R1.4 Entropy and Spontaneity

1. Define entropy (S). Explain why gases generally have higher entropy than liquids, and liquids higher than solids.

Entropy (S) is a thermodynamic quantity that measures the degree of disorder or randomness in a system, or more precisely, the number of possible ways in which particles and energy can be distributed. A highly ordered system has low entropy, whereas a highly disordered system has high entropy.

The physical state of matter strongly affects entropy. Solids have the lowest entropy because particles are fixed in a rigid lattice, with very limited freedom of movement and only vibrational motion allowed. Liquids have higher entropy because particles are close together but able to move past one another in a random fashion, giving rise to more possible arrangements. Gases have the highest entropy because particles are far apart, move freely in all directions, and occupy an enormous number of possible microstates.

Consequence: This means that physical changes such as melting and vaporisation are usually accompanied by an increase in entropy, which contributes to their thermodynamic feasibility.

2. Predict and explain whether entropy increases or decreases:

(a) Ice melting
Entropy increases. A solid lattice of water molecules (ice) has low disorder. When it melts, the molecules can move around each other randomly in the liquid, creating many more possible arrangements.

(b) CaCO₃(s) → CaO(s) + CO₂(g)
Entropy increases. The formation of a gas molecule (CO₂) introduces a state with far greater disorder and more microstates than the solid reactant.

(c) Condensation of steam
Entropy decreases. A gas has maximum disorder and freedom of movement. When it condenses to form liquid water, particle motion becomes more restricted and the number of possible arrangements decreases.

3. Explain why exothermic reactions are usually spontaneous, while endothermic reactions often require a positive entropy change to occur.

Exothermic reactions (ΔH < 0) release energy into the surroundings. This release of heat increases the entropy of the surroundings because the energy becomes more dispersed. As a result, the overall entropy change of the universe (system + surroundings) is usually positive, so the Gibbs free energy (ΔG = ΔH – TΔS) becomes negative, meaning the reaction is spontaneous.

In contrast, endothermic reactions (ΔH > 0) absorb energy from the surroundings, which lowers the entropy of the surroundings. For such reactions to occur spontaneously, there must be a compensating increase in the entropy of the system itself (ΔS > 0). If this increase is large enough, and especially at high temperature (where the TΔS term is magnified), the overall Gibbs free energy can still become negative.

Consequence: This explains why many endothermic processes, such as the dissolution of salts or thermal decomposition, occur only when entropy increases sufficiently to drive them.

4. Calculate ΔS° for a reaction using tabulated entropy values.

The standard entropy change for a reaction is calculated using:

ΔS∘=∑S∘(products)−∑S∘(reactants)

Each standard molar entropy value (S°) is taken from tabulated data and multiplied by its stoichiometric coefficient.

1. Define Gibbs free energy (ΔG). Explain how it combines enthalpy and entropy to determine spontaneity.

Gibbs free energy (ΔG) is a thermodynamic quantity that determines whether a reaction is feasible (spontaneous) under constant temperature and pressure. It combines the effects of enthalpy (ΔH), which reflects energy change due to bond breaking and forming, and entropy (ΔS), which reflects the degree of disorder.

The relationship is given by:

ΔG=ΔH−TΔS\Delta G = \Delta H - T \Delta SΔG=ΔH−TΔS

• If ΔG < 0, the reaction is spontaneous.

• If ΔG = 0, the system is at equilibrium.

• If ΔG > 0, the reaction is non-spontaneous under those conditions.

Thus, ΔG balances both the drive to minimise energy (favourable ΔH) and the drive to maximise disorder (favourable ΔS).

2. Use the equation ΔG° = ΔH° – TΔS° to determine feasibility.

Model Answer (Method):

1. Convert all values into consistent units (ΔH in kJ mol⁻¹, ΔS in J K⁻¹ mol⁻¹ → ÷1000 to convert into kJ).

2. Insert into the Gibbs free energy equation.

3. A negative ΔG° indicates the reaction is thermodynamically feasible at the specified temperature.

3. Explain how temperature affects spontaneity.

(a) ΔH < 0, ΔS > 0 → Both terms favour spontaneity. ΔG is always negative, so the reaction is spontaneous at all temperatures.

(b) ΔH < 0, ΔS < 0 → Enthalpy favours spontaneity but entropy does not. At low temperatures, the –TΔS term is small, so ΔG remains negative. At high temperatures, the entropy term dominates and makes ΔG positive. Reaction is only feasible at low temperatures.

(c) ΔH > 0, ΔS > 0 → Enthalpy opposes spontaneity, entropy favours it. At low temperatures, ΔH dominates so ΔG is positive. At high temperatures, the TΔS term becomes large enough to overcome ΔH, making ΔG negative. Reaction is only feasible at high temperatures.

(d) ΔH > 0, ΔS < 0 → Both terms oppose spontaneity. ΔG is always positive, so the reaction is never spontaneous at any temperature.

4. Rearrange Gibbs free energy equation to calculate the temperature at which a reaction becomes spontaneous.

At the threshold of spontaneity, ΔG = 0.

0=ΔH−TΔS0 = \Delta H - T \Delta S0

Here ΔH must be in kJ mol⁻¹ and ΔS in kJ K⁻¹ mol⁻¹. This gives the minimum temperature required for an endothermic, entropy-driven process to become spontaneous.

1. Explain why ΔG = 0 at equilibrium.

At equilibrium, the forward and reverse reactions occur at the same rate, so there is no net change in the composition of the system. This means the system has reached a state of maximum stability, where no more useful work can be extracted.

From the Gibbs free energy perspective, ΔG measures the driving force for change. If ΔG < 0, the forward reaction is spontaneous; if ΔG > 0, the reverse is spontaneous. At equilibrium, neither forward nor reverse is favoured, so ΔG = 0.

Consequence: The condition ΔG = 0 defines equilibrium in thermodynamic terms, linking Gibbs free energy directly to the equilibrium constant, K.

3. Explain how the equilibrium constant, K, relates to Gibbs free energy.

(a) Large K value
If K ≫ 1, the equilibrium position lies far to the right, strongly favouring products. From ΔG° = –RT lnK, this corresponds to a large negative ΔG°, meaning the reaction is highly spontaneous under standard conditions.

(b) Small K value
If K ≪ 1, the equilibrium position lies far to the left, strongly favouring reactants. lnK is negative and large in magnitude, so ΔG° is positive. This means the forward reaction is not spontaneous under standard conditions.

Examples:

• Large K: Combustion of methane (K extremely large, ΔG° very negative).

• Small K: Thermal decomposition of CaCO₃ at room temperature (K very small, ΔG° positive).

4. Sketch a graph of Gibbs free energy versus reaction progress.

Model Answer (description):

• On the y-axis: Gibbs free energy (ΔG).

• On the x-axis: reaction progress from reactants to products.

• Curve: ΔG decreases steeply from the reactants side, reaches a minimum, then rises again toward the products.

• The minimum corresponds to equilibrium (ΔG = 0).

• To the left of equilibrium: ΔG < 0, forward reaction spontaneous.

• To the right of equilibrium: ΔG > 0, reverse reaction spontaneous.

Consequence: The graph visually demonstrates that equilibrium is the state of lowest Gibbs free energy.

1. Explain why entropy change is usually positive when a reaction produces a gas. Give an example.

When a reaction produces a gas, entropy almost always increases because gases have the greatest freedom of movement and the largest number of possible microstates. Even if the number of moles stays constant, converting solids or liquids to gases introduces far more disorder.

Example:
CaCO₃(s) → CaO(s) + CO₂(g). The production of gaseous CO₂ creates a large increase in entropy.

Consequence: This explains why many decomposition reactions producing gases are driven by entropy at higher temperatures.

2. Describe and explain the entropy change when ammonium nitrate dissolves in water.

When NH₄NO₃ dissolves in water, it dissociates into NH₄⁺(aq) and NO₃⁻(aq). Although the ions are surrounded by ordered shells of water molecules (hydration), the increase in the number of particles and the dispersal of ions throughout the solvent lead to a net increase in entropy.

Consequence: This explains why dissolution of salts like NH₄NO₃ can occur spontaneously even though the process is endothermic — the positive entropy change drives the reaction.

1. Explain why some endothermic reactions are spontaneous. Use Gibbs free energy to support your answer.

Endothermic reactions absorb heat (ΔH > 0), which on its own would make them non-spontaneous. However, if the entropy change of the system (ΔS) is strongly positive, the TΔS term in the Gibbs free energy equation can outweigh the positive ΔH:

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS

At sufficiently high temperature, the –TΔS term dominates, making ΔG negative and the reaction spontaneous.

Example: The dissolution of some salts (e.g., NH₄NO₃) is endothermic but still spontaneous because the entropy increase is large.

3. Explain the significance of ΔG = 0 at equilibrium, and how this links to the expression ΔG° = –RT lnK.

ΔG = 0 at equilibrium because there is no net driving force for the reaction to move in either direction. The system has reached maximum stability. Substituting into the general Gibbs equation,

ΔG=ΔG∘+RTln K

and setting Q = K gives:

ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln KΔG∘=−RTlnK

Thus, equilibrium is the point where the Gibbs free energy of the system is minimised and directly connected to the equilibrium constant.

4. Describe how the magnitude of K relates to the sign and size of ΔG°. Give examples.

• Large K (≫1): Equilibrium lies far to the right. lnK is positive and large, so ΔG° = –RT lnK is strongly negative. The reaction is highly spontaneous. Example: combustion of hydrocarbons.

• Small K (≪1): Equilibrium lies far to the left. lnK is negative, so ΔG° is positive. The reaction is not spontaneous under standard conditions. Example: decomposition of CaCO₃ at room temperature.

• K ≈ 1: lnK ≈ 0, so ΔG° ≈ 0. Reaction has no strong thermodynamic drive in either direction.

Why is the entropy of a perfect crystal at 0 K predicted to be zero?

According to the Third Law of Thermodynamics, the entropy of a perfect crystal at absolute zero (0 K) is zero. At this temperature, all particle motion ceases, and there is only one possible arrangement of the particles in the perfect lattice. Since entropy measures the number of possible microstates, and there is only one, S = 0.

Reactivity 2.3 — What is the likely composition of an equilibrium mixture when ΔG° is positive?

If ΔG° is positive, the forward reaction is not thermodynamically favourable. The equilibrium constant K is therefore much less than 1, meaning the equilibrium mixture will contain mostly reactants with very little product formed.

1. State the Second Law of Thermodynamics and explain its significance for spontaneity.

The Second Law of Thermodynamics states that for any spontaneous process, the total entropy of the universe (system + surroundings) must increase.

WHY. Entropy is a measure of disorder. Natural processes tend to move toward more probable states, which correspond to higher disorder. For example, gases diffuse to fill a container because there are vastly more arrangements of particles in a spread-out state than in a confined one.

HOW. In a chemical reaction, the entropy change of the system may be positive or negative. However, if the entropy of the surroundings changes enough to compensate, the overall entropy of the universe can still increase.

CONSEQUENCES. This law underpins the Gibbs free energy equation:

ΔG=ΔH−TΔS

A reaction is spontaneous if ΔG < 0, which is equivalent to the total entropy of the universe increasing.

2. Explain why the entropy of a perfect crystal at 0 K is defined as zero.

According to the Third Law of Thermodynamics, the entropy of a perfect crystal at absolute zero (0 K) is zero.

WHY. Entropy measures the number of possible microstates. At 0 K, all particle motion ceases and the crystal is in a single, perfectly ordered arrangement with only one microstate.

HOW. Since entropy is defined as S=kln⁡WS = k \ln WS=klnW, where WWW is the number of microstates, if W=1W = 1W=1, then S=0S = 0S=0.

CONSEQUENCES. This law provides a universal reference point, allowing absolute values of entropy to be calculated and compared across different substances.

3. Describe and explain the trend in standard molar entropy values (S°) across different states of matter.

Model Answer:
The standard molar entropy of a substance increases in the order:

WHY. Solids have particles fixed in a lattice with minimal freedom, so their entropy values are lowest. Liquids have greater freedom of movement, so their entropy values are higher. Gases have maximum freedom and occupy the largest number of possible microstates, so their entropy values are highest.

HOW (example). For H₂O at 298 K:

• S°(ice) = 48 J K⁻¹ mol⁻¹

• S°(liquid water) = 70 J K⁻¹ mol⁻¹

• S°(steam) = 189 J K⁻¹ mol⁻¹

CONSEQUENCES. This explains why phase changes such as melting and vaporisation involve significant entropy increases and play a key role in spontaneity.

4. Explain why entropy changes can be used to predict the feasibility of reactions producing a different number of gaseous moles.

Model Answer:
In reactions involving gases, changes in the number of moles strongly influence entropy.

WHY. Entropy increases when the number of gas molecules increases, because more molecules means more possible arrangements and higher disorder. Conversely, entropy decreases if gas molecules are consumed.

HOW (examples).

• N₂(g) + 3H₂(g) → 2NH₃(g). Here, four gas molecules become two, so entropy decreases (ΔS < 0).

• CaCO₃(s) → CaO(s) + CO₂(g). Gas molecules are produced from a solid, so entropy increases (ΔS > 0).

CONSEQUENCES. Such considerations are crucial in predicting feasibility: reactions with ΔS < 0 may still occur if ΔH is negative enough, while reactions with ΔS > 0 are more easily driven at high temperatures.

5. Why can an endothermic reaction become spontaneous at high temperatures?

Endothermic reactions (ΔH > 0) absorb heat, which would normally make them non-spontaneous. However, spontaneity depends on Gibbs free energy:

ΔG=ΔH−TΔS

WHY. If ΔS is positive, the –TΔS term becomes increasingly large and negative as temperature rises.

HOW. At sufficiently high temperature, the –TΔS term can outweigh the positive ΔH, making ΔG negative.

EXAMPLE. The thermal decomposition of calcium carbonate:

CaCO3(s)→CaO(s)+CO2(g)CaCO₃(s) \to CaO(s) + CO₂(g)CaCO3​(s)→CaO(s)+CO2​(g)

is endothermic, but the large entropy gain from gas formation makes it spontaneous at high T.

CONSEQUENCE. This principle explains why industrial processes like decomposition and dissolution are temperature-dependent.

6. Explain why the dissolution of many salts in water is spontaneous even when it is endothermic.

Some salts, like ammonium nitrate, dissolve endothermically (ΔH > 0).

WHY. Dissolution greatly increases entropy because an ordered crystal lattice is broken down into freely moving hydrated ions. The increase in system entropy (ΔS > 0) is usually large enough to make ΔG negative.

HOW. For NH₄NO₃(s) → NH₄⁺(aq) + NO₃⁻(aq), ΔH ≈ +25 kJ mol⁻¹ but ΔS is strongly positive. At room temperature, the –TΔS term outweighs ΔH, so ΔG < 0.

CONSEQUENCES. This explains why endothermic salts dissolve readily and why dissolution is used in cold packs.

7. Explain why ΔG° provides only a thermodynamic criterion for spontaneity and does not determine reaction rate.

ΔG° tells us whether a reaction is thermodynamically feasible (ΔG° < 0 means spontaneous).

WHY. However, spontaneity does not mean the reaction will be fast. The rate depends on kinetic factors, such as activation energy and the availability of a suitable pathway.

HOW (example). The decomposition of diamond to graphite has ΔG° < 0, so it is thermodynamically spontaneous. Yet, the reaction is extremely slow under normal conditions because the activation energy is very high.

CONSEQUENCES. A negative ΔG° guarantees feasibility but not speed. Both thermodynamics (ΔG) and kinetics (Ea) must be considered in predicting reaction behaviour.