Some processes lower the system's entropy and happen anyway: water freezes, and ammonia forms from a reaction with ΔS° = −198 J/K. The second law explains this by counting the surroundings too.
Key idea
Second law: a process is thermodynamically favored (spontaneous) when the entropy of the universe increases: $\Delta S_\text{univ} = \Delta S_\text{sys} + \Delta S_\text{surr} > 0$. At constant temperature and pressure, the surroundings gain entropy from the heat the system releases: $\Delta S_\text{surr} = -\dfrac{\Delta H_\text{sys}}{T}$.
Why it works
When a reaction releases heat, that energy spreads into the surroundings and gives them more ways to share energy, raising their entropy. The same amount of heat matters more when the surroundings are cold: adding 1 kJ to a cold body raises its entropy more than adding it to a hot one, because the cold body had fewer ways to share energy to begin with. So the entropy gain is the heat divided by $T$. The minus sign appears because heat released by the system ($\Delta H < 0$) is heat gained by the surroundings.
Worked examples
Worked example: Water freezing
For water, $\Delta H_\text{fus} = 6.01\ \text{kJ/mol}$ and $\Delta S_\text{fus} = 22.0\ \text{J/(mol·K)}$. Is freezing favored at $-10\ ^\circ\text{C}$ and at $+10\ ^\circ\text{C}$?
- Freezing: $\Delta H_\text{sys} = -6010\ \text{J}$, $\Delta S_\text{sys} = -22.0\ \text{J/K}$ per mole.
- At $263\ \text{K}$: $\Delta S_\text{surr} = \dfrac{6010}{263} = +22.9\ \text{J/K}$; $\Delta S_\text{univ} = -22.0 + 22.9 = +0.9\ \text{J/K}$. Favored.
- At $283\ \text{K}$: $\Delta S_\text{surr} = \dfrac{6010}{283} = +21.2\ \text{J/K}$; $\Delta S_\text{univ} = -22.0 + 21.2 = -0.8\ \text{J/K}$. Not favored: ice melts instead.
- At $273\ \text{K}$ the two cancel: ice and water are in equilibrium at the melting point.
Answer: Favored at −10 °C, not at +10 °C.
Worked example: Making ammonia
At $298\ \text{K}$, $\mathrm{N_2 + 3H_2 \to 2NH_3}$ has $\Delta H^\circ = -91.8\ \text{kJ}$ and $\Delta S^\circ = -198.1\ \text{J/K}$. Is it favored?
- $\Delta S_\text{surr} = -\dfrac{-91\,800\ \text{J}}{298\ \text{K}} = +308.1\ \text{J/K}$.
- $\Delta S_\text{univ} = -198.1 + 308.1 = +110.0\ \text{J/K}$. Positive: favored.
Answer: Yes: $\Delta S_\text{univ} = +110\ \text{J/K}$.
Common mistake
Judging whether a process is favored from the system's entropy alone. Freezing and ammonia synthesis both lower the system's entropy, yet each can be favored because the heat released raises the surroundings' entropy more. Check: always add $-\Delta H/T$ before deciding.
"Spontaneous" in thermodynamics means favored, not fast. Whether it happens at a noticeable rate is a question of kinetics.
The Science Lab's Entropy and the Second Law chapter lets you run a gas backward and see why the reverse never happens on its own.