Chemistry (9701) | Entropy change, ΔS

Welcome to Entropy: The Science of Disorder!

Hello! This chapter takes you deeper into Chemical Energetics. You already know about Enthalpy Change (\(\Delta H\)), which tells you if a reaction releases or absorbs heat. But $\Delta H$ alone isn't enough to predict if a reaction will actually happen (if it's feasible).

To truly understand why some reactions occur spontaneously (all by themselves!) and others don't, we need to introduce a new, fundamental concept: Entropy (\(S\)). Don't worry if it sounds complicated—we'll break down this "chaos factor" piece by piece.

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1. Defining Entropy (S) and Entropy Change (\(\Delta S\))

1.1 What is Entropy?

The most straightforward way to think about entropy is as a measure of disorder or the degree of randomness in a system.

The formal scientific definition is:
Entropy (\(S\)) is the number of possible ways of arranging the particles and their energy in a given system.

Analogy: Your Room vs. Your Backpack

Imagine you have 10 books.

• If they are all stacked perfectly (low disorder), there is only 1 way to arrange them. (Low Entropy)
• If they are scattered randomly across the floor (high disorder), there are millions of ways they could be arranged. (High Entropy)

In chemistry, a system naturally tends towards the state with the highest entropy (most disorder).

1.2 Entropy Change (\(\Delta S\))

The Entropy Change (\(\Delta S\)) is the change in disorder when a reaction or process occurs.

• If \(\Delta S\) is positive (\(\Delta S > 0\)): The system becomes more disordered (more chaotic). This is favourable.
• If \(\Delta S\) is negative (\(\Delta S < 0\)): The system becomes less disordered (more ordered). This is unfavourable.

The standard units for entropy ($S$) are typically Joules per Kelvin per mole (\(\text{J K}^{-1} \text{mol}^{-1}\)).

Key Takeaway for Section 1

Entropy = Disorder. Systems prefer positive $\Delta S$ (more disorder).

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2. Predicting the Sign of Entropy Change (\(\Delta S\))

You must be able to predict whether $\Delta S$ will be positive or negative by looking at a reaction or process. This is often called the Chaos Rule: look for increases in freedom and movement!

2.1 Changes in State (Phase Changes)

Entropy increases as particles gain more freedom of movement.

Melting (Solid \(\rightarrow\) Liquid): \(\Delta S\) is positive. (e.g., Ice melting)
Explanation: Particles in a solid are fixed in a lattice (highly ordered). In a liquid, they can move and slide past each other (more arrangements possible).

Boiling (Liquid \(\rightarrow\) Gas): \(\Delta S\) is very positive.
Explanation: Gas particles are completely free and occupy a huge volume, allowing for maximum randomness. This change results in the largest increase in entropy.

The reverse processes (freezing, condensation) have negative \(\Delta S\).

2.2 Dissolving

Dissolving a solid or liquid in a solvent: \(\Delta S\) is usually positive.
Explanation: When a salt dissolves, the highly ordered crystal lattice breaks down, and the ions become mobile and surrounded by solvent molecules (hydration). This drastically increases the disorder.

2.3 Changes in the Number of Gaseous Molecules

In chemical reactions, the biggest factor determining the sign of \(\Delta S\) is the change in the number of moles of gas.

If the number of moles of gas increases: \(\Delta S\) is positive.
Example: \(2\text{SO}_3(\text{g}) \rightarrow 2\text{SO}_2(\text{g}) + \text{O}_2(\text{g})\)
(2 moles of gas $\rightarrow$ 3 moles of gas. More gas means more disorder.)

If the number of moles of gas decreases: \(\Delta S\) is negative.
Example: \(\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g})\)
(4 moles of gas $\rightarrow$ 2 moles of gas. Less gas means more order.)

2.4 Temperature Change

Increasing temperature: \(\Delta S\) is positive.
Explanation: Higher temperature means particles have more kinetic energy. This energy can be distributed among the particles in more ways, increasing the number of possible energy arrangements (higher entropy).

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3. Calculating the Entropy Change for a Reaction (\(\Delta S^\theta\))

Just like standard enthalpy change (\(\Delta H^\theta\)) can be calculated using $\Delta H_f^\theta$ values, we can calculate the standard entropy change for a reaction (\(\Delta S^\theta\)) using the standard entropies ($S^\theta$) of reactants and products.

3.1 Standard Entropies (\(S^\theta\))

The Standard Entropy (\(S^\theta\)) is the absolute entropy content of one mole of a substance under standard conditions (298 K and 101 kPa).

Note: Unlike standard enthalpy of formation ($\Delta H_f^\theta$), which is zero for elements in their standard state, $S^\theta$ for all substances (including elements) is always positive. Why? Because everything, even a perfectly ordered crystal at 298 K, has some degree of particle movement and energy dispersal.

3.2 The Calculation Formula

To find the entropy change for a reaction, we sum the entropies of the products and subtract the sum of the entropies of the reactants.

\[ \Delta S^\theta = \sum S^\theta (\text{products}) - \sum S^\theta (\text{reactants}) \]

Step-by-Step Calculation Guide

1. Balance the Equation: Ensure the stoichiometry is correct.
2. List Standard Entropies: Multiply each $S^\theta$ value by its coefficient in the balanced equation.
3. Calculate Total Product Entropy: Sum the weighted $S^\theta$ values for all products.
4. Calculate Total Reactant Entropy: Sum the weighted $S^\theta$ values for all reactants.
5. Subtract: Products minus Reactants.

Did you know? Since $\Delta S^\theta$ is usually given in J K^-1 mol^-1, this calculation gives an answer in J K^-1 mol^-1. Remember this for the next section!

Key Takeaway for Section 3

Calculate $\Delta S^\theta$: Products minus Reactants. All $S^\theta$ values are positive.

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4. Gibbs Free Energy (\(\Delta G\)): The Criterion for Feasibility

4.1 Why Enthalpy ($\Delta H$) is not enough

We learned in AS Chemistry that exothermic reactions (\(\Delta H < 0\)) are favourable. But wait! Ice melting (\(\text{H}_2\text{O}(\text{s}) \rightarrow \text{H}_2\text{O}(\text{l})\)) is endothermic ($\Delta H > 0$) but happens spontaneously above $0^\circ\text{C}$.

This tells us that two factors determine if a process is feasible (can happen spontaneously):

1. Enthalpy (\(\Delta H\)): Seeking lower energy (exothermic, \(\Delta H < 0\)).
2. Entropy (\(\Delta S\)): Seeking greater disorder (positive \(\Delta S\)).

4.2 The Gibbs Free Energy Equation

The Gibbs Free Energy Change (\(\Delta G\)) combines these two driving forces, giving us the ultimate measure of feasibility.

This is one of the most important equations in A-Level Chemistry. Learn it, live it, love it!

\[ \Delta G^\theta = \Delta H^\theta - T \Delta S^\theta \]

Where:

• \(\Delta G^\theta\): Standard Gibbs Free Energy Change (kJ mol^-1)
• \(\Delta H^\theta\): Standard Enthalpy Change (kJ mol^-1)
• \(T\): Absolute Temperature (in Kelvin, K)
• \(\Delta S^\theta\): Standard Entropy Change (J K^-1 mol^-1)

4.3 Feasibility (Spontaneity)

The sign of \(\Delta G\) determines whether a reaction is thermodynamically feasible:

• If \(\Delta G\) is negative (\(\Delta G < 0\)): The reaction is feasible (spontaneous) under the given conditions.
• If \(\Delta G\) is positive (\(\Delta G > 0\)): The reaction is non-feasible (non-spontaneous). It won't happen naturally.
• If \(\Delta G = 0\): The reaction is at equilibrium.

Memory Aid: Feasibility is Negative! If you want something to happen, you want a negative value for $\Delta G$.

Key Takeaway for Section 4

$\Delta G$ dictates feasibility. $\Delta G = \Delta H - T \Delta S$. Remember to convert $\Delta S$ units (J $\rightarrow$ kJ).

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5. The Effect of Temperature on Feasibility

The term \(T\Delta S^\theta\) represents the energy contributed by the increased disorder. Because temperature (\(T\)) is always positive (in Kelvin), the effect of $T$ depends entirely on the sign of $\Delta S$.

By examining the signs of $\Delta H$ and $\Delta S$, we can predict how temperature affects feasibility.

5.1 The Four Scenarios

This table summarises how the reaction feasibility depends on temperature. You need to be able to predict and explain all four outcomes.

\(\Delta H\) Sign	\(\Delta S\) Sign	\(\Delta G = \Delta H - T\Delta S\)	Feasibility Prediction
Negative (\(-\)) (Exothermic)	Positive (\(+\)) (More Disorder)	\(\Delta G\) is always Negative.	Feasible at ALL temperatures. (Both factors are favourable.)
Positive (\(+\)) (Endothermic)	Negative (\(-\)) (Less Disorder)	\(\Delta G\) is always Positive.	Non-feasible at ALL temperatures. (Both factors are unfavourable.)
Negative (\(-\)) (Exothermic)	Negative (\(-\)) (Less Disorder)	Becomes positive at high T.	Feasible only at LOW temperatures. (T must be small so \(|T\Delta S| < |\Delta H|\)).
Positive (\(+\)) (Endothermic)	Positive (\(+\)) (More Disorder)	Becomes negative at high T.	Feasible only at HIGH temperatures. (T must be large so \(|T\Delta S| > |\Delta H|\)).

5.2 Finding the Crossover Temperature (\(T\))

For the scenarios where feasibility depends on temperature (the last two rows), there is a temperature where the reaction switches from feasible to non-feasible, or vice versa. This occurs when the system is at equilibrium, where \(\Delta G = 0\).

Step-by-step to find T:

1. Set the Gibbs equation to zero:
\[ 0 = \Delta H^\theta - T \Delta S^\theta \]
2. Rearrange to solve for T:
\[ T = \frac{\Delta H^\theta}{\Delta S^\theta} \]

Remember the Units AGAIN: Ensure that $\Delta H^\theta$ and $\Delta S^\theta$ are in consistent units (both J or both kJ) before dividing! Since $\Delta H^\theta$ is usually given in $\text{kJ mol}^{-1}$, it is easiest to convert $\Delta S^\theta$ from $\text{J K}^{-1} \text{mol}^{-1}$ to $\text{kJ K}^{-1} \text{mol}^{-1}$ (divide by 1000) before performing the division to find $T$.

The calculated value of $T$ is the equilibrium temperature.

• If $\Delta H$ is positive and $\Delta S$ is positive, the reaction is feasible when $T$ is above this value.
• If $\Delta H$ is negative and $\Delta S$ is negative, the reaction is feasible when $T$ is below this value.