Entropy, Lattice Energy and Born-Haber Cycles

Q: What is entropy, and what are its standard SI units?

Entropy (\f$S\f$) is a measure of the disorder, randomness, or dispersal of matter and energy among particles in a system, measured in $\text{J mol}^{-1}\text{ K}^{-1}$.

Practice

Chemistry (YCH11) · Pearson Edexcel International A Level · 14 min read · Updated: 22 Oct 2026

Welcome to Entropy, Lattice Energy and Born-Haber Cycles

In IAS Chemistry (Unit 2), you explored enthalpy changes ($\Delta H$) and learned that exothermic reactions release heat while endothermic reactions absorb heat. But have you ever wondered why ice melts spontaneously at room temperature even though melting is endothermic? Or why ammonium nitrate dissolves in water and cools the beaker down to near freezing all by itself?

The answer lies beyond enthalpy alone. In this chapter, we explore entropy (nature's tendency toward disorder and energy dispersal), feasibility, and the energetics of ionic bonding through lattice energy and Born-Haber cycles. Don't worry if this seems like a lot at first—we will break down each calculation and concept step-by-step!

---

Part 1: Understanding Entropy ($S$)

What is Entropy?

Entropy (symbol $S$) is a measure of the disorder, randomness, or dispersal of matter and energy among the particles in a system. The units of entropy are $\text{J mol}^{-1}\text{ K}^{-1}$. Notice that entropy is measured in joules, whereas enthalpy is usually quoted in kilojoules!

The Third Law of Thermodynamics: Absolute Zero

At absolute zero ($0\text{ K}$), a perfect, pure crystalline substance has completely stationary particles arranged in a perfect lattice. There is only one possible arrangement of its particles and energy. Therefore, a perfect crystal at $0\text{ K}$ has an entropy of zero ($S = 0\text{ J mol}^{-1}\text{ K}^{-1}$). As the temperature rises above $0\text{ K}$, particles gain thermal energy, vibrate, rotate, and move, so their entropy is always positive ($S > 0$).

Factors that Change the Entropy of a System ($\Delta S_{\text{system}}$)

Whenever matter or energy becomes more dispersed, entropy increases ($\Delta S_{\text{system}} > 0$). Entropy increases significantly in the following situations:

1. Changes of State:

Particles in a solid are locked in fixed positions with low disorder. In a liquid, particles can slide past one another, increasing disorder. In a gas, particles move rapidly and randomly throughout the entire volume, resulting in huge disorder. Therefore:

$S(\text{solid}) < S(\text{liquid}) \ll S(\text{gas})$

Boiling produces a much larger increase in entropy than melting because particles transition from being close together in a liquid to occupying a vastly larger volume with complete spatial freedom as a gas.

2. Dissolving a Solid in a Liquid:

When an ionic lattice dissolves in water, the highly ordered crystal breaks apart into freely moving aqueous ions dispersed throughout the solvent. This dispersal usually leads to an overall increase in entropy ($\Delta S_{\text{system}} > 0$).

3. Change in the Number of Moles of Gas:

Gases have vastly higher standard molar entropies than solids or liquids. If a chemical reaction produces more moles of gas than it consumes, the entropy of the system increases dramatically:

$\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})\quad (\Delta S_{\text{system}} \text{ is strongly positive because } 0\text{ moles of gas} \rightarrow 1\text{ mole of gas})$

Conversely, if gas moles decrease, entropy decreases:

$\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g})\quad (\Delta S_{\text{system}} \text{ is strongly negative because } 4\text{ moles of gas} \rightarrow 2\text{ moles of gas})$

Calculating $\Delta S^{\ominus}_{\text{system}}$

Standard molar entropies ($S^{\ominus}$) are tabulated at standard conditions ($100\text{ kPa}$ and $298\text{ K}$). You can calculate the entropy change of the system by taking the sum of the standard entropies of the products minus the sum of the standard entropies of the reactants:

$\Delta S^{\ominus}_{\text{system}} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})$

Key Takeaway for Part 1: Entropy measures energy and matter dispersal. Solids have low entropy, liquids intermediate, and gases very high entropy. An increase in the number of gaseous particles always gives a large positive $\Delta S_{\text{system}}$.

---

Part 2: Total Entropy Change ($\Delta S_{\text{total}}$) and Reaction Feasibility

The Surroundings and $\Delta S_{\text{surroundings}}$

A reaction doesn't happen in isolation—it exchanges heat with its surroundings (the rest of the universe). When an exothermic reaction occurs ($\Delta H$ is negative), heat flows from the system into the surroundings. This thermal energy causes the particles in the surroundings to move faster, increasing their dispersal and disorder. Thus, exothermic reactions increase the entropy of the surroundings!

The mathematical relationship is:

$\Delta S_{\text{surroundings}} = -\frac{\Delta H}{T}$

Where:

• $\Delta H$ is the enthalpy change of the reaction (must be converted to $\text{J mol}^{-1}$ by multiplying $\text{kJ mol}^{-1}$ by $1000$).
• $T$ is the absolute temperature in Kelvin ($\text{K}$).

The Criterion for Feasibility: $\Delta S_{\text{total}}$

For any process or chemical reaction to be spontaneous or thermodynamically feasible, the total entropy change must be positive:

$\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0$

Substituting the expression for $\Delta S_{\text{surroundings}}$:

$\Delta S_{\text{total}} = \Delta S_{\text{system}} - \frac{\Delta H}{T}$

Note on Gibbs Energy: The Edexcel specification notes that the Gibbs free energy equation ($\Delta G = \Delta H - T\Delta S_{\text{system}}$) is permitted, but the primary language and focus of Unit 4 is total entropy. Notice the direct connection: $\Delta G = -T\Delta S_{\text{total}}$. A negative $\Delta G$ corresponds to a positive $\Delta S_{\text{total}}$.

Temperature and Feasibility

Because $\Delta S_{\text{surroundings}} = -\frac{\Delta H}{T}$, changing the temperature influences the balance between system entropy and surroundings entropy:

• Exothermic reactions with negative $\Delta S_{\text{system}}$ (e.g., $2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{SO}_3(\text{g})$):
Here $\Delta H$ is negative, so $\Delta S_{\text{surroundings}}$ is positive. At low temperatures, $\Delta S_{\text{surroundings}}$ is large and positive, making $\Delta S_{\text{total}} > 0$. At high temperatures, $\Delta S_{\text{surroundings}}$ becomes smaller, and the negative $\Delta S_{\text{system}}$ dominates, making $\Delta S_{\text{total}} < 0$. The reaction is feasible only at lower temperatures.

• Endothermic reactions with positive $\Delta S_{\text{system}}$ (e.g., thermal decomposition of $\text{CaCO}_3$):
Here $\Delta H$ is positive, so $\Delta S_{\text{surroundings}}$ is negative. At low temperatures, $\Delta S_{\text{total}} < 0$ (not feasible). As $T$ increases, $-\frac{\Delta H}{T}$ becomes less negative until $\Delta S_{\text{total}}$ becomes positive.

Finding the temperature at which a reaction becomes feasible:
A reaction is just on the verge of feasibility when $\Delta S_{\text{total}} = 0$:

$0 = \Delta S_{\text{system}} - \frac{\Delta H}{T} \implies \Delta S_{\text{system}} = \frac{\Delta H}{T} \implies T = \frac{\Delta H}{\Delta S_{\text{system}}}$

Crucial Examination Warning: Ensure $\Delta H$ and $\Delta S_{\text{system}}$ are in consistent units! If $\Delta H$ is in $\text{kJ mol}^{-1}$, multiply it by $1000$ to match $\text{J mol}^{-1}\text{ K}^{-1}$.

Thermodynamic versus Kinetic Stability

Have you ever asked yourself: If diamond converting to graphite has a positive $\Delta S_{\text{total}}$, why don't diamond rings turn into pencil lead? Or why methane doesn't spontaneously burn in air right now?

• Thermodynamic feasibility tells us whether a reaction can occur based on energy and entropy ($\Delta S_{\text{total}} > 0$).
• Kinetic stability depends on the activation energy ($E_{\text{a}}$). If the activation energy is very high, the reaction occurs at an immeasurably slow rate at room temperature.

A mixture of petrol and air is thermodynamically unstable (feasible, $\Delta S_{\text{total}} \gg 0$) but kinetically stable at room temperature until a spark provides enough energy to overcome the high activation energy barrier.

Key Takeaway for Part 2: A reaction is feasible when $\Delta S_{\text{total}} > 0$. A reaction that is thermodynamically feasible may still fail to occur under standard conditions if it is kinetically stable due to a high activation energy.

---

Part 3: Lattice Energy and Born-Haber Cycles

What is Lattice Energy?

Lattice energy ($\Delta_{\text{latt}}H^{\ominus}$) is the standard enthalpy change when one mole of an ionic solid is formed from its gaseous ions under standard conditions:

$\text{M}^+(\text{g}) + \text{X}^-(\text{g}) \rightarrow \text{MX}(\text{s})$

By this definition (lattice formation), lattice energy is always negative (exothermic) because strong electrostatic forces of attraction form between oppositely charged gaseous ions.

Note: Lattice energy cannot be measured directly by experiment. Instead, we use an indirect thermochemical cycle called a Born-Haber cycle, based on Hess's Law.

Definitions Needed for Born-Haber Cycles

To construct a Born-Haber cycle, you need the following standard enthalpy definitions:

• Standard Enthalpy of Formation ($\Delta_{\text{f}}H^{\ominus}$): The enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions.
Example: $\text{Na}(\text{s}) + \frac{1}{2}\text{Cl}_2(\text{g}) \rightarrow \text{NaCl}(\text{s})$

• Standard Enthalpy of Atomisation ($\Delta_{\text{at}}H^{\ominus}$): The enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions (always endothermic, positive).
Example for sodium: $\text{Na}(\text{s}) \rightarrow \text{Na}(\text{g})$
Example for chlorine: $\frac{1}{2}\text{Cl}_2(\text{g}) \rightarrow \text{Cl}(\text{g})$

• First Ionisation Energy ($\Delta_{\text{IE1}}H$): The energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous $1+$ ions (endothermic, positive).
Example: $\text{Na}(\text{g}) \rightarrow \text{Na}^+(\text{g}) + \text{e}^-$
Second Ionisation Energy ($\Delta_{\text{IE2}}H$) applies to Group 2 metals: $\text{M}^+(\text{g}) \rightarrow \text{M}^{2+}(\text{g}) + \text{e}^-$.

• First Electron Affinity ($\Delta_{\text{EA1}}H$): 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.
Example: $\text{Cl}(\text{g}) + \text{e}^- \rightarrow \text{Cl}^-(\text{g})$
First electron affinity is usually exothermic (negative) because the incoming electron is attracted by the positive nucleus.

• Second Electron Affinity ($\Delta_{\text{EA2}}H$): The enthalpy change when one electron is added to each ion in one mole of gaseous $1-$ ions to form one mole of gaseous $2-$ ions.
Example: $\text{O}^-(\text{g}) + \text{e}^- \rightarrow \text{O}^{2-}(\text{g})$
Second electron affinity is always endothermic (positive) because energy must be supplied to overcome electrostatic repulsion between the negatively charged ion and the incoming negative electron.

Constructing a Born-Haber Cycle (Example: $\text{NaCl}$)

Think of a Born-Haber cycle as an energy ladder. We start at zero enthalpy level with the elements in their standard states: $\text{Na}(\text{s}) + \frac{1}{2}\text{Cl}_2(\text{g})$.

1. Downwards step to compound: $\Delta_{\text{f}}H^{\ominus}[\text{NaCl}(\text{s})]$
2. Upwards steps to gaseous atoms:
    • Atomise metal: $+\Delta_{\text{at}}H^{\ominus}[\text{Na}] \implies \text{Na}(\text{g}) + \frac{1}{2}\text{Cl}_2(\text{g})$
    • Atomise non-metal: $+\Delta_{\text{at}}H^{\ominus}[\text{Cl}] \implies \text{Na}(\text{g}) + \text{Cl}(\text{g})$
3. Upwards step to ionise metal:
    • First ionisation energy: $+\Delta_{\text{IE1}}H[\text{Na}] \implies \text{Na}^+(\text{g}) + \text{e}^- + \text{Cl}(\text{g})$
4. Downwards step for electron affinity of non-metal:
    • Electron affinity: $+\Delta_{\text{EA1}}H[\text{Cl}] \implies \text{Na}^+(\text{g}) + \text{Cl}^-(\text{g})$
5. Final downward step: Lattice energy ($\Delta_{\text{latt}}H^{\ominus}$) connects the gaseous ions to the solid lattice.

By Hess's Law, traveling the direct formation path equals traveling the indirect route:

$\Delta_{\text{f}}H^{\ominus} = \Delta_{\text{at}}H^{\ominus}(\text{metal}) + \sum \Delta_{\text{IE}}H(\text{metal}) + \Delta_{\text{at}}H^{\ominus}(\text{non-metal}) + \sum \Delta_{\text{EA}}H(\text{non-metal}) + \Delta_{\text{latt}}H^{\ominus}$

Rearranging to find lattice energy:

$\Delta_{\text{latt}}H^{\ominus} = \Delta_{\text{f}}H^{\ominus} - \left( \Delta_{\text{at}}H^{\ominus}(\text{metal}) + \sum \Delta_{\text{IE}}H(\text{metal}) + \Delta_{\text{at}}H^{\ominus}(\text{non-metal}) + \sum \Delta_{\text{EA}}H(\text{non-metal}) \right)$

Common Pitfall: Pay close attention to stoichiometry! For magnesium chloride ($\text{MgCl}_2$), you need the sum of the first AND second ionisation energies of magnesium, TWO times the atomisation enthalpy of chlorine (or the bond enthalpy of $\text{Cl}_2$), and TWO times the electron affinity of chlorine because two moles of $\text{Cl}^-$ ions are formed.

Key Takeaway for Part 3: Born-Haber cycles use Hess's Law to calculate experimental lattice energies from formation, atomisation, ionisation energy, and electron affinity data.

---

Part 4: Ionic Model vs. Covalent Character

The Theoretical (Purely Ionic) Model

Lattice energies can also be calculated theoretically using physics (electrostatic equations based on Coulomb's law). The purely ionic model makes two fundamental assumptions:

1. The ions are point charges or perfectly spherical.
2. The bonding between ions is $100\%$ ionic with purely electrostatic attraction and no covalent character.

Comparing Experimental (Born-Haber) and Theoretical Values

When we compare experimental lattice energies from Born-Haber cycles with theoretical values from the ionic model, we observe two main outcomes:

• Good agreement (e.g., $\text{NaCl}$, $\text{KBr}$): The experimental value is almost identical to the theoretical value (e.g., within $1\%$ or $2\%$). This proves that these compounds are very close to being purely ionic, with spherical ions and negligible covalent character.

• Significant discrepancy (e.g., $\text{AgCl}$, $\text{AgI}$, $\text{MgI}_2$): The experimental lattice energy is significantly more exothermic (more negative) than the theoretical value. This occurs because the compound has significant covalent character, which provides additional bonding strength beyond simple electrostatic attraction.

Polarisation and Fajans' Rules

Why does covalent character occur in ionic compounds? It results from polarisation:

• A small, highly charged cation has high polarising power (it exerts a strong positive pull on nearby electron clouds).
• A large anion has high polarisability (its valence electron cloud is far from its own nucleus and easily distorted).
• When a polarising cation distorts the electron cloud of a polarisable anion, electron density is drawn into the region between the nuclei, creating partial sharing of electrons—that is, covalent character.

Factors Increasing Covalent Character:

• High charge on cation (e.g., $\text{Al}^{3+} > \text{Mg}^{2+} > \text{Na}^+$)
• Small ionic radius of cation (e.g., $\text{Li}^+ > \text{Na}^+ > \text{K}^+$)
• Large ionic radius of anion (e.g., $\text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-$)

For example, in $\text{AgI}$, the large iodide ion is readily polarised by the silver ion, resulting in significant covalent character. Consequently, the experimental lattice energy of $\text{AgI}$ is much more exothermic than the theoretical value calculated by the ionic model.

Key Takeaway for Part 4: Perfect agreement between experimental and theoretical lattice energy confirms a purely ionic structure. A higher experimental value indicates covalent character caused by cation polarising power and anion polarisability.

---

Part 5: Enthalpy of Solution and Hydration

Definitions of Solution Energetics

When an ionic lattice dissolves in water, two competing processes take place:

1. Standard Enthalpy of Solution ($\Delta_{\text{sol}}H^{\ominus}$): The enthalpy change when one mole of an ionic solid dissolves in water to form infinitely dilute aqueous ions under standard conditions:
$\text{NaCl}(\text{s}) + \text{aq} \rightarrow \text{Na}^+(\text{aq}) + \text{Cl}^-(\text{aq})$

2. Standard Enthalpy of Hydration ($\Delta_{\text{hyd}}H^{\ominus}$): The enthalpy change when one mole of specified gaseous ions dissolves in water to form infinitely dilute aqueous ions under standard conditions:
$\text{M}^+(\text{g}) + \text{aq} \rightarrow \text{M}^+(\text{aq})$
Hydration is always exothermic (negative) because ion-dipole attractions form between ions and polar water molecules (the $\delta-$ oxygen of water attracts cations, and the $\delta+$ hydrogen attracts anions).

The Enthalpy Cycle for Dissolving

We can relate enthalpy of solution, lattice energy, and enthalpies of hydration using Hess's Law:

To dissolve a solid, we can imagine two theoretical steps:
Step 1: Break the lattice apart into gaseous ions. This is the reverse of lattice formation, so it requires enthalpy equal to $-\Delta_{\text{latt}}H^{\ominus}$ (lattice dissociation, endothermic).
Step 2: Hydrate the gaseous ions by surrounding them with water molecules ($\sum \Delta_{\text{hyd}}H^{\ominus}$, exothermic).

Therefore:

$\Delta_{\text{sol}}H^{\ominus} = \sum \Delta_{\text{hyd}}H^{\ominus}(\text{cations and anions}) - \Delta_{\text{latt}}H^{\ominus}(\text{formation})$

Alternative phrasing: If you define lattice energy as lattice dissociation ($\text{MX}(\text{s}) \rightarrow \text{M}^+(\text{g}) + \text{X}^-(\text{g})$), then $\Delta_{\text{sol}}H^{\ominus} = \Delta_{\text{latt,diss}}H + \sum \Delta_{\text{hyd}}H^{\ominus}$. Both give identical numerical answers, but remember that in Edexcel, lattice energy is defined as formation (exothermic).

Factors Affecting Lattice Energy and Hydration Enthalpy

Both lattice energy and hydration enthalpy depend on charge density, governed by two variables:

1. Ionic Charge: Higher charge gives much stronger electrostatic attraction. For example, $\text{Mg}^{2+}$ ions form much stronger attractions to water molecules and counter-ions than $\text{Na}^+$ ions. Higher charge leads to a more exothermic lattice energy and a more exothermic hydration enthalpy.
2. Ionic Radius: Smaller ions allow their centers of positive/negative charge to get closer to water molecules or counter-ions. Smaller radius leads to a more exothermic lattice energy and a more exothermic hydration enthalpy.

Explaining Solubility Trends

Whether a substance dissolves easily depends on the balance between $\Delta H_{\text{sol}}$ and $\Delta S_{\text{system}}$:

• If $\Delta_{\text{sol}}H^{\ominus}$ is exothermic or only slightly endothermic, the substance is usually soluble because $\Delta S_{\text{total}}$ is positive.
• If $\Delta_{\text{sol}}H^{\ominus}$ is very endothermic, the reaction requires too much energy, making $\Delta S_{\text{surroundings}} = -\frac{\Delta H}{T}$ very negative, which often overcomes $\Delta S_{\text{system}}$ and makes $\Delta S_{\text{total}} < 0$ (insoluble).

Why do Group 2 hydroxide solubilities increase down the group, while Group 2 sulfate solubilities decrease?
As you descend Group 2 from $\text{Mg}^{2+}$ to $\text{Ba}^{2+}$, the cation radius increases, so both lattice energy and hydration enthalpy become less exothermic (weaker attractions).

• For Hydroxides ($\text{OH}^-$): The hydroxide ion is very small. As you go down the group, the percentage increase in cation size makes a relatively large change to the lattice spacing. As a result, lattice energy decreases faster than hydration enthalpy. Consequently, $\Delta_{\text{sol}}H^{\ominus}$ becomes more exothermic (or less endothermic) down the group, so solubility increases down Group 2.

• For Sulfates ($\text{SO}_4^{2-}$): The sulfate ion is very large. Because the sulfate ion already dominates the interionic distance, the increasing size of the cation produces only a small percentage change in lattice spacing, so lattice energy decreases slowly down the group. However, the hydration enthalpy of the cation drops significantly as its charge density falls. Here, hydration enthalpy decreases faster than lattice energy. Consequently, $\Delta_{\text{sol}}H^{\ominus}$ becomes more endothermic down the group, and solubility decreases down Group 2.

Key Takeaway for Part 5: Dissolving balances breaking the lattice against hydrating the ions. High charge and small radius make both lattice energy and hydration enthalpy more exothermic.

---

Quick Summary & Revision Checklist

Before heading into past paper questions, ensure you can:

• State that entropy is zero at $0\text{ K}$ for a perfect crystal.
• Predict the sign of $\Delta S_{\text{system}}$ using states of matter and gas moles.
• Calculate $\Delta S_{\text{system}}$ from standard molar entropies ($\sum S_{\text{products}} - \sum S_{\text{reactants}}$).
• Calculate $\Delta S_{\text{surroundings}}$ using $-\frac{\Delta H}{T}$, ensuring you convert $\Delta H$ to $\text{J mol}^{-1}$.
• Determine feasibility using $\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0$.
• Calculate the minimum temperature for feasibility using $T = \frac{\Delta H}{\Delta S_{\text{system}}}$.
• Explain why a feasible reaction might not happen at room temperature (kinetic stability due to high $E_{\text{a}}$).
• Define lattice energy (formation), atomisation, ionisation energy, electron affinity, solution, and hydration.
• Construct a Born-Haber cycle and calculate an unknown value.
• Explain differences between experimental and theoretical lattice energies in terms of polarisation and covalent character.
• Relate lattice energy and hydration enthalpy to ionic charge and radius, and use them to explain Group 2 solubility trends.

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is entropy, and what are its standard SI units?

Entropy (\f$S $) is a measure of the disorder, randomness, or dispersal of matter and energy among particles in a system, measured in $\text{J mol}^{-1}\text{ K}^{-1}$.

What is the entropy of a perfect, pure crystal at absolute zero according to the Third Law of Thermodynamics?

The entropy of a perfect, pure crystal at $0\text{ K}$ is zero ($S = 0\text{ J mol}^{-1}\text{ K}^{-1}$).

How do changes of state and changes in gas moles affect the entropy of a system ($\Delta S_{\text{system}}$)?

$\Delta S_{\text{system}}$ increases when changing from solid to liquid to gas, or when a reaction produces more moles of gas than it consumes.

How is the standard entropy change of a system ($\Delta S^{\ominus}_{\text{system}}$) calculated from standard molar entropies?

$\Delta S^{\ominus}_{\text{system}} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})$

How do you calculate the entropy change of the surroundings ($\Delta S_{\text{surroundings}}$) from enthalpy change and temperature?

$\Delta S_{\text{surroundings}} = -\frac{\Delta H}{T}$, where $\Delta H$ is in $\text{J mol}^{-1}$ and $T$ is in Kelvin.

What is the condition for a chemical reaction or physical process to be thermodynamically feasible?

A process is thermodynamically feasible if the total entropy change is positive: $\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0$.

How do you calculate the minimum temperature at which a reaction with an endothermic enthalpy and positive system entropy becomes feasible?

$T = \frac{\Delta H}{\Delta S_{\text{system}}}$, with $\Delta H$ converted to $\text{J mol}^{-1}$.

Why might a thermodynamically feasible reaction ($\Delta S_{\text{total}} > 0$) fail to occur spontaneously at room temperature?

The reaction is kinetically stable due to a high activation energy ($E_{\text{a}}$), making the rate of reaction imperceptibly slow.

How is standard lattice energy ($\Delta_{\text{latt}}H^{\ominus}$) defined?

It is the standard enthalpy change when one mole of an ionic solid is formed from its constituent gaseous ions under standard conditions.

Why is the first electron affinity exothermic while the second electron affinity is always endothermic?

First electron affinity is exothermic due to attraction between the incoming electron and the positive nucleus; second electron affinity is endothermic because energy is required to overcome electrostatic repulsion from the negative ion.

What does a significant discrepancy between experimental and theoretical lattice energies reveal about bonding in an ionic solid?

It indicates that the compound possesses significant covalent character, making the experimental lattice energy more exothermic than predicted by the purely ionic model.

What ionic features promote high polarising power in cations and high polarisability in anions according to Fajans' rules?

Cations have high polarising power if they are small and highly charged; anions have high polarisability if they are large and highly charged.

How are enthalpy of solution, lattice energy, and enthalpy of hydration related via Hess's Law?

$\Delta_{\text{sol}}H^{\ominus} = \sum \Delta_{\text{hyd}}H^{\ominus} - \Delta_{\text{latt}}H^{\ominus}(\text{formation})$

Why does the solubility of Group 2 hydroxides increase down the group while sulfate solubility decreases?

Down Group 2, lattice energy decreases faster than hydration enthalpy for hydroxides (making $\Delta_{\text{sol}}H^{\ominus}$ more exothermic), whereas hydration enthalpy decreases faster than lattice energy for sulfates.

Ready to test yourself?

10 practice questions on this topic, with worked solutions. Multiple choice is marked as you go. No account needed.

Practise This Topic

More Chemistry (YCH11) chapters

* The content provided by thinka is generated by AI and may not always be accurate or up-to-date. Please use it as a supplementary resource and verify with official materials.

Put These Notes into Practice

Reading the notes is a great start. Now practise with unlimited AI-generated questions and get instant feedback. 100,000+ students are already improving their grades.

Start Practising Now View Pricing

Done reading? Test yourself with AI practice questions

Practise This Topic Now