Chemical Thermodynamics · Lesson 8 of 10

Spontaneity, Entropy, and the Thermodynamic Laws

“Explain the preferred direction of change using the entropy of both system and surroundings.”

Learning Objectives

• Distinguish thermodynamic spontaneity from speed. • Explain why enthalpy alone is not a universal direction criterion. • Predict and calculate entropy changes in simple processes. • Use total entropy to apply the second law. • Explain the third law and the role of absolute entropy data.

Energy conservation does not choose a direction

The first law accounts for energy, but it does not say whether heat will flow from cold to hot by itself. We observe the reverse: heat flows spontaneously from a warmer body to a cooler one. A gas spreads into available space and two different gases mix when a partition is removed. The reverse changes require intervention even though energy conservation need not forbid them.

Definition

Spontaneous process

A process that has a thermodynamic tendency to proceed in the stated direction under specified conditions. Spontaneity does not specify how rapidly it proceeds or rule out an initial activation requirement.

A mixture of hydrogen and oxygen may persist at room temperature because the reaction is extremely slow until initiated. A tendency toward lower Gibbs energy is not a prediction of immediate visible reaction. A spontaneous process is irreversible in its natural direction; restoring both system and surroundings requires an external change.

Exothermic changes often proceed spontaneously because heat released increases the entropy of the surroundings. But an enthalpy decrease is not the complete criterion. Melting above a substance’s melting temperature and many dissolving processes are endothermic yet can occur spontaneously. Temperature and the state of the surroundings matter.

Formation of ammonia, hydrogen chloride, and liquid water from their reference elements gives negative reaction enthalpies in the source examples. Formation of nitrogen dioxide and liquid carbon disulphide gives positive values: approximately $+ 33.2$ and $+ 128.5 kJ mo l^{- 1}$ , respectively. These values illustrate both signs of enthalpy; neither sign by itself establishes spontaneity at an unspecified temperature. A complete direction test must include entropy and the stated conditions.

Enthalpy levels for opposite heat effects— These are endpoint enthalpy diagrams, not activation-energy graphs. Spontaneity must also account for entropy.

Entropy and the distribution of energy

Imagine two gases initially kept on opposite sides of a removable partition in an isolated container. Removing it lets each gas spread throughout the available volume. The combined state is more dispersed and has many more possible microscopic arrangements. For ideal gases at the same temperature, this mixing can occur with no enthalpy change.

Definition

Entropy

A state function, S, associated with the dispersal of energy and the number of microscopic arrangements compatible with the macroscopic state. “Disorder” is a useful first picture, but entropy is a quantitative thermodynamic property, not a judgment about visual untidiness.

A gas-mixing thought experiment— Initially each gas is restricted to half the vessel. After mixing, either gas can be found throughout the same total volume.

For the same substance under comparable conditions, a gas generally has higher molar entropy than its liquid, and a liquid higher than an ordered crystal. Heating normally increases entropy by making more energy arrangements accessible. Producing more gas particles often increases entropy; eliminating gas particles often decreases it. These trends guide predictions, while tabulated entropy values provide quantitative tests.

Example — Predicting entropy changes

Problem
Predict the entropy trend for crystallising a liquid, warming a crystal from near 0 K to 115 K, decomposing solid sodium hydrogen carbonate to a solid and gases, and dissociating hydrogen gas into atoms.

1.Crystallisation restricts the arrangements available to particles, so the system entropy decreases.
2.Warming the crystal increases accessible thermal motions and energy arrangements, so entropy increases. Zero-temperature order does not mean that every form of microscopic motion or energy vanishes.
3.In $2NaHC O_{3} (s) \to N a_{2} C O_{3} (s) + C O_{2} (g) + H_{2} O (g)$ , gaseous products are formed from solids, so a substantial entropy increase is expected.
4.For $H_{2} (g) \to 2H (g)$ , the number of freely moving gaseous particles increases; the entropy change is expected to be positive.

Calculating entropy and applying the second law

The same amount of heat changes entropy more when transferred reversibly at a lower temperature than at a higher one. The temperature divisor is essential. For an isothermal reversible transfer, entropy change equals reversible heat divided by absolute temperature. For a changing temperature, the contributions must instead be accumulated as an integral.

Entropy changeLaTeX

d S = \frac{δ q _{rev}}{T}, Δ S = \int_{1}^{2} \frac{δ q _{rev}}{T}

For a constant-temperature reversible process this becomes ΔS = q_rev/T. T must be in kelvin; entropy has units J K−1, or J mol−1 K−1 for a molar value.

Entropy is a state function, even though actual heat transfer is path dependent. To calculate the entropy difference for an irreversible process, imagine a reversible route with the same initial and final states. Do not divide the irreversible heat by temperature and assume it is the system entropy change.

Total entropy criterionLaTeX

Δ S_{total} = Δ S_{sys} + Δ S_{surr} \geq 0

For an isolated combined system, a spontaneous irreversible change gives a positive total; the reversible limit gives zero. At stable equilibrium no allowed spontaneous change increases total entropy further.

This is the entropy statement of the second law. An isolated system can undergo a spontaneous change while its energy remains constant: internal energy alone cannot distinguish the direction. Its entropy increases until it reaches the maximum permitted by the constraints. A decrease in the entropy of a non-isolated system is allowed if the surroundings increase by a larger amount.

Example — Reversible expansion versus free expansion

Problem
One mole of ideal gas doubles its volume isothermally. Compare its entropy change for a reversible route and an adiabatic free expansion.

1.For a reversible isothermal route, $q_{rev} = n R T ln (V_{f} / V_{i})$ . Divide by T to get $Δ S_{sys} = n R ln 2 = (1) (8.314) (0.6931) = 5.76 J K^{- 1}$ .
2.The free expansion has the same ideal-gas temperature and volume endpoints, so its system entropy change is also $+ 5.76 J K^{- 1}$ , despite its actual $q = 0$ .
3.In reversible heat exchange with a reservoir at the same limiting temperature, the surroundings lose $5.76 J K^{- 1}$ . The total is zero.
4.For the insulated free expansion there is no heat transfer to the surroundings, so their entropy change is zero and the total is $+ 5.76 J K^{- 1}$ .
5.The same system state change can therefore be achieved reversibly or irreversibly with different entropy production in system plus surroundings.

Example — Iron oxidation despite decreasing system entropy

Problem
For $4Fe (s) + 3 O_{2} (g) \to 2F e_{2} O_{3} (s)$ at 298 K, use $Δ S_{sys} = - 549.4 J K^{- 1} mo l^{- 1}$ and $Δ H_{sys} = - 1648 \times 1 0^{3} J mo l^{- 1}$ . Is its total entropy change positive?

1.At constant pressure the reaction transfers its released heat to the surroundings. A large surrounding reservoir at 298 K receives $+ 1648 \times 1 0^{3} J$ per written reaction amount.
2.Its entropy change is $Δ S_{surr} = - Δ H_{sys} / T = 1648000/298 \approx 5530.2 J K^{- 1} mo l^{- 1}$ .
3.Add the system change: $Δ S_{total} \approx 5530.2 - 549.4 = 4980.8 J K^{- 1} mo l^{- 1}$ .
4.The total is positive, so the thermodynamic direction is spontaneous under these conditions even though the system becomes more ordered.

The third law and absolute entropy

The third law supplies a reference for entropy rather than merely its differences. For a perfectly ordered pure crystal with a unique ground state, entropy approaches zero as temperature approaches absolute zero. Imperfect ordering, mixtures, or frozen-in disorder need not satisfy that ideal limiting description.

Starting from this reference, reversible heating contributions and phase-transition contributions can be added to calculate an absolute molar entropy at a higher temperature. In a single phase, small contributions have the form $C_{p} d T / T$ ; a reversible transition at its equilibrium temperature contributes $Δ H_{transition} / T$ . Standard reaction entropy is then found by a products-minus-reactants sum of standard molar entropies.

Standard reaction entropyLaTeX

Δ_{r} S^{\circ} = i \sum a_{i} S_{m, i}^{\circ} (products) - j \sum b_{j} S_{m, j}^{\circ} (reactants)

Unlike standard formation enthalpies of reference elements, standard molar entropies of elements at ordinary temperatures are not set to zero.

Common mistake

The system entropy does not have to increase in every spontaneous process; total entropy is the criterion. Also, the third law does not assign zero entropy to ordinary-temperature elements, and the formula q_rev/T must not be applied to arbitrary irreversible heat transfer.

Quick check

What does spontaneity tell us?

Quick check

Which change is expected to increase the system entropy?

Quick check

An isothermal reversible process absorbs 600 J at 300 K. What is its entropy change?

Quick check

A spontaneous process decreases system entropy by 10 J/K. Which surrounding change is compatible with the second law?

Quick check

Why can an irreversible ideal-gas free expansion have positive system entropy change when q is zero?

Quick check

To which limiting state does the zero-entropy statement of the third law apply?

Practice Problems

Explain the difference between reaction spontaneity and reaction rate using a slowly reacting gas mixture.
Predict the entropy signs for melting, condensation, heating a solid, and producing gas from a solid; give a reason in each case.
Find the entropy change for reversible melting of one mole of ice at 273 K using a fusion enthalpy of 6.00 kJ/mol.
A system has entropy change −30 J/K and releases 12.0 kJ at constant pressure to a reservoir at 300 K. Calculate the surrounding and total entropy changes.
Explain why an isolated system may have $Δ U = 0$ while its entropy increases, and state what changes at equilibrium.
Describe how thermal measurements and the third-law reference can produce an absolute entropy value without setting ordinary-temperature elemental entropies to zero.

Key Takeaways

Spontaneity describes direction, not speed, and is not determined by enthalpy alone. Entropy is a state function calculated using a reversible heat route. The second law concerns the total entropy of an isolated system plus surroundings. The third law supplies a zero-temperature reference for a perfectly ordered pure crystal and permits absolute entropy calculations.

Previous · Lesson 7

Lattice, Solution, and Dilution Enthalpies

Next · Lesson 9

Gibbs Energy, Temperature, and Equilibrium

Energy conservation does not choose a direction

Entropy and the distribution of energy

Calculating entropy and applying the second law

The third law and absolute entropy

Quiz

Practice Problems

Key Takeaways