Entropy and second law

Spontaneous processes need a second criterion beyond $\Delta H$. A spontaneous process happens on its own, without continuous outside push, once it starts — spontaneity says nothing about how fast it happens, only that it can. Many spontaneous processes are exothermic (rusting, combustion), which tempted early chemists to think "exothermic means spontaneous." But ice melting above $0\ ^\circ\text{C}$ and many salts dissolving are spontaneous and endothermic. Enthalpy alone cannot be the whole story; a second quantity, entropy, is needed.

Entropy measures disorder, and it is a state function. $S$ depends only on the current state of the system, not on how it got there, so for any process

where: $S_{final}$ and $S_{initial}$ are the entropies of the final and initial states. Because $S$ is a state function, $\Delta S$ for a reaction can be tracked just like $\Delta H$, without worrying about path. Entropy rises whenever particles gain freedom to spread into more arrangements (more "microstates"):

• melting (solid to liquid) or vaporisation (liquid to gas) — molecules gain freedom of motion;
• a solid or liquid dissolving into solution — an ordered lattice disperses among solvent molecules;
• the number of moles of gas increasing in a reaction — more independent gas particles, more ways to arrange them;
• heating a substance — more thermal motion, more microstates become accessible.

The reverse of any of these — condensation, crystallisation, decreasing gas moles, cooling — tends to lower entropy.

The second law: total entropy of the universe increases for any spontaneous process.

where: $\Delta S_{system}$ is the entropy change of the reacting substances, $\Delta S_{surroundings}$ is the entropy change of everything outside the system (the rest of the universe, in practice the immediate thermal environment), and $\Delta S_{total}$ is their sum. Equality holds only for an idealised reversible process at equilibrium; every real, spontaneous process makes $\Delta S_{total}$ strictly positive. This is the rigorous test for spontaneity — not $\Delta S_{system}$ alone. A system's own entropy can fall in a spontaneous process (water freezing below $0\ ^\circ\text{C}$: the system becomes more ordered) provided the surroundings gain enough entropy to compensate (freezing releases heat, warming and disordering the surroundings by more than the system ordered itself).

Linking $\Delta S_{surroundings}$ to heat flow. At constant pressure and temperature $T$, the entropy the surroundings gain is tied directly to the heat the system releases:

where: $\Delta H_{system}$ is the enthalpy change of the system (negative for exothermic reactions) and $T$ is the absolute temperature in kelvin at which heat transfer occurs. An exothermic reaction ($\Delta H_{system} < 0$) dumps heat into the surroundings, making $\Delta S_{surroundings}$ positive; an endothermic reaction pulls heat from the surroundings, cooling and ordering them, so $\Delta S_{surroundings}$ is negative. This one equation is what turns "entropy" from a qualitative disorder story into a number you can add to $\Delta S_{system}$ and compare against zero — the whole engine behind Worked example 2 below, and the qualitative seed of the later Gibbs free energy topic, which packages $\Delta H$ and $\Delta S$ of the system into a single criterion without needing $\Delta S_{surroundings}$ explicitly.

For a reaction at $298\ \text{K}$, $\Delta H_{system} = -180\ \text{kJ mol}^{-1}$ and $\Delta S_{system} = -150\ \text{J K}^{-1}\text{mol}^{-1}$. Is it spontaneous at $298\ \text{K}$?

Step 1. Convert $\Delta H_{system}$ to joules so units match $\Delta S$: $\Delta H_{system} = -180\ \text{kJ mol}^{-1} = -180{,}000\ \text{J mol}^{-1}$.

Step 2. Find $\Delta S_{surroundings}$ using $\Delta S_{surroundings} = -\Delta H_{system}/T$: $\Delta S_{surroundings} = -(-180{,}000)/298 = 604.0\ \text{J K}^{-1}\text{mol}^{-1}$.

Step 3. Add to get $\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} = -150 + 604.0 = 454.0\ \text{J K}^{-1}\text{mol}^{-1}$.

Step 4. Since $\Delta S_{total} > 0$, the reaction is spontaneous at $298\ \text{K}$, even though $\Delta S_{system} < 0$ — the system itself becomes more ordered, but the large exothermic heat release drives up the surroundings' entropy by far more, and the total still increases. Answer: spontaneous, with $\Delta S_{total} = +454.0\ \text{J K}^{-1}\text{mol}^{-1}$.

Competitive stretch

At $T = 1200\ \text{K}$ instead, $\Delta S_{surroundings} = -(-180{,}000)/1200 = 150.0\ \text{J K}^{-1}\text{mol}^{-1}$, so $\Delta S_{total} = -150 + 150.0 = 0\ \text{J K}^{-1}\text{mol}^{-1}$ — the borderline of equilibrium. Raising $T$ shrinks the favourable $-\Delta H_{system}/T$ contribution, so a reaction comfortably spontaneous at $298\ \text{K}$ can approach non-spontaneity at high $T$ if $\Delta S_{system}$ is negative — the temperature sensitivity the later Gibbs free energy criterion, $\Delta G = \Delta H - T\Delta S$, tracks directly.

• Judging spontaneity from $\Delta S_{system}$ alone instead of $\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings}$ — a system can lose entropy in a spontaneous process if the surroundings gain more.
• Confusing "spontaneous" with "fast." Spontaneity only says a process can occur on its own eventually; rate is a separate question belonging to chemical kinetics.
• Assuming all exothermic reactions are spontaneous or all endothermic ones are not — both $\Delta H$ and $\Delta S$ of the system, weighed together through $\Delta S_{total}$, decide it.
• Dropping or flipping the sign in $\Delta S_{surroundings} = -\Delta H_{system}/T$ — an exothermic system ($\Delta H_{system}<0$) must give a positive $\Delta S_{surroundings}$, not negative.
• Forgetting to convert kJ to J (or vice versa) before adding $\Delta S_{surroundings}$ (in J K⁻¹mol⁻¹) to $\Delta S_{system}$ — mismatched units silently corrupt the sum.

Ask the tutor: Want to work through more $\Delta S_{total}$ calculations, or check your sign conventions on a tricky exothermic-versus-endothermic problem?