enthalpy changes

When expressing enthalpy changes, we write the balanced chemical equation followed by the $\Delta H$ value. For example, for the combustion of methane:

$\text{CH}_4(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{g}) \quad \Delta H = -890 \text{ kJ mol}^{-1}$

The term $\Delta H$ is sometimes called the heat of reaction.

Specifying amounts - the importance of "per mole"

The definition of $\Delta H$ includes the phrase "per mole of specified reactant or product" because chemical equations can be written in different ways. This specification is crucial to avoid ambiguity.

For example, consider the formation of water from hydrogen and oxygen. We could write:

$2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O}(\text{l})$

or

$\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l})$

These equations describe the same reaction but in different amounts. To be clear, we must specify which substance we're referring to:

• $\Delta H = -572$ kJ per mole of oxygen
• $\Delta H = -286$ kJ per mole of water (or per mole of hydrogen)

Alternatively, we write the specific equation:

$\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l}) \quad \Delta H = -286 \text{ kJ mol}^{-1}$

Here, "per mole" means per mole of the reaction as written - in this case, per mole of hydrogen or per mole of water formed.

The importance of physical states

Changes of state (solid to liquid, liquid to gas, etc.) involve absorption or release of energy. Therefore, enthalpy changes depend on the physical state of both reactants and products.

Enthalpy of solution

Definition

The molar enthalpy of solution (or heat of solution), $\Delta H_{\text{soln}}$, is the heat absorbed when one mole of a substance dissolves in a large excess of water.

If heat is absorbed during dissolution, $\Delta H_{\text{soln}}$ is positive and the process is endothermic. If heat is released, $\Delta H_{\text{soln}}$ is negative and the process is exothermic.

Dissolution of different substances

Covalent molecular substances (such as glucose, ethanol or iodine) dissolve by forming individual molecules that move independently through the solution. For glucose:

$\text{C}_6\text{H}_{12}\text{O}_6(\text{s}) \rightarrow \text{C}_6\text{H}_{12}\text{O}_6(\text{aq}) \quad \Delta H_{\text{soln}} = +11 \text{ kJ mol}^{-1}$

Ionic substances dissolve by breaking up into separate ions, which then move freely and independently through the solution. This process is called dissociation. For potassium nitrate:

$\text{KNO}_3(\text{s}) \rightarrow \text{K}^+(\text{aq}) + \text{NO}_3^-(\text{aq}) \quad \Delta H_{\text{soln}} = 35 \text{ kJ mol}^{-1}$

Measuring enthalpy changes experimentally

Enthalpy changes for many chemical reactions can be measured by mixing known amounts of reactants at the same temperature in a well-insulated container and measuring the temperature after the reaction is complete.

Experimental setup

A simple calorimeter can be made using polystyrene foam cups, which provide good insulation and have negligible heat capacity:

The key features of this setup are:

• Nested polystyrene cups for insulation to minimise heat loss
• A thermometer to measure temperature changes
• A small beaker to add one reactant to the other
• Water or solution as one of the reactants

Calculation procedure

To calculate molar enthalpy changes from experimental data, follow these steps:

Step 1: Calculate the amount of heat released or absorbed using:

$q = mc\Delta T$

where:

• $q$ is the heat absorbed or released (in joules)
• $m$ is the mass of the final solution (in kilograms)
• $c$ is the specific heat capacity of the solution (typically assumed to be $4.18 \times 10^3$ J K$^{-1}$ kg$^{-1}$, the value for water)
• $\Delta T$ is the temperature change (final temperature − initial temperature, in °C or K)

Step 2: Calculate the number of moles that reacted:

$\text{number of moles} = \frac{\text{mass}}{\text{molar mass}}$

Step 3: Calculate the heat released or absorbed per mole:

$\text{heat per mole} = \frac{\text{total heat released or absorbed}}{\text{number of moles that reacted}}$

Step 4: Determine the sign of $\Delta H$:

• $\Delta H$ is positive for heat absorbed
• $\Delta H$ is negative for heat released

Worked example: combustion reactions

Worked example: reaction in solution

Important assumptions in calorimetry

When performing these calculations, we typically make several assumptions:

Remember!