Chemistry (9620) | Thermodynamics

Welcome to A2 Thermodynamics!

Hello! You've successfully navigated the Energetics section (AS), and now we dive deeper into Thermodynamics. This chapter explains the 'why' and 'how far' of chemical reactions. Why does one reaction happen naturally (spontaneously) while another needs a massive energy push?

We will link energy changes ($\Delta H$) with the concepts of disorder ($\Delta S$, entropy) to determine the ultimate driving force for a reaction: the Gibbs Free Energy ($\Delta G$). This is the cornerstone of understanding chemical stability and feasibility. Don't worry if it seems challenging; we'll break down these powerful ideas into simple, manageable steps!

3.1.8.1 Born-Haber Cycles: Building Ionic Compounds

What is Lattice Enthalpy?

When discussing ionic compounds (like NaCl or MgO), a critical value is the Lattice Enthalpy. This measures the strength of the ionic bonds holding the crystal structure together.

It can be defined in two ways:
1. Lattice Dissociation Enthalpy (\(\Delta H_{latt, diss}^\theta\)): The enthalpy change required to break 1 mole of an ionic solid into its separate gaseous ions. (Always positive/endothermic.)
2. Lattice Formation Enthalpy (\(\Delta H_{latt, form}^\theta\)): The enthalpy change when 1 mole of an ionic solid is formed from its separate gaseous ions. (Always negative/exothermic.)
Tip: These values are numerically equal but have opposite signs. Focus on the definition you are asked for in the exam!

Why We Need Born-Haber Cycles (Hess's Law in Action)

We cannot measure lattice enthalpy directly. Instead, we use Hess's Law, which states that the total enthalpy change for a reaction is independent of the route taken. A Born-Haber Cycle is simply a visual application of Hess's Law for forming an ionic compound, using a series of measurable steps.

To build the cycle for forming M⁺X^- (s) from M(s) and 1/2 X\(_2\)(g), we need these specific enthalpy steps:

Key Enthalpy Definitions for the Cycle

• Standard Enthalpy of Formation ($\Delta H_f^\theta$): The enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions (100 kPa and a stated temperature, usually 298 K). Example: Na(s) + 1/2 Cl\(_2\)(g) \(\to\) NaCl(s)
• Enthalpy of Atomisation ($\Delta H_{at}^\theta$): The enthalpy change required to produce 1 mole of gaseous atoms from the element in its standard state. (Always positive/endothermic.) Example: Na(s) \(\to\) Na(g)
• First Ionisation Energy (IE\(_1\)): The energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of gaseous 1+ ions. (Always positive/endothermic.) Example: Na(g) \(\to\) Na$^+$(g) + e$^-$
• First Electron Affinity (EA\(_1\)): The enthalpy change when 1 mole of electrons is added to 1 mole of gaseous atoms to form 1 mole of gaseous 1- ions. (Usually negative/exothermic, as the atom gains stability.) Example: Cl(g) + e$^-$ \(\to\) Cl$^-$ (g)
• Bond Enthalpy: For diatomic molecules (like Cl\(_2\)), this is used to break the bond to form gaseous atoms. (This is related to the atomisation of non-metals.) Example: 1/2 Cl\(_2\)(g) \(\to\) Cl(g)

Using the Cycle

Since the formation of the ionic solid from elements ($\Delta H_f^\theta$) must equal the sum of all steps via the gaseous ion route (Hess's Law):
\( \Delta H_f^\theta = \Delta H_{at}(\text{metal}) + \text{IE} + \Delta H_{at}(\text{non-metal}) + \text{EA} + \Delta H_{latt, form}^\theta \)

You will typically be given all values except one (usually the lattice enthalpy) and asked to calculate it.

Lattice Enthalpy and Covalent Character

Lattice enthalpies can also be calculated theoretically assuming the ions are perfectly spherical and purely ionic (no covalent character).

If the value calculated via the Born-Haber cycle (the experimental value) is significantly more negative (more exothermic) than the theoretical value, it provides evidence for covalent character within the ionic compound.
Why? If the cation distorts the anion (polarisation), the bond gains some covalent character, making the bond stronger and releasing more energy upon formation, thus making the lattice formation enthalpy more negative.

Enthalpies of Solution and Hydration

Born-Haber-style cycles can also be used to look at dissolution:

• Enthalpy of Solution ($\Delta H_{sol}^\theta$): The enthalpy change when 1 mole of an ionic solid dissolves completely in a large amount of water. Example: NaCl(s) \(\to\) Na$^+$(aq) + Cl$^-$ (aq)
• Enthalpy of Hydration ($\Delta H_{hyd}^\theta$): The enthalpy change when 1 mole of specified gaseous ions dissolves in water to form 1 mole of aqueous ions. (Always negative/exothermic, as water stabilises the ions.) Example: Na$^+$(g) \(\to\) Na$^+$(aq)

These three terms link via a cycle:
\( \Delta H_{sol}^\theta = \Delta H_{latt, diss}^\theta + \sum \Delta H_{hyd}^\theta (\text{cations}) + \sum \Delta H_{hyd}^\theta (\text{anions}) \)

Key Takeaway: Born-Haber cycles use Hess’s Law and a series of defined, measurable energy steps (like IE and EA) to calculate the non-measurable Lattice Enthalpy. Comparing this value with theory reveals how much the ionic bond deviates towards being covalent.

3.1.8.2 Gibbs Free-Energy Change ($\Delta G$) and Entropy Change ($\Delta S$)

The Concept of Disorder: Entropy ($\Delta S$)

You might have learned that reactions with a negative $\Delta H$ (exothermic) are usually spontaneous. But wait—ice melts (endothermic), and many compounds decompose (endothermic) spontaneously! Clearly, enthalpy ($\Delta H$) alone is not enough to explain feasibility.

We need Entropy ($\Delta S$). Entropy is a measure of the disorder or the degree of randomness in a system.

• If $\Delta S$ is positive, the system becomes more disordered (more random).
• If $\Delta S$ is negative, the system becomes more ordered.

How to Spot an Increase in Entropy (\(\Delta S > 0\))

Disorder usually increases when:

• A solid turns into a liquid, or a liquid turns into a gas (e.g., evaporation). Gases have much higher entropy than solids.
• The number of moles of gaseous products is greater than the moles of gaseous reactants.
• Complex molecules break down into simpler ones.
• Analogy: Think of your room. When you tidy it, you decrease entropy ($\Delta S$ is negative). When you let clutter build up, entropy increases!

Calculating Entropy Change

We can calculate the standard entropy change for a reaction ($\Delta S^\theta$) using the absolute entropy values ($S^\theta$) for the reactants and products (these values are typically looked up in a data booklet):
\[ \Delta S_{reaction}^\theta = \sum S^\theta(\text{products}) - \sum S^\theta(\text{reactants}) \]

The Feasibility Test: Gibbs Free Energy ($\Delta G$)

The Gibbs Free-Energy Change ($\Delta G$) combines both the enthalpy ($\Delta H$) and the entropy ($\Delta S$) to give us the true criterion for reaction feasibility (spontaneity).

This is often called the 'chemical compass' because it points in the direction of a spontaneous process.

The core relationship (which you must know, but you don't need to derive):
\[ \Delta G = \Delta H - T\Delta S \]
Note: T must always be in Kelvin (K).

Criterion for Feasibility

For a reaction to be feasible (to occur spontaneously under the given conditions), the Gibbs Free-Energy change must be zero or negative:
$$ \Delta G \le 0 $$
A strongly negative $\Delta G$ means the reaction is highly likely to happen.

How Temperature Affects Feasibility

Because the temperature ($T$) term multiplies the entropy term ($\Delta S$), temperature becomes crucial in determining whether a reaction is feasible.

We can examine the four scenarios based on the signs of $\Delta H$ and $\Delta S$:

• Case 1: Favourable (Exothermic and Increased Disorder)
  $\Delta H$ is Negative, $\Delta S$ is Positive.
  $\Delta G = (\text{negative}) - T(\text{positive}) = \text{Always Negative}$
  Result: The reaction is always feasible at all temperatures. (This is the ideal spontaneous reaction!)
• Case 2: Unfavourable (Endothermic and Increased Order)
  $\Delta H$ is Positive, $\Delta S$ is Negative.
  $\Delta G = (\text{positive}) - T(\text{negative}) = \text{Always Positive}$
  Result: The reaction is never feasible (not spontaneous) at any temperature.
• Case 3: Enthalpy-Driven (Feasible at Low T)
  $\Delta H$ is Negative, $\Delta S$ is Negative.
  $\Delta G$ is negative only if \( |\Delta H| > |T\Delta S| \). The reaction is feasible only at low temperatures.
• Case 4: Entropy-Driven (Feasible at High T)
  $\Delta H$ is Positive, $\Delta S$ is Positive.
  $\Delta G$ is negative only if \( |T\Delta S| > |\Delta H| \). The reaction is feasible only at high temperatures. (e.g., endothermic decomposition reactions).

Determining the Feasibility Temperature

For Case 3 and Case 4, there is a specific temperature ($T$) where the reaction switches from feasible to non-feasible. At this point, the reaction is at equilibrium, and $\Delta G = 0$.

We can set the equation to zero:
\[ 0 = \Delta H - T\Delta S \]
Rearranging to find the transition temperature ($T$):
\[ T = \frac{\Delta H}{\Delta S} \]

If you calculate this temperature, you can determine the feasibility range. For example, if $T=500\text{ K}$ and the reaction is entropy-driven ($\Delta H$ positive, $\Delta S$ positive), the reaction is feasible at any temperature above 500 K.

Key Takeaway: Gibbs Free Energy ($\Delta G$) is the ultimate judge of spontaneity. A reaction is only feasible if $\Delta G$ is negative. Temperature (T) influences feasibility, especially when both $\Delta H$ and $\Delta S$ share the same sign.

Keep practicing those calculations, and you'll master Thermodynamics!