The Power of Energy: Your Guide to Chemical Energetics (\(9701\) AS & A Level)
Welcome to Chemical Energetics! This chapter is all about understanding the energy changes that happen during chemical reactions. Why is this important? Because energy determines whether a reaction is practical, safe, or even possible! It’s the driving force behind everything from cooking food to generating electricity.
Don't worry if you find terms like enthalpy and entropy confusing at first. We will break them down step-by-step, using simple analogies to make sure you master this essential area of Physical Chemistry!
Section 1: Fundamental Concepts of Enthalpy (\(\Delta H\)) (AS Level)
What is Enthalpy?
Enthalpy, symbol \(H\), is essentially the total heat content of a chemical system at constant pressure. We cannot measure the absolute enthalpy of a system, but we can measure the Enthalpy Change, \(\Delta H\), which is the heat energy absorbed or released during a reaction.
The standard unit for enthalpy change is kilojoules per mole (\(kJ \ mol^{-1}\)).
Exothermic and Endothermic Reactions
Chemical reactions can release heat or absorb heat. This is classified using the sign of \(\Delta H\).
1. Exothermic Reactions
Definition: Reactions that release thermal energy (heat) to the surroundings.
Feeling: The surroundings get warmer.
Sign of \(\Delta H\): Negative (\(\Delta H < 0\)). This means the products have less energy than the reactants (energy was lost from the system).
Example: Combustion (burning fuel), Neutralisation reactions.
🔥 Memory Aid: EXO means EXIT. Energy Exits the system.
2. Endothermic Reactions
Definition: Reactions that absorb thermal energy (heat) from the surroundings.
Feeling: The surroundings get colder.
Sign of \(\Delta H\): Positive (\(\Delta H > 0\)). This means the products have more energy than the reactants (energy was gained by the system).
Example: Thermal decomposition, Melting ice, Photosynthesis (requires sunlight).
❄️ Memory Aid: ENDO means ENTER. Energy Enters the system.
Reaction Pathway Diagrams
These diagrams visually represent the energy changes during a reaction. They show two key things: the Enthalpy Change (\(\Delta H\)) and the Activation Energy (\(E_a\)).
The Activation Energy (\(E_a\)) is the minimum energy required to start a reaction (the energy needed to break the initial bonds).
Analogy: Think of a rollercoaster.
- You must push the cart (reactants) up the first hill (\(E_a\)).
- The height difference between the start and end point is \(\Delta H\).
Exothermic Diagram Sketch Features:
- Reactants (start) level is higher than Product (end) level.
- \(\Delta H\) arrow points downwards (negative).
- \(E_a\) is the height from reactants to the transition state (peak).
Endothermic Diagram Sketch Features:
- Reactants (start) level is lower than Product (end) level.
- \(\Delta H\) arrow points upwards (positive).
- \(E_a\) is the height from reactants to the transition state (peak).
The sign of \(\Delta H\) tells you where the heat goes. Negative \(\Delta H\) means heat is released (Exothermic). Pathway diagrams show both the overall energy change (\(\Delta H\)) and the energy barrier to start the reaction (\(E_a\)).
Section 2: Definitions and Standard Enthalpy Changes (AS Level)
When we quote enthalpy values, they must refer to Standard Conditions, indicated by the symbol \(\theta\).
Standard Conditions (\(\theta\))
This syllabus assumes standard conditions are:
- Temperature: \(298 \text{ K}\) (\(25^\circ\text{C}\))
- Pressure: \(101 \text{ kPa}\) (1 atmosphere)
- Concentration: \(1.0 \text{ mol}\ \text{dm}^{-3}\) (for solutions)
- Substances in their standard physical states (the state in which they exist naturally at $298\text{ K}$ and $101\text{ kPa}$).
Key Definitions of Standard Enthalpy Changes (\(\Delta H^{\theta}\))
You must know the definitions precisely. They usually involve the formation or reaction of one mole of a specific substance under standard conditions.
1. Standard Enthalpy Change of Reaction (\(\Delta H_r^{\theta}\))
The enthalpy change when reaction occurs in the molar quantities shown in the chemical equation, under standard conditions.
2. Standard Enthalpy Change of Formation (\(\Delta H_f^{\theta}\))
The enthalpy change when one mole of a compound is formed from its constituent elements in their standard states, under standard conditions.
Example: \(\text{C}(\text{graphite}) + 2\text{H}_2(\text{g}) \rightarrow \text{CH}_4(\text{g})\)
Did you know? The \(\Delta H_f^{\theta}\) of any element in its standard state (like \(\text{O}_2(\text{g})\) or \(\text{Na}(\text{s})\)) is, by definition, zero.
3. Standard Enthalpy Change of Combustion (\(\Delta H_c^{\theta}\))
The enthalpy change when one mole of a substance is completely burnt in oxygen, under standard conditions.
Example: \(\text{CH}_4(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})\)
4. Standard Enthalpy Change of Neutralisation (\(\Delta H_{neut}^{\theta}\))
The enthalpy change when one mole of water (\(\text{H}_2\text{O}\)) is formed from the reaction between an acid and an alkali under standard conditions.
Example: \(\text{H}^{+}(\text{aq}) + \text{OH}^{-}(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})\)
Always ensure your equation is balanced correctly so that the substance being defined (e.g., the compound formed for \(\Delta H_f^{\theta}\) or the substance combusted for \(\Delta H_c^{\theta}\)) has a coefficient of 1 mole.
Section 3: Calculating Enthalpy Changes from Experiments (AS Level)
Calorimetry is the experimental process used to measure enthalpy changes directly. This usually involves measuring the temperature change (\(\Delta T\)) of a known mass of water or solution.
The Calorimetry Formula: \(q = mc\Delta T\)
The heat energy (\(q\)) absorbed or released by the *surroundings* (usually water or solution) is calculated using:
\[q = mc\Delta T\]
- \(q\): Heat energy transferred (Joules, J).
- \(m\): Mass of the substance (solution or water) being heated/cooled (grams, g).
- \(c\): Specific heat capacity of the substance (\(J \ g^{-1} \ K^{-1}\) or \(J \ g^{-1} \ ^\circ\text{C}^{-1}\)). (We usually assume the specific heat capacity of the solution is the same as water: \(4.18 \ J \ g^{-1} \ K^{-1}\)).
- \(\Delta T\): Temperature change (Kelvin, K, or Celsius, $^\circ\text{C}$).
Calculating \(\Delta H\) for the Reaction
Once \(q\) is found (in J), we convert it to the enthalpy change of the reaction, \(\Delta H\) (in \(kJ \ mol^{-1}\)).
\[\Delta H = -\frac{q}{(n \times 1000)}\]
The syllabus provides the formula:
\[\Delta H = -\frac{mc\Delta T}{n}\]
Where:
- \(n\): Moles of the reactant that caused the temperature change (the limiting reagent).
- The negative sign is crucial: it converts the heat change measured in the surroundings ($q$, positive if heated) into the enthalpy change of the system ($\Delta H$, negative if exothermic).
Step-by-Step Calculation Process:
- Calculate \(q\): Find the heat transferred using \(q = m \times c \times \Delta T\). (Result in J).
- Convert \(q\) to kJ: Divide by 1000.
- Calculate moles (\(n\)): Find the moles of the reactant that reacted.
- Calculate \(\Delta H\): Divide the heat transferred (in kJ) by the moles (\(n\)). Remember to apply the correct sign (negative for exothermic, positive for endothermic).
- \(q\) (measured by thermometer) is for the surroundings.
- \(\Delta H\) is for the system (reaction).
- If \(T\) increases (\(q\) is positive), the reaction is exothermic, so \(\Delta H\) must be negative. Hence the mandatory minus sign in the final formula.
Section 4: Calculating \(\Delta H\) using Bond Energies (AS Level)
Chemical reactions involve breaking existing bonds (requiring energy) and forming new bonds (releasing energy).
Bond Breaking vs. Bond Making
- Bond Breaking: Requires energy input. This is always an endothermic process (\(\Delta H\) is positive).
- Bond Making: Releases energy. This is always an exothermic process (\(\Delta H\) is negative).
The bond energy (\(\Delta H\)) is the energy required to break one mole of a specific covalent bond in the gaseous state.
Note: The syllabus specifies that some bond energies are exact (for diatomic molecules) and some are averages (for polyatomic molecules, as the energy required for a bond changes slightly based on its environment).
Calculation using Bond Energies
The overall enthalpy change of a reaction is calculated by summing the energy required to break bonds and the energy released when forming new bonds.
\[\Delta H_r = \sum (\text{Energy used for Bond Breaking}) - \sum (\text{Energy released from Bond Making})\]
Since bond energies are always quoted as positive (energy required to break the bond):
\[\Delta H_r = \sum (\text{Bond energies of Reactants}) - \sum (\text{Bond energies of Products})\]
Step-by-Step Calculation Process:
- Draw the structures: Sketch the displayed formula of all reactants and products to count the number of each type of bond.
- Calculate Energy In (Reactants): Sum the energies of all bonds broken (Reactants side). (This sum is positive).
- Calculate Energy Out (Products): Sum the energies of all bonds formed (Products side). (This sum is positive, but represents energy released).
- Find \(\Delta H_r\): Subtract Energy Out from Energy In.
Example: For the reaction $\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$
- Energy In: Break $1 \times (\text{H}-\text{H})$ bond and $1 \times (\text{Cl}-\text{Cl})$ bond.
- Energy Out: Form $2 \times (\text{H}-\text{Cl})$ bonds.
- \(\Delta H_r = (E_{\text{H}-\text{H}} + E_{\text{Cl}-\text{Cl}}) - (2 \times E_{\text{H}-\text{Cl}})\)
Remember the order: R - P (Reactants minus Products). If the energy released (bond formation) is greater than the energy required (bond breaking), the result is negative and the reaction is exothermic.
Section 5: Hess's Law and Enthalpy Cycles (AS Level)
Definition of Hess's Law
Hess's Law states that the total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final conditions are the same.
Analogy: Imagine climbing a hill. The overall change in height (enthalpy change) depends only on where you start and where you finish, not whether you took a direct path or a winding path.
Application: Using Enthalpy of Formation (\(\Delta H_f^{\theta}\))
Hess's Law allows us to calculate an unknown \(\Delta H_r\) if we know the \(\Delta H_f^{\theta}\) values for all reactants and products. We construct a cycle involving the formation of all substances from their elements.
Cycle Structure (Formation):
\(\sum \text{Reactants} \xrightarrow{\Delta H_r^{\theta}} \sum \text{Products}\)
\(\downarrow \sum \Delta H_f^{\theta}(\text{Reactants}) \quad \uparrow \sum \Delta H_f^{\theta}(\text{Products})\)
\(\sum \text{Elements}\)
Using the cyclic rule (sum of \(\Delta H\) in a closed loop is zero):
\[\Delta H_r^{\theta} = \sum \Delta H_f^{\theta}(\text{Products}) - \sum \Delta H_f^{\theta}(\text{Reactants})\]
Application: Using Enthalpy of Combustion (\(\Delta H_c^{\theta}\))
If we know the \(\Delta H_c^{\theta}\) values, we construct a cycle involving the combustion of all substances to form common combustion products (like \(\text{CO}_2\) and \(\text{H}_2\text{O}\)).
Cycle Structure (Combustion):
\(\sum \text{Reactants} \xrightarrow{\Delta H_r^{\theta}} \sum \text{Products}\)
\(\downarrow \sum \Delta H_c^{\theta}(\text{Reactants}) \quad \uparrow \sum \Delta H_c^{\theta}(\text{Products})\)
\(\sum \text{Combustion Products}\)
Using the cyclic rule (note that the reactant combustion arrow is in the same direction as \(\Delta H_r^{\theta}\)):
\[\Delta H_r^{\theta} = \sum \Delta H_c^{\theta}(\text{Reactants}) - \sum \Delta H_c^{\theta}(\text{Products})\]
Always draw the cycle clearly. When reversing an arrow's direction, remember to reverse the sign of its \(\Delta H\) value. When using Formation data, it is P - R (Products minus Reactants). When using Combustion data, it is R - P (Reactants minus Products).
Section 6: A Level Advanced Energetics: Ionic Equilibria and Lattice Energy
This section explores the energy changes involved when forming and dissolving ionic compounds. These concepts require understanding of enthalpy definitions for processes involving ions and atoms.
6.1 Advanced Definitions for Ionic Cycles
These terms are necessary steps in forming an ionic lattice from standard elements:
- Standard Enthalpy Change of Atomisation (\(\Delta H_{at}\)): The enthalpy change when one mole of gaseous atoms is formed from the element in its standard state, under standard conditions. (Always positive, endothermic).
Example: \(\frac{1}{2}\text{Cl}_2(\text{g}) \rightarrow \text{Cl}(\text{g})\) - First Ionisation Energy (\(IE_1\)): The energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous \(1+\) ions. (Always positive, endothermic).
Example: \(\text{Na}(\text{g}) \rightarrow \text{Na}^{+}(\text{g}) + \text{e}^{-}\) - First Electron Affinity (\(EA_1\)): The enthalpy change when one electron is added to each atom in one mole of gaseous atoms to form one mole of gaseous \(1-\) ions. (Usually negative/exothermic, as the attractive force of the nucleus dominates).
Example: \(\text{Cl}(\text{g}) + \text{e}^{-} \rightarrow \text{Cl}^{-}(\text{g})\) - Lattice Energy (\(\Delta H_{latt}\)): The enthalpy change when one mole of a solid ionic lattice is formed from its constituent gaseous ions. (Always highly negative, highly exothermic).
Example: \(\text{Na}^{+}(\text{g}) + \text{Cl}^{-}(\text{g}) \rightarrow \text{NaCl}(\text{s})\)
6.2 The Born-Haber Cycle
The Born-Haber cycle is a specific application of Hess's Law used to determine the lattice energy of an ionic compound, which cannot be measured directly. It links \(\Delta H_f^{\theta}\) to all the steps necessary to turn standard elements into gaseous ions, then into the solid lattice.
Cycle Steps (Example: NaCl):
- Formation of the lattice from elements: \(\text{Na}(\text{s}) + \frac{1}{2}\text{Cl}_2(\text{g}) \xrightarrow{\Delta H_f^{\theta}} \text{NaCl}(\text{s})\)
- Atomisation of Na: \(\text{Na}(\text{s}) \xrightarrow{+\Delta H_{at}(\text{Na})} \text{Na}(\text{g})\)
- Atomisation of Cl: \(\frac{1}{2}\text{Cl}_2(\text{g}) \xrightarrow{+\Delta H_{at}(\text{Cl})} \text{Cl}(\text{g})\)
- Ionisation of Na: \(\text{Na}(\text{g}) \xrightarrow{+IE_1} \text{Na}^{+}(\text{g}) + \text{e}^{-}\)
- Electron affinity of Cl: \(\text{Cl}(\text{g}) + \text{e}^{-} \xrightarrow{+EA_1} \text{Cl}^{-}(\text{g})\)
- Lattice Formation (the unknown step): \(\text{Na}^{+}(\text{g}) + \text{Cl}^{-}(\text{g}) \xrightarrow{\Delta H_{latt}} \text{NaCl}(\text{s})\)
By Hess's Law:
\[\Delta H_f^{\theta} = \Delta H_{at}(\text{Na}) + \Delta H_{at}(\text{Cl}) + IE_1(\text{Na}) + EA_1(\text{Cl}) + \Delta H_{latt}\]
6.3 Factors Affecting Lattice Energy Magnitude
Lattice energy is an indicator of the strength of the ionic bond. A more negative (more exothermic) lattice energy means a stronger bond.
The magnitude of lattice energy is affected by:
- Ionic Charge: Higher charges lead to stronger electrostatic attraction. \(\text{Mg}^{2+}\text{O}^{2-}\) has a much more negative \(\Delta H_{latt}\) than \(\text{Na}^{+}\text{Cl}^{-}\).
- Ionic Radius: Smaller ions mean the charges are closer together, leading to stronger electrostatic attraction. Smaller radii result in a more negative \(\Delta H_{latt}\).
Born-Haber cycles use a sequence of measurable enthalpy changes to calculate the non-measurable Lattice Energy. High charge and small radius lead to high (more negative) lattice energy.
Section 7: Enthalpies of Solution and Hydration (A Level)
When an ionic solid dissolves, it involves two energy steps: breaking the lattice and hydrating the ions.
7.1 Key Definitions
- Standard Enthalpy Change of Hydration (\(\Delta H_{hyd}^{\theta}\)): The enthalpy change when one mole of specified gaseous ions dissolves in water to form an infinitely dilute solution. (Always negative, exothermic).
Example: \(\text{Na}^{+}(\text{g}) \xrightarrow{\Delta H_{hyd}^{\theta}} \text{Na}^{+}(\text{aq})\) - Standard Enthalpy Change of Solution (\(\Delta H_{sol}^{\theta}\)): The enthalpy change when one mole of an ionic solid dissolves in water to form an infinitely dilute solution. (Can be positive or negative).
Example: \(\text{NaCl}(\text{s}) \xrightarrow{\Delta H_{sol}^{\theta}} \text{Na}^{+}(\text{aq}) + \text{Cl}^{-}(\text{aq})\)
7.2 The Enthalpy of Solution Cycle
We use Hess's Law to link the lattice energy (breaking the solid) and the hydration energy (stabilising the ions in water).
Cycle Structure:
\(\text{MX}(\text{s}) \xrightarrow{\Delta H_{sol}^{\theta}} \text{M}^{+}(\text{aq}) + \text{X}^{-}(\text{aq})\)
\(\uparrow -\Delta H_{latt} \quad \downarrow \Delta H_{hyd}^{\theta}(\text{M}^+) + \Delta H_{hyd}^{\theta}(\text{X}^-)\)
\(\text{M}^{+}(\text{g}) + \text{X}^{-}(\text{g})\)
Note: We use the negative of \(\Delta H_{latt}\) because the arrow shown (solid $\rightarrow$ gas ions) is the reverse of the Lattice Energy definition (gas ions $\rightarrow$ solid).
\[\Delta H_{sol}^{\theta} = (-\Delta H_{latt}) + (\Delta H_{hyd}^{\theta}(\text{M}^+) + \Delta H_{hyd}^{\theta}(\text{X}^-))\]
7.3 Factors Affecting Hydration Energy Magnitude
Hydration energy reflects the attractive forces between the ion and the polar water molecules. A more negative (more exothermic) \(\Delta H_{hyd}^{\theta}\) means the ion is more strongly hydrated.
Similar to lattice energy, hydration energy is governed by:
- Ionic Charge: Higher charge leads to stronger attraction to water dipoles. Higher charge means more negative \(\Delta H_{hyd}^{\theta}\).
- Ionic Radius: Smaller radius means a higher charge density (charge concentrated over a small area). Higher charge density means stronger attraction to water dipoles, resulting in a more negative \(\Delta H_{hyd}^{\theta}\).
Dissolving is a competition between the energy needed to break the lattice (\(-\Delta H_{latt}\), endothermic) and the energy gained from surrounding ions with water (\(\Delta H_{hyd}^{\theta}\), exothermic). For solubility, \(\Delta H_{hyd}\) usually needs to be large enough to offset the large \(\Delta H_{latt}\).
Section 8: Entropy (\(\Delta S\)) (A Level)
While enthalpy ($\Delta H$) tells us about heat flow, it doesn't fully determine whether a reaction will happen spontaneously. We also need Entropy.
8.1 Definition of Entropy (\(S\))
Entropy, \(S\), is a measure of the degree of disorder or the number of possible arrangements of the particles and their energy in a given system. Higher disorder means higher entropy.
Units for entropy are typically \(J \ K^{-1} \ mol^{-1}\).
Analogy: Imagine your room. If it is tidy, there is low entropy (few arrangements). If it is messy, there is high entropy (many possible arrangements for things). Nature prefers high entropy (messy systems)!
8.2 Predicting the Sign of Entropy Change (\(\Delta S\))
The sign of the Entropy Change of System (\(\Delta S_{sys}\)) tells us if the disorder increased (positive) or decreased (negative).
\(\Delta S\) is positive (disorder increases) when:
- Changing state: Solid \(\rightarrow\) Liquid \(\rightarrow\) Gas (Gases are highly disordered). (e.g., Melting or Boiling).
- Dissolving: Solid solute \(\rightarrow\) Aqueous ions (particles become spread out).
- Temperature increases: Particles move faster, increasing their random arrangements.
- Gas moles increase: Reactions where the total number of moles of gaseous products is greater than gaseous reactants.
Example of positive \(\Delta S\): \(\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})\) (One mole of gas is produced from a solid, major increase in disorder).
8.3 Calculating Standard Entropy Change (\(\Delta S^{\theta}\))
We can calculate the standard entropy change for a reaction using the standard entropy values (\(S^{\theta}\)) of the products and reactants:
\[\Delta S^{\theta} = \sum S^{\theta}(\text{products}) - \sum S^{\theta}(\text{reactants})\]
Entropy measures disorder. Look for an increase in the number of gaseous molecules or a change from solid to liquid/gas to determine if \(\Delta S\) is positive or negative.
Section 9: Gibbs Free Energy (\(\Delta G\)) and Reaction Feasibility (A Level)
For a process to occur spontaneously (to be feasible), the overall disorder of the universe must increase. This is governed by the Gibbs Free Energy Change, \(\Delta G\).
9.1 The Gibbs Equation
Gibbs free energy combines the roles of enthalpy and entropy at a specific temperature ($T$ in Kelvin):
\[\Delta G^{\theta} = \Delta H^{\theta} - T\Delta S^{\theta}\]
Where:
- \(\Delta G^{\theta}\): Standard Gibbs Free Energy Change (\(kJ \ mol^{-1}\)).
- \(\Delta H^{\theta}\): Standard Enthalpy Change (\(kJ \ mol^{-1}\)).
- \(T\): Absolute Temperature (Kelvin, K).
- \(\Delta S^{\theta}\): Standard Entropy Change (\(J \ K^{-1} \ mol^{-1}\)).
! Important Note on Units: \(\Delta H\) is usually in kJ, while \(\Delta S\) is usually in J. You MUST convert \(\Delta S\) to kJ/K/mol (divide by 1000) before using the equation, or convert \(\Delta H\) to J.
9.2 Predicting Reaction Feasibility
A reaction is considered feasible (thermodynamically possible) if the Gibbs Free Energy Change is negative.
Feasibility Rule: If \(\Delta G^{\theta} \le 0\), the reaction is feasible.
9.3 The Effect of Temperature on Feasibility
Temperature ($T$) plays a crucial role, especially when \(\Delta H\) and \(\Delta S\) have the same sign. We examine four scenarios:
| \(\Delta H\) Sign | \(\Delta S\) Sign | \(\Delta G = \Delta H - T\Delta S\) | Feasibility |
|---|---|---|---|
| Negative (Exothermic) | Positive (Disorder increases) | \(\text{Negative} - (\text{Positive}) = \text{Always Negative}\) | Always Feasible (Spontaneous at all temperatures) |
| Positive (Endothermic) | Negative (Disorder decreases) | \(\text{Positive} - (\text{Negative}) = \text{Always Positive}\) | Never Feasible (Non-spontaneous at all temperatures) |
| Negative (Exothermic) | Negative (Disorder decreases) | Feasible only if \(T\Delta S\) is small. | Feasible at Low Temperatures |
| Positive (Endothermic) | Positive (Disorder increases) | Feasible only if \(T\Delta S\) is large. | Feasible at High Temperatures |
Determining the Crossover Temperature:
If feasibility depends on temperature (Scenarios 3 & 4), you can find the temperature ($T$) at which the reaction changes from feasible to non-feasible. This is the point where \(\Delta G = 0\):
If \(\Delta G = 0\):
\[0 = \Delta H^{\theta} - T\Delta S^{\theta}\]
\[T = \frac{\Delta H^{\theta}}{\Delta S^{\theta}}\]
If the required temperature is exceeded (or falls below this threshold), the reaction will become feasible (or non-feasible).
A reaction is feasible if \(\Delta G\) is negative. Enthalpy and Entropy both contribute to feasibility, with Temperature controlling the magnitude of the entropy term (\(T\Delta S\)).