Study Notes: Enthalpy Change, \(\Delta H\) (Chemical Energetics)

Welcome to one of the most fundamental and fascinating topics in A-Level Chemistry: Chemical Energetics! Don't worry if the calculations look daunting; we will break down how to measure and predict the energy changes that drive every reaction around us, from burning fuel to how your body digests food. Understanding $\Delta H$ is key to knowing why reactions happen.

Part 1: The Basics of Enthalpy and Energy Transfer (AS Level)

1.1 What is Enthalpy (\(H\)) and Enthalpy Change (\(\Delta H\))?

The Enthalpy (\(H\)) of a substance is essentially the total heat energy stored inside it (chemical potential energy) at constant pressure. Since we can't measure the absolute enthalpy of a substance, we focus on the Enthalpy Change (\(\Delta H\)), which is the heat energy change during a reaction.

The standard equation for calculating enthalpy change is:

$$\Delta H = H_{\text{products}} - H_{\text{reactants}}$$

Key Definitions: Exothermic vs. Endothermic

Chemical reactions can either release energy or absorb energy from the surroundings:

  • Exothermic Reaction: Releases heat energy into the surroundings. The surroundings get hotter.

    Example: Combustion (burning fuel), Neutralisation reactions, Setting of cement.

    • Since energy is lost by the system (chemicals), the products have less energy than the reactants.

    • Sign Convention: \(\Delta H\) is negative (e.g., \(-500\text{ kJ mol}^{-1}\)).

  • Endothermic Reaction: Absorbs heat energy from the surroundings. The surroundings get colder.

    Example: Photosynthesis, Thermal decomposition, Dissolving certain salts (like ammonium nitrate) in water.

    • Since energy is gained by the system, the products have more energy than the reactants.

    • Sign Convention: \(\Delta H\) is positive (e.g., \(+100\text{ kJ mol}^{-1}\)).

Memory Aid (Sign Convention): EXO is EXITING (energy leaving), so \(\Delta H\) is negative (like debt). ENDO is ENTERING (energy needed), so \(\Delta H\) is positive.

1.2 Reaction Pathway Diagrams

These diagrams show the energy changes during the course of a reaction. They help us understand both $\Delta H$ and the $E_a$.

  • Activation Energy (\(E_a\)): The minimum energy required by colliding particles to start a reaction. It's the "hill" you have to climb to get the reaction going.
  • The overall enthalpy change, $\Delta H$, is the difference between the starting energy (reactants) and the ending energy (products).
Quick Review: Diagrams

In an exothermic reaction, the peak (transition state) is higher than the reactants, but the products level is lower than the reactants (negative $\Delta H$). In an endothermic reaction, the products level is higher than the reactants (positive $\Delta H$).

1.3 Standard Conditions

When comparing reactions, scientists must use a fixed set of conditions called Standard Conditions, denoted by the symbol $\ominus$ (theta).

  • Temperature (\(T\)): \(298\text{ K}\) (or $25\text{ }^{\circ}\text{C}$)
  • Pressure (\(P\)): \(101\text{ kPa}\) (standard atmospheric pressure)
  • Concentration: \(1.0\text{ mol dm}^{-3}\) for solutions.

Note: The standard state is the most stable state of the substance under these standard conditions (e.g., the standard state of water is liquid, \(\text{H}_2\text{O}(l)\), not gas or solid).

Key Takeaway (Part 1): \(\Delta H\) tells you if energy is released (Exothermic, \(\Delta H < 0\)) or absorbed (Endothermic, \(\Delta H > 0\)). The $E_a$ determines how fast the reaction happens, but $\Delta H$ determines the overall energy change.

Part 2: Standard Enthalpy Definitions (AS Level)

It is vital to be able to define and write equations for the following standard enthalpy changes. Remember, all definitions must relate to one mole of the specified substance being formed, burned, or reacted, under standard conditions ($\ominus$).

2.1 Standard Enthalpy Change of Formation (\(\Delta H_f^{\ominus}\))

Definition: The enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.

Example: Formation of liquid water.

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

Note: The $\Delta H_f^{\ominus}$ of any element in its standard state is, by definition, zero (e.g., $\Delta H_f^{\ominus}(\text{O}_2(g)) = 0\text{ kJ mol}^{-1}$).

2.2 Standard Enthalpy Change of Combustion (\(\Delta H_c^{\ominus}\))

Definition: The enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions.

Example: Combustion of methane.

$$\text{CH}_4(g) + 2\text{O}_2(g) \longrightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$$

Note: Combustion reactions are almost always exothermic (\(\Delta H_c^{\ominus}\) is negative).

2.3 Standard Enthalpy Change of Neutralisation (\(\Delta H_{neut}^{\ominus}\))

Definition: The enthalpy change when one mole of water is formed from the reaction between an acid and an alkali under standard conditions.

Example: Reaction between $\text{NaOH}$ and $\text{HCl}$.

$$\text{H}^{+}(aq) + \text{OH}^{-}(aq) \longrightarrow \text{H}_2\text{O}(l)$$

Note: Neutralisation is always exothermic, and the value is nearly constant (\(\approx -57\text{ kJ mol}^{-1}\)) for strong acids/strong bases because the reaction is always the same ionic equation.

Key Takeaway (Part 2): Always define enthalpy changes based on ONE MOLE of the substance being formed/combusted/neutralised, using standard states and standard conditions.

Part 3: Calculating Enthalpy Changes Experimentally (AS Level)

The core experimental method for finding $\Delta H$ is called Calorimetry, which involves measuring the heat transferred to or from a substance (usually water).

3.1 Heat Transfer Calculation (\(q\))

The heat energy (\(q\)) absorbed or released by the substance (e.g., water in a calorimeter) is calculated using the formula:

$$q = mc\Delta T$$

Where:

  • \(q\) = Heat energy change (in Joules, J).
  • \(m\) = Mass of the substance absorbing/releasing heat (in grams, g). Often, this is the mass of water/solution.
  • \(c\) = Specific heat capacity (in \(\text{J g}^{-1}\text{ K}^{-1}\)). Usually, we use the value for water, \(4.18\text{ J g}^{-1}\text{ K}^{-1}\).
  • \(\Delta T\) = Temperature change (\(T_{\text{final}} - T_{\text{initial}}\)) (in Kelvin, K, or $^{\circ}\text{C}$).

3.2 Enthalpy Change Calculation (\(\Delta H\))

Once you have \(q\), you must convert it into the molar enthalpy change ($\Delta H$) using the number of moles ($n$) that reacted. Remember the key formula link:

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

  • The negative sign: This is crucial! $\Delta H$ is the energy change of the reaction (system), while $q$ is the energy change of the solution/water (surroundings). If the surroundings heated up (\(q\) is positive), the reaction must have been exothermic ($\Delta H$ is negative).
  • Units: $\Delta H$ is usually expressed in \(\text{kJ mol}^{-1}\), so you must convert $q$ from J to kJ (divide by 1000).
Step-by-Step Calorimetry Calculation
  1. Measure mass $m$ (of water/solution) and temperature change $\Delta T$.
  2. Calculate heat change in surroundings: $$q = m \times c \times \Delta T$$
  3. Calculate moles $n$ of the reactant used or product formed (whichever defines the specific $\Delta H$).
  4. Calculate the molar enthalpy change: $$\Delta H = -\frac{q}{n}$$ (Remember to divide $q$ by 1000 for kJ units).
Common Mistake Alert!

Students often forget the negative sign! If the temperature RISES, the reaction is exothermic ($\Delta H$ is negative).

Also, make sure the units match: if $c$ is in $\text{J g}^{-1}\text{ K}^{-1}$, $q$ will be in J. You must convert $q$ to kJ before calculating $\Delta H\text{ (kJ mol}^{-1})$.

Key Takeaway (Part 3): Calorimetry measures the energy of the surroundings ($q = mc\Delta T$). We then reverse the sign and divide by moles to find the molar enthalpy of the system ($\Delta H = -q/n$).

Part 4: Calculating Enthalpy Changes using Bond Energies (AS Level)

Chemical reactions involve the breaking of old bonds and the formation of new ones. Energy is required to break bonds, and energy is released when bonds are formed.

4.1 Bond Breaking and Bond Making

  • Bond Breaking: Requires energy input. This is an endothermic process ($\Delta H$ is positive).
  • Bond Making: Releases energy. This is an exothermic process ($\Delta H$ is negative).

We can calculate the overall reaction enthalpy ($\Delta H_r$) using bond energies ($\text{BE}$) with this formula:

$$\Delta H_r = \sum (\text{Energy required to break bonds}) - \sum (\text{Energy released in forming bonds})$$

$$\Delta H_r = \sum (\text{BE}_{\text{reactants}}) - \sum (\text{BE}_{\text{products}})$$

If the answer is negative, the reaction is exothermic overall (more energy released by making bonds than absorbed by breaking them).

4.2 Average vs. Exact Bond Energies

  • Exact Bond Energies: Used for diatomic molecules (like \(\text{H-H}\) or \(\text{Cl-Cl}\)). These values are precise because the bond is in the exact same environment every time.
  • Average Bond Energies: Used for polyatomic molecules (like \(\text{C-H}\) in methane). The actual energy of a \(\text{C-H}\) bond varies slightly depending on which molecule it's in. Therefore, the values used in calculations are averages taken from many different compounds.

Because average bond energies are used, calculations based on them are often estimates, not exact values, which is important to mention in evaluations.

Key Takeaway (Part 4): Calculations using bond energies are typically quick estimates. Remember: Energy In (Breaking, +ve) minus Energy Out (Making, -ve).

Part 5: Hess’s Law and Enthalpy Cycles (AS Level)

5.1 The Principle of Hess's Law

Definition: Hess's Law states that the total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.

This law is extremely useful because many reactions cannot be measured directly (e.g., they might be too slow, too dangerous, or produce multiple side products). Hess's law allows us to find these unknown $\Delta H$ values by constructing an energy cycle using other known $\Delta H$ values.

5.2 Constructing Energy Cycles

We use two main types of cycles: Formation cycles and Combustion cycles.

A. Cycles using Enthalpy of Formation (\(\Delta H_f^{\ominus}\))

These cycles involve elements. The key is that the reactants and products can both be broken down into their constituent elements.

$$\text{Reactants} \xrightarrow{\Delta H_r^{\ominus}} \text{Products}$$

$$\text{Reactants} \xleftarrow{\sum \Delta H_f^{\ominus}} \text{Elements} \xrightarrow{\sum \Delta H_f^{\ominus}} \text{Products}$$

Using Hess’s Law (the path via the elements must equal the direct path):

$$\Delta H_r^{\ominus} = \sum \Delta H_f^{\ominus}(\text{Products}) - \sum \Delta H_f^{\ominus}(\text{Reactants})$$

B. Cycles using Enthalpy of Combustion (\(\Delta H_c^{\ominus}\))

These cycles are used for organic compounds that combust easily. The common intermediary substance is the mixture of complete combustion products ($\text{CO}_2$ and $\text{H}_2\text{O}$).

$$\text{Reactants} \xrightarrow{\Delta H_r^{\ominus}} \text{Products}$$

$$\text{Reactants} \xrightarrow{\sum \Delta H_c^{\ominus}} \text{Products of Combustion} \xleftarrow{\sum \Delta H_c^{\ominus}} \text{Products}$$

Using Hess’s Law (the path via combustion products must equal the direct path):

$$\Delta H_r^{\ominus} = \sum \Delta H_c^{\ominus}(\text{Reactants}) - \sum \Delta H_c^{\ominus}(\text{Products})$$

Trick for Using Cycles: Follow the arrows! If you go against an arrow, change the sign of that $\Delta H$ value.

Did you know?
The field of thermochemistry relies heavily on Hess's Law. It means we can measure the enthalpy of something like the combustion of carbon monoxide ($\text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g)$) even though it's impossible to carry out cleanly in a lab without getting $\text{CO}_2$! We just use the enthalpies of formation of $\text{CO}_2$ and $\text{CO}$ instead.
Key Takeaway (Part 5): Hess's Law allows us to calculate unknown enthalpy changes by summing up known enthalpy changes from different reaction routes (cycles).

Part 6: Lattice Energy and Born-Haber Cycles (A Level Extension)

This section applies Hess's Law to understand the energy changes involved in forming ionic solids, linking gas-phase ions to the solid crystal lattice.

6.1 Definitions for Ionic Compounds

To build an ionic lattice from scratch, we need specific energy terms:

  1. Enthalpy Change of Atomisation (\(\Delta H_{at}^{\ominus}\)): The enthalpy change when one mole of gaseous atoms is formed from an element in its standard state. (Always endothermic, +ve).

    Example: \(\frac{1}{2}\text{Cl}_2(g) \longrightarrow \text{Cl}(g)\)

  2. First Ionisation Energy (IE\(_1\)): Energy required to remove one electron from each atom in one mole of gaseous atoms. (Always endothermic, +ve).

    Example: \(\text{Na}(g) \longrightarrow \text{Na}^{+}(g) + \text{e}^{-}\)

  3. First Electron Affinity (\(EA_1\)): The enthalpy change when one electron is added to each atom in one mole of gaseous atoms. (Usually exothermic, -ve).

    Example: \(\text{Cl}(g) + \text{e}^{-} \longrightarrow \text{Cl}^{-}(g)\)

  4. Lattice Energy (\(\Delta H_{latt}^{\ominus}\)): The enthalpy change when one mole of a solid ionic compound is formed from its gaseous ions. (Always highly exothermic, -ve).

    Example: \(\text{Na}^{+}(g) + \text{Cl}^{-}(g) \longrightarrow \text{NaCl}(s)\)

6.2 The Born-Haber Cycle

The Born-Haber cycle uses Hess's Law to link the standard enthalpy of formation ($\Delta H_f^{\ominus}$) of an ionic compound with all the individual steps needed to turn the elements into gaseous ions, and then into the lattice.

It typically shows that the direct route ($\Delta H_f^{\ominus}$) equals the sum of the energies of the indirect route (atomisation, ionisation, electron affinity, and lattice formation).

$$\Delta H_f^{\ominus} = \Delta H_{at}^{\ominus}(\text{metal}) + \Delta H_{at}^{\ominus}(\text{non-metal}) + \text{IE} + \text{EA} + \Delta H_{latt}^{\ominus}$$

6.3 Factors Affecting Lattice Energy Magnitude

Lattice energy is a measure of the strength of the electrostatic forces between ions in the solid lattice. The magnitude of $\Delta H_{latt}^{\ominus}$ depends on two factors:

  1. Ionic Charge: Higher ionic charges (e.g., $\text{Mg}^{2+}$ vs $\text{Na}^{+}$) result in stronger attraction between the ions. This leads to a more negative (larger magnitude) lattice energy.
  2. Ionic Radius: Smaller ions (smaller radius) can approach each other more closely, leading to stronger attraction (Coulombic forces). This results in a more negative (larger magnitude) lattice energy.

Analogy: Imagine two magnets. Lattice energy is like the strength of the stick. Stronger magnets (higher charge) stick better. Closer magnets (smaller radius) stick better.

Key Takeaway (Part 6): Born-Haber cycles use Hess's law to calculate Lattice Energy, a crucial value derived from five distinct enthalpy terms. Lattice energy magnitude increases with increasing ionic charge and decreasing ionic radius.

Part 7: Enthalpies of Solution and Hydration (A Level Extension)

When an ionic solid dissolves in water, two major energy changes occur: the lattice breaks apart (endothermic) and the ions are surrounded by water molecules (exothermic).

7.1 Definitions for Dissolving

  1. Enthalpy Change of Solution (\(\Delta H_{sol}^{\ominus}\)): The enthalpy change when one mole of an ionic solid dissolves in a large amount of water to form an infinitely dilute solution under standard conditions.

    Example: \(\text{NaCl}(s) + aq \longrightarrow \text{Na}^{+}(aq) + \text{Cl}^{-}(aq)\)

  2. Enthalpy Change of Hydration (\(\Delta H_{hyd}^{\ominus}\)): The enthalpy change when one mole of gaseous ions dissolves in a large amount of water to form an infinitely dilute solution. (Always exothermic, -ve).

    Example: \(\text{Na}^{+}(g) + aq \longrightarrow \text{Na}^{+}(aq)\)

7.2 The Solution Energy Cycle

We can link these terms using Hess's Law:

$$\text{Solid Lattice} \xrightarrow{\Delta H_{sol}^{\ominus}} \text{Aqueous Ions}$$

$$\text{Solid Lattice} \xrightarrow{-\Delta H_{latt}^{\ominus}} \text{Gaseous Ions} \xrightarrow{\sum \Delta H_{hyd}^{\ominus}} \text{Aqueous Ions}$$

This gives the relationship:

$$\Delta H_{sol}^{\ominus} = (-\Delta H_{latt}^{\ominus}) + \sum \Delta H_{hyd}^{\ominus}$$

Note: $-\Delta H_{latt}^{\ominus}$ is the energy required to break the lattice (Lattice Dissociation Energy).

7.3 Factors Affecting Hydration Enthalpy Magnitude

Hydration occurs because water molecules (polar) are attracted to the charged ions. A greater attraction results in a more negative $\Delta H_{hyd}^{\ominus}$ (more energy released).

  • Ionic Charge: Higher ionic charges (e.g., $\text{Mg}^{2+}$ vs $\text{Na}^{+}$) create stronger attraction to water dipoles. This leads to a more negative (larger magnitude) hydration enthalpy.
  • Ionic Radius: Smaller ions result in a higher charge density, creating stronger electric fields. This leads to a more negative (larger magnitude) hydration enthalpy.
Key Takeaway (Part 7): Solubility depends on the balance between the energy required to break the lattice (Lattice Energy) and the energy released when ions are surrounded by water (Hydration Energy). Both terms are affected by the charge and size of the ions in the same way (higher charge, smaller radius = larger energy magnitude).

Part 8: Entropy (\(\Delta S\)) and Gibbs Free Energy (\(\Delta G\)) (A Level Extension)

Enthalpy ($\Delta H$) only tells us about heat transfer. To determine if a reaction is truly possible (feasible), we need to consider Entropy and Gibbs Free Energy.

8.1 Entropy (\(S\)) and Entropy Change (\(\Delta S\))

Definition: Entropy (\(S\)) is a measure of the number of possible arrangements (disorder or randomness) of the particles and their energy in a given system.

  • A system with high disorder (e.g., a gas) has high entropy.
  • A system with low disorder (e.g., a solid crystal) has low entropy.

Predicting the Sign of \(\Delta S\):

The entropy change ($\Delta S$) is positive when disorder increases, and negative when disorder decreases. Disorder typically increases when:

  1. A solid or liquid turns into a gas (\(s \rightarrow l\), \(l \rightarrow g\)). Example: Boiling water.
  2. A substance dissolves or mixes (\(s \rightarrow aq\)). Example: Dissolving salt.
  3. The number of moles of gaseous particles increases during a reaction. Example: \(2\text{SO}_3(g) \rightarrow 2\text{SO}_2(g) + \text{O}_2(g)\) (2 moles of gas become 3 moles of gas, \(\Delta S\) is positive).

Calculating \(\Delta S^{\ominus}\) for a reaction:

$$\Delta S^{\ominus} = \sum S^{\ominus}(\text{Products}) - \sum S^{\ominus}(\text{Reactants})$$

8.2 Gibbs Free Energy (\(\Delta G\)) and Feasibility

The Gibbs free energy change ($\Delta G$) combines enthalpy and entropy to determine the feasibility (spontaneity) of a reaction. A feasible reaction can occur without continuous external intervention.

The Gibbs Equation:

$$\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}$$

Where \(T\) is the absolute temperature in Kelvin (K).

Feasibility Criteria

For a reaction or process to be feasible, the Gibbs Free Energy change ($\Delta G^{\ominus}$) must be negative.

  • If $\Delta G < 0$: The reaction is feasible (spontaneous).
  • If $\Delta G > 0$: The reaction is not feasible (non-spontaneous).
  • If $\Delta G = 0$: The reaction is at equilibrium.
Effect of Temperature on Feasibility

Temperature ($T$) plays a vital role because it multiplies the entropy term ($\Delta S$) in the Gibbs equation. This means temperature can change whether a reaction is feasible or not.

\(\Delta H\) Sign \(\Delta S\) Sign \(\Delta G = \Delta H - T\Delta S\) Feasibility
Negative (Exothermic) Positive (More disorder) Always Negative Always feasible (Spontaneous at all temperatures)
Positive (Endothermic) Negative (Less disorder) Always Positive Never feasible (Non-spontaneous at all temperatures)
Negative (Exothermic) Negative (Less disorder) Negative only if \(T\Delta S\) is small. Feasible only at Low Temperatures
Positive (Endothermic) Positive (More disorder) Negative only if \(T\Delta S\) is large. Feasible only at High Temperatures
Key Takeaway (Part 8): Feasibility is determined by $\Delta G$. A reaction is possible if $\Delta G$ is negative. Always ensure $T$ is in Kelvin and that $\Delta H$ and $T\Delta S$ are in the same units (usually kJ).