Entropy, S, is a measure of energy and matter dispersal (or dilution). A system that is a perfect, pure crystal at 0 K is defined as having zero entropy.

The entropy of a reversible process can be related to the amount of heat exchanged at a particular temperature for that process such that

\[\Delta S = \dfrac{q_p}{T}\]

or equivalently,

\[\Delta S = \dfrac{\Delta H}{T}\]

Entropy is usually given in units of J K–1, and therefore, T is in K in the equations above.


A process is said to be spontaneous if it increases the entropy of the universe (ΔSuniverse > 0).

\[\Delta S_{\mathrm{univ}} = \Delta S_{\mathrm{surr}} + \Delta S_{\mathrm{sys}}\]

\[\begin{align*} \Delta S_{\mathrm{univ}} > 0 &\quad \mathrm{for~a~spontaneous~process}\\ \Delta S_{\mathrm{univ}} < 0 &\quad \mathrm{for~a~nonspontaneous~process}\\ \Delta S_{\mathrm{univ}} = 0 &\quad \mathrm{for~a~system~at~equilibrium} \end{align*}\]

Example: Melting of Ice

The entropy change for the process below is 22.1 J K–1 and requires 6.025 kJ of heat from the surroundings.

\[\mathrm{H_2O}(s) \longrightarrow \mathrm{H_2O}(l)\]

Is the process spontaneous at –10.00 °C? Is it spontaneous at 10.00 °C?

At –10.00 °C

\[\begin{align*} \Delta S_{\mathrm{univ}} &= \Delta S_{\mathrm{surr}} + \Delta S_{\mathrm{sys}} \\[1.5ex] &= \dfrac{-6.025\times 10^{3}~\mathrm{J}}{263.15~\mathrm{K}} + \dfrac{22.1~\mathrm{J}}{\mathrm{K}} \\[1.5ex] &= -0.8~\mathrm{J~K^{-1}} \end{align*}\]

No, it is not spontaneous at –10.00 °C.

At +10.00 °C

\[\begin{align*} \Delta S_{\mathrm{univ}} &= \Delta S_{\mathrm{surr}} + \Delta S_{\mathrm{sys}} \\[1.5ex] &= \dfrac{-6.025\times 10^{3}~\mathrm{J}}{283.15~\mathrm{K}} + \dfrac{22.0~\mathrm{J}}{\mathrm{K}} \\[1.5ex] &= 0.8~\mathrm{J~K^{-1}} \end{align*}\]

Yes, it is spontaneous at 10.00 °C.

Predicting Entropy

We can predict if entropy increases or decreases with some straightforward rules of thumb. For example, entropy increases

  1. as you go from a solid -> liquid -> gas
  2. as you increase the number of moles of gas
  3. with molecular weight (due to more electrons) and complexity

Example 1: Entropy is positive for the following process.

\[\mathrm{H_2O}(l) \longrightarrow \mathrm{H_2O}(g)\]

Example 2: Entropy is negative for the following process

\[3\mathrm{H_2}(g) + \mathrm{N_2}(g) \longrightarrow \mathrm{2NH_3}(g)\]

since we are reducing the number of moles of gas from 4 to 2.

Example 3: Which has larger entropy?

  • He(g)
  • Xe(g)

Entropy is larger Xenon gas as it has a larger molecular weight (and, consequently, more electrons).

Practice Problems

Problem 1

A process takes place at 100 °C and requires 44.01 kJ worth of heat from the surroundings. What is the entropy (in J K–1) of this process?


\[\begin{align*} \Delta S &= \dfrac{\Delta H}{T}\\ &= \dfrac{44.01~\mathrm{kJ}}{373.15~\mathrm{K}}\\ &= \dfrac{44,010~\mathrm{J}}{373.15~\mathrm{K}}\\ &= 117.9~\mathrm{J~K^{-1}} \end{align*}\]