Introduction: Why Do Chemical Reactions Happen?
Welcome to one of the most exciting topics in A2 Chemistry! Have you ever wondered why an ice cube melts at room temperature, why iron rusts all on its own, or why baking soda and vinegar react instantly without any heating? In this chapter, we explore the driving forces of the universe: Enthalpy (energy), Entropy (disorder), and Free Energy (the ultimate decider of whether a reaction can happen).
Don't worry if these terms sound a bit formal right now. We will break down every single idea step by step with clear everyday analogies, simple calculations, and handy tips to help you ace your exams!
Section 1: Lattice Enthalpy & Born–Haber Cycles
What is Lattice Enthalpy?
Ionic compounds like sodium chloride (\(\text{NaCl}\)) form giant three-dimensional crystalline lattices. Oppositely charged ions are held together by powerful electrostatic forces of attraction. Lattice Enthalpy is a measure of the strength of these ionic bonds.
Standard Lattice Enthalpy of Formation (\(\Delta_{\text{latt}}H^{\ominus}\)):
The enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions (\(298\text{ K}\) and \(100\text{ kPa}\)).
Example: \(\text{Na}^+(g) + \text{Cl}^-(g) \rightarrow \text{NaCl}(s)\)
Because strong bonds are being formed, lattice formation is always exothermic (the value of \(\Delta_{\text{latt}}H^{\ominus}\) is always negative).
Note: Sometimes questions define lattice enthalpy in terms of lattice dissociation (\(\text{NaCl}(s) \rightarrow \text{Na}^+(g) + \text{Cl}^-(g)\)), which is endothermic and has an equal positive value. Always check which direction your question is referring to!
Key Enthalpy Terms You Need for Born–Haber Cycles
To calculate lattice enthalpy indirectly, we use an energy cycle called a Born–Haber cycle (an application of Hess's Law). Here are the building blocks you must know:
• Standard Enthalpy of Formation (\(\Delta_f H^{\ominus}\)): The enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.
Example: \(\text{Na}(s) + \frac{1}{2}\text{Cl}_2(g) \rightarrow \text{NaCl}(s)\)
• Standard Enthalpy of Atomisation (\(\Delta_{\text{at}}H^{\ominus}\)): The enthalpy change when one mole of gaseous atoms is formed from an element in its standard state.
Example for metal: \(\text{Na}(s) \rightarrow \text{Na}(g)\)
Example for non-metal: \(\frac{1}{2}\text{Cl}_2(g) \rightarrow \text{Cl}(g)\)
Atomisation is always endothermic (positive) because you must break bonds.
• First Ionisation Energy (\(\Delta_{\text{IE1}}H^{\ominus}\)): The enthalpy change when one mole of electrons is removed from one mole of gaseous atoms to form one mole of gaseous \(1+\) ions.
Example: \(\text{Na}(g) \rightarrow \text{Na}^+(g) + \text{e}^-\) (Always endothermic)
• First Electron Affinity (\(\Delta_{\text{EA1}}H^{\ominus}\)): The enthalpy change when one mole of electrons is added to one mole of gaseous atoms to form one mole of gaseous \(1-\) ions.
Example: \(\text{Cl}(g) + \text{e}^- \rightarrow \text{Cl}^-(g)\) (Exothermic for the 1st electron affinity because the incoming electron is attracted to the positive nucleus).
• Second Electron Affinity (\(\Delta_{\text{EA2}}H^{\ominus}\)): The enthalpy change when one mole of electrons is added to one mole of gaseous \(1-\) ions to form one mole of gaseous \(2-\) ions.
Example: \(\text{O}^-(g) + \text{e}^- \rightarrow \text{O}^{2-}(g)\)
Common Mistake to Avoid: Second electron affinities are always endothermic (positive) because energy is needed to overcome the repulsion between a negative ion and a negative electron!
Constructing a Born–Haber Cycle Step-by-Step
Think of a Born–Haber cycle as climbing up and down an energy ladder:
1. Start at the baseline: Elements in their standard states (e.g., \(\text{Na}(s) + \frac{1}{2}\text{Cl}_2(g)\)).
2. Climb UP (Endothermic steps):
- Atomise the metal: \(\text{Na}(s) \rightarrow \text{Na}(g)\)
- Atomise the non-metal: \(\frac{1}{2}\text{Cl}_2(g) \rightarrow \text{Cl}(g)\)
- Ionise the metal: \(\text{Na}(g) \rightarrow \text{Na}^+(g) + \text{e}^-\)
3. Climb DOWN (Exothermic steps):
- Add electron to non-metal (1st EA): \(\text{Cl}(g) + \text{e}^- \rightarrow \text{Cl}^-(g)\)
- Form the lattice (\(\Delta_{\text{latt}}H^{\ominus}\)): \(\text{Na}^+(g) + \text{Cl}^-(g) \rightarrow \text{NaCl}(s)\)
4. Direct route: The direct formation of \(\text{NaCl}(s)\) from elements is \(\Delta_f H^{\ominus}\).
Using Hess's Law (Total energy going clockwise = Total energy going anticlockwise):
\(\Delta_f H^{\ominus} = \Delta_{\text{at}}H^{\ominus}(\text{metal}) + \text{IE}(\text{metal}) + \Delta_{\text{at}}H^{\ominus}(\text{non-metal}) + \text{EA}(\text{non-metal}) + \Delta_{\text{latt}}H^{\ominus}\)
Factors Affecting the Size of Lattice Enthalpy
Lattice enthalpy depends directly on the strength of electrostatic attraction between ions. This is governed by two factors:
• Ionic Charge: Greater charge means stronger attraction. For example, \(\text{Mg}^{2+}\) attracts \(\text{O}^{2-}\) much more strongly than \(\text{Na}^+\) attracts \(\text{Cl}^-\), leading to a far more exothermic lattice enthalpy.
• Ionic Radius: Smaller ions can pack closer together, which increases the electrostatic attraction and makes the lattice enthalpy more exothermic.
Quick Review: The most exothermic lattice enthalpies come from small, highly charged ions (e.g., \(\text{Al}_2\text{O}_3\) or \(\text{MgO}\)).
Theoretical vs. Experimental Lattice Enthalpy (Covalent Character)
Scientists can calculate a theoretical lattice enthalpy using the purely ionic model (assuming ions are perfect, hard spheres). When compared to the experimental value from the Born–Haber cycle:
• Purely Ionic Compounds (e.g., \(\text{NaCl}\)): Theoretical and experimental values are almost identical.
• Compounds with Covalent Character (e.g., \(\text{AgI}\)): The experimental value is significantly more exothermic than the theoretical value.
Why? A small, highly charged cation polarises (distorts) the electron cloud of a large anion. This electron sharing gives the compound covalent character, making the bonding stronger than purely ionic forces alone.
Key Takeaway for Section 1: Born–Haber cycles use Hess's Law to calculate lattice enthalpy. Stronger ionic bonding (smaller ions, higher charges) leads to more exothermic lattice enthalpies, and polarisation adds extra stability via covalent character.
Section 2: Enthalpy of Solution & Hydration
What Happens When an Ionic Solid Dissolves?
When you drop table salt into water, two processes happen simultaneously:
1. The ionic lattice is broken apart into gaseous ions (requires energy = lattice dissociation enthalpy).
2. Water molecules surround the separated ions (releases energy = hydration enthalpy).
Definitions
• Enthalpy of Solution (\(\Delta_{\text{sol}}H^{\ominus}\)): The enthalpy change when one mole of an ionic solid dissolves completely in sufficient water to form an infinitely dilute solution under standard conditions.
Example: \(\text{NaCl}(s) + \text{aq} \rightarrow \text{Na}^+(aq) + \text{Cl}^-(aq)\) (Can be endothermic or exothermic).
• Enthalpy of Hydration (\(\Delta_{\text{hyd}}H^{\ominus}\)): The enthalpy change when one mole of specified gaseous ions dissolves in water to form an infinitely dilute solution under standard conditions.
Example: \(\text{Na}^+(g) + \text{aq} \rightarrow \text{Na}^+(aq)\)
Hydration is always exothermic because bonds (ion-dipole attractions) are formed between ions and polar water molecules.
The Solution Cycle Equation
We can link these terms using Hess's Law:
\(\Delta_{\text{sol}}H^{\ominus} = \sum \Delta_{\text{hyd}}H^{\ominus}(\text{cations}) + \sum \Delta_{\text{hyd}}H^{\ominus}(\text{anions}) - \Delta_{\text{latt}}H^{\ominus}(\text{formation})\)
Or alternatively, in terms of lattice breaking (dissociation):
\(\Delta_{\text{sol}}H^{\ominus} = \text{Lattice Dissociation Enthalpy} + \sum \Delta_{\text{hyd}}H^{\ominus}\)
Did you know? If dissolving an ionic solid is endothermic (like \(\text{NH}_4\text{NO}_3\) in instant cold packs), the solution absorbs heat from the surroundings and turns freezing cold. But why does it dissolve if it's endothermic? The answer lies in our next topic: entropy!
Key Takeaway for Section 2: Dissolving is a balance between breaking the lattice (endothermic) and hydrating the ions (exothermic).
Section 3: Entropy (\(S^{\ominus}\))
What is Entropy?
Think of your bedroom. Does it stay tidy on its own? No! Without effort, it naturally becomes messy and disordered. In chemistry, nature behaves the exact same way.
Entropy (\(S^{\ominus}\)) is a measure of the degree of disorder or randomness in a system, or the number of ways that particles and energy quanta can be arranged. Its units are \(\text{J K}^{-1}\text{mol}^{-1}\).
Factors that Increase Entropy
• Change of State: Moving from solid \(\rightarrow\) liquid \(\rightarrow\) gas increases entropy dramatically because particles become much more free to move randomly.
\(\text{Solid (lowest entropy)} < \text{Liquid} \ll \text{Gas (highest entropy)}\)
• Number of Moles: Reactions that produce more moles of substance (especially gas molecules) lead to an increase in entropy (\(\Delta S > 0\)).
Example: \(\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)\)
Here, \(1\text{ mole of solid} \rightarrow 1\text{ mole of solid} + 1\text{ mole of gas}\). A gas is produced, so disorder increases (\(\Delta S\) is positive).
• Temperature: Increasing temperature gives particles more kinetic energy, increasing disorder and therefore entropy.
Calculating the Standard Entropy Change (\(\Delta S^{\ominus}\))
You can calculate the entropy change of a chemical reaction using standard entropy values:
\(\Delta S^{\ominus} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})\)
Worked Example:
Calculate \(\Delta S^{\ominus}\) for: \(\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)\)
Given \(S^{\ominus}\) values: \(\text{N}_2 = 192\), \(\text{H}_2 = 131\), \(\text{NH}_3 = 193\text{ J K}^{-1}\text{mol}^{-1}\).
• \(\sum S^{\ominus}(\text{products}) = 2 \times 193 = 386\text{ J K}^{-1}\text{mol}^{-1}\)
• \(\sum S^{\ominus}(\text{reactants}) = 192 + (3 \times 131) = 192 + 393 = 585\text{ J K}^{-1}\text{mol}^{-1}\)
• \(\Delta S^{\ominus} = 386 - 585 = -199\text{ J K}^{-1}\text{mol}^{-1}\)
Does this make sense? Yes! We went from \(4\text{ moles of gas}\) to only \(2\text{ moles of gas}\). The system became more ordered, so \(\Delta S^{\ominus}\) is negative.
Key Takeaway for Section 3: Entropy measures disorder. Gases have much higher entropy than solids. \(\Delta S^{\ominus} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})\).
Section 4: Gibbs Free Energy (\(\Delta G^{\ominus}\)) & Feasibility
What Determines if a Reaction Can Happen?
For a reaction to occur spontaneously (meaning it is feasible without continuous outside intervention), the overall entropy of the universe must increase. In the lab, we combine enthalpy (\(\Delta H\)) and entropy (\(\Delta S\)) into a single master equation: Gibbs Free Energy.
The Gibbs Equation:
\(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}\)
Where:
• \(\Delta G^{\ominus}\) = Gibbs Free Energy change (\(\text{kJ mol}^{-1}\))
• \(\Delta H^{\ominus}\) = Enthalpy change (\(\text{kJ mol}^{-1}\))
• \(T\) = Temperature in Kelvin (\(\text{K}\)), where \(\text{K} = ^{\circ}\text{C} + 273\)
• \(\Delta S^{\ominus}\) = Entropy change (\(\text{J K}^{-1}\text{mol}^{-1}\))
The Golden Rule of Feasibility
A reaction is thermodynamically feasible only when:
\(\Delta G^{\ominus} \le 0\)
(That is, \(\Delta G^{\ominus}\) must be negative or zero).
🚨 The #1 Exam Trap: The Unit Mismatch!
Look at the units carefully: \(\Delta H^{\ominus}\) is in \(\text{kJ mol}^{-1}\), but \(\Delta S^{\ominus}\) is given in \(\text{J K}^{-1}\text{mol}^{-1}\)!
Before using the Gibbs equation, you must divide \(\Delta S^{\ominus}\) by \(1000\) to convert it into \(\text{kJ K}^{-1}\text{mol}^{-1}\).
How Temperature Affects Feasibility
Because \(T\) is always positive in Kelvin, the signs of \(\Delta H\) and \(\Delta S\) determine how temperature influences \(\Delta G\):
1. \(\Delta H\) is Negative (Exothermic) and \(\Delta S\) is Positive (More Disordered):
\(\Delta G\) will always be negative at any temperature. The reaction is feasible at all temperatures.
2. \(\Delta H\) is Positive (Endothermic) and \(\Delta S\) is Negative (More Ordered):
\(\Delta G\) will always be positive at any temperature. The reaction is never feasible at any temperature.
3. \(\Delta H\) is Positive (Endothermic) and \(\Delta S\) is Positive (More Disordered):
Feasible only at high temperatures (where the \(T\Delta S\) term becomes large enough to outweigh \(\Delta H\)). Example: Thermal decomposition of limestone.
4. \(\Delta H\) is Negative (Exothermic) and \(\Delta S\) is Negative (More Ordered):
Feasible only at low temperatures (where the \(-T\Delta S\) term remains small). Example: Freezing water.
Calculating the Temperature at which a Reaction Becomes Feasible
A reaction just becomes feasible when \(\Delta G^{\ominus} = 0\). Setting the equation to zero gives:
\(0 = \Delta H^{\ominus} - T\Delta S^{\ominus}\)
\(\implies T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}}\)
Worked Example:
A reaction has \(\Delta H^{\ominus} = +178\text{ kJ mol}^{-1}\) and \(\Delta S^{\ominus} = +161\text{ J K}^{-1}\text{mol}^{-1}\). Calculate the minimum temperature for feasibility.
1. Convert \(\Delta S^{\ominus}\) to \(\text{kJ}\): \(\Delta S^{\ominus} = \frac{161}{1000} = 0.161\text{ kJ K}^{-1}\text{mol}^{-1}\)
2. Apply the formula: \(T = \frac{178}{0.161} = 1105.6\text{ K}\)
3. In Celsius: \(1105.6 - 273 = 832.6^{\circ}\text{C}\)
Thermodynamic Feasibility vs. Kinetic Stability
Exam favourite question: "Why does a reaction with a negative \(\Delta G^{\ominus}\) not happen immediately at room temperature?"
Answer: While \(\Delta G^{\ominus}\) tells us if a reaction is thermodynamically feasible, it tells us nothing about the rate of the reaction! If the reaction has a very high activation energy (\(E_a\)), it will occur too slowly to be observed (it is kinetically stable / kinetically inert). A catalyst or spark may be needed to get it started.
Key Takeaway for Section 4: A reaction is feasible when \(\Delta G^{\ominus} \le 0\). Use \(\Delta G = \Delta H - T\Delta S\), and remember to convert \(\Delta S\) by dividing by \(1000\)! Feasible reactions may still be slow if activation energy is high.
Quick Summary & Revision Checklist
Before sitting your exam, make sure you can:
• Define lattice formation enthalpy and state the factors affecting it (ionic charge and ionic radius).
• Construct a full Born–Haber cycle and calculate an unknown value like lattice enthalpy.
• Explain the difference between theoretical and experimental lattice enthalpies using covalent character and polarisation.
• Define hydration enthalpy and solution enthalpy, and connect them with a Hess's Law cycle.
• Explain entropy as disorder and calculate \(\Delta S^{\ominus} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})\).
• Use \(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}\) with correct unit conversions.
• Find the temperature at which a reaction becomes feasible (\(T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}}\)).
• Distinguish clearly between thermodynamic feasibility (\(\Delta G \le 0\)) and kinetic stability (high activation energy).