The Second Law of Thermodynamics

Core Concept

The Second Law of Thermodynamics states that for any spontaneous process, the total entropy of the universe increases: ΔS_universe > 0. The universe is the system plus its surroundings, so ΔS_universe = ΔS_system + ΔS_surroundings. A spontaneous process may decrease the system's entropy (water freezing) as long as the surroundings' entropy increases by more. At equilibrium, ΔS_universe = 0. This is the fundamental, universal test of spontaneity — more fundamental than ΔG, which is a convenience derived from it.

Key Ideas

• Universe = system + surroundings. The surroundings are everything outside the system that can exchange heat with it.
• The decisive quantity. A process is spontaneous if and only if ΔS_universe > 0; it is at equilibrium when ΔS_universe = 0.
• Heat transfer drives ΔS_surroundings. For a process at constant temperature, ΔS_surroundings = −ΔH_system / T.
• Exothermic favors spontaneity. When ΔH_system < 0, the surroundings gain heat, so ΔS_surroundings > 0 — an exothermic reaction pushes the universe toward higher entropy.
• Temperature moderates the effect. The same heat transfer produces a larger surroundings-entropy change at low T than at high T (dividing by T).

Equations and Variables

ΔS_univ = ΔS_sys + ΔS_surr, with ΔS_surr = −ΔH_sys / T.

How It Works

1. Any process transfers energy and redistributes matter between the system and surroundings.
2. Compute the system's entropy change (from tables or ΔS = q_rev/T for reversible paths).
3. Compute the surroundings' change: heat −ΔH_sys leaves or enters the surroundings, changing its entropy by −ΔH_sys/T.
4. Add the two. If the sum is positive, the process is spontaneous; if zero, equilibrium; if negative, the process is nonspontaneous (its reverse is spontaneous).

Worked Example

Is ice melting spontaneous at 25 °C? Use ΔS_universe.

Melting: H₂O(s) → H₂O(l), with ΔH_fus = +6.01 kJ/mol = +6010 J/mol.

• ΔS_sys = ΔH_fus / T_m = 6010 J/mol ÷ 273 K = +22.0 J/(mol·K) (entropy increases on melting).
• ΔS_surr = −ΔH_sys / T = −6010 J/mol ÷ 298 K = −20.2 J/(mol·K) (surroundings lose heat to the endothermic melting).
• ΔS_univ = +22.0 + (−20.2) = +1.8 J/(mol·K) > 0, so melting is spontaneous at 25 °C.

At exactly 0 °C (273 K), ΔS_surr = −6010/273 = −22.0 J/(mol·K), making ΔS_univ = 0 — the melting point is the equilibrium (phase-transition) temperature.

How it works

1. Any process transfers energy and redistributes matter between the system and surroundings.
2. Compute the system's entropy change (from tables or ΔS = q_rev/T for reversible paths).
3. Compute the surroundings' change: heat −ΔH_sys leaves or enters the surroundings, changing its entropy by −ΔH_sys/T.
4. Add the two. If the sum is positive, the process is spontaneous; if zero, equilibrium; if negative, the process is nonspontaneous (its reverse is spontaneous).

Common confusions

• "The entropy of the system must increase for any spontaneous process." — Wrong. Only ΔS_universe must increase; water freezing is spontaneous with ΔS_sys < 0.
• "ΔS_surroundings = ΔH/T." — Wrong sign. Heat leaving the system (exothermic) increases surroundings entropy, so ΔS_surr = −ΔH_sys/T.
• "Temperature can be in °C." — Wrong. Must use Kelvin; using °C gives nonsense (and division by zero near 0 °C).
• "ΔS_universe = 0 means nothing is happening." — Wrong. It means the system is at equilibrium — forward and reverse processes balance.
• "The Second Law forbids local order." — Wrong. Local order (crystallization, life) can and does form; it is always paid for by a larger entropy increase elsewhere.

Quick review

• Second Law: spontaneous ⇒ ΔS_universe > 0; equilibrium ⇒ ΔS_universe = 0.
• ΔS_univ = ΔS_sys + ΔS_surr.
• ΔS_surr = −ΔH_sys/T (Kelvin).
• Exothermic helps spontaneity by raising surroundings entropy.
• A decrease in system entropy can still be spontaneous if surroundings entropy rises more.