Entropy and Gibbs Free Energy

What is it?

Some reactions happen on their own once started (they are spontaneous); others need a continuous input of energy. Enthalpy alone cannot decide which: ice melts spontaneously at room temperature even though melting is endothermic. A second quantity is needed.

Entropy, $S$ , measures how spread out energy and matter are: the number of ways the particles and their energy can be arranged. Gases have high entropy, liquids less, and solids the least.

The second law of thermodynamics: in a spontaneous process, the total entropy of the universe increases.

Chemists combine enthalpy and entropy into the Gibbs free energy change:

\Delta G = \Delta H - T\Delta S

where $T$ is the temperature in kelvin.

Key idea

At constant temperature and pressure, a reaction is spontaneous when $\Delta G$ is negative, at equilibrium when $\Delta G = 0$ , and non-spontaneous (the reverse is spontaneous) when $\Delta G$ is positive. “Spontaneous” says nothing about speed: diamond turning into graphite is spontaneous, but immeasurably slow.

Why does it matter?

Predicting reactions. $\Delta G$ tells chemists whether a reaction can go at all, and at what temperature it becomes possible, before they try it.
Industry. Lime ( $\ce{CaO}$ ) is made by heating limestone above about 850 °C, the temperature at which $\Delta G$ for the decomposition becomes negative.
Life and batteries. Cells couple reactions with negative $\Delta G$ (such as using ATP) to drive ones with positive $\Delta G$ ; a battery’s voltage is a direct measure of $\Delta G$ ( $\Delta G = -nFE$ ).

How does it work?

1. Predicting the sign of ΔS

Entropy increases ( $\Delta S$ positive) when:

a solid melts or a liquid boils (solid → liquid → gas);
the number of moles of gas increases, for example $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$ ;
a solid dissolves to form ions in solution (usually);
the temperature rises.

Entropy decreases when gas molecules combine into fewer molecules, for example $\ce{N2(g) + 3H2(g) -> 2NH3(g)}$ (4 mol of gas → 2 mol).

2. Calculating ΔS°

Unlike enthalpies of formation, absolute standard entropies, $S°$ , can be measured for every substance, including elements (they are not zero). Then:

\begin{aligned} &\Delta S° = \textstyle\sum n S°(\text{products}) \\[2pt] &\qquad - \textstyle\sum n S°(\text{reactants}) \end{aligned}

$S°$ values are in J mol⁻¹ K⁻¹, so $\Delta S°$ comes out in J/K. Convert it to kJ/K before combining it with $\Delta H°$ in kJ.

3. The four cases

ΔH	ΔS	ΔG = ΔH − TΔS	Spontaneous?
−	+	always negative	at all temperatures
+	−	always positive	never (the reverse is)
−	−	negative when T is low	at low temperature
+	+	negative when T is high	at high temperature

In the last two cases, the reaction switches at the temperature where $\Delta G = 0$ :

T = \frac{\Delta H}{\Delta S}

4. ΔG° and the equilibrium constant

The standard free energy change is linked to the equilibrium constant:

\Delta G° = -RT \ln K

with $R$ = 8.314 J mol⁻¹ K⁻¹. A large negative $\Delta G°$ means a large $K$ (products favoured); a positive $\Delta G°$ means $K$ is less than 1.

Think of it like this

Think of a tidy bedroom. Left alone, it gets messier, because there are far more messy arrangements than tidy ones: that is entropy. Tidying it needs energy (an enthalpy cost). Whether the room ends up tidy depends on both: how much energy you put in, and how strongly “messiness” wins. Temperature sets how much weight entropy gets in the balance.

More precisely

$\Delta G$ is the maximum useful (non-expansion) work a process can do at constant temperature and pressure. The relation $\Delta G = \Delta H - T\Delta S$ comes from the total entropy change: $\Delta S_\text{universe} = \Delta S_\text{system} - \Delta H/T$ , so $\Delta G = -T\Delta S_\text{universe}$ , and a negative $\Delta G$ is the same as an increase in the entropy of the universe. Using $\Delta H°$ and $\Delta S°$ at 298 K for other temperatures is an approximation, because both change a little with temperature.

Visualise it

A graph of ΔG° in kJ against temperature in kelvin from 0 to 1400 K, with the region below zero shaded as spontaneous. The line for N2 + 3H2 → 2NH3 (ΔH° and ΔS° both negative) starts at −91.8 kJ and rises, crossing zero at 463 K: spontaneous when cool. The line for CaCO3 → CaO + CO2 (ΔH° and ΔS° both positive) starts at 179.2 kJ and falls, crossing zero at 1119 K: spontaneous when hot. Crossing point: T = ΔH° divided by ΔS°. — When ΔH and ΔS have the same sign, temperature decides: the line crosses ΔG° = 0 at T = ΔH°/ΔS°.

Worked example

Worked example: ΔS°, ΔG° and the switching temperature for ammonia synthesis

Question: For $\ce{N2(g) + 3H2(g) -> 2NH3(g)}$ , $\Delta H°$ = −91.8 kJ. Using $S°$ (J mol⁻¹ K⁻¹): $\ce{N2}$ 191.6, $\ce{H2}$ 130.7, $\ce{NH3}$ 192.8, find $\Delta S°$ , $\Delta G°$ at 298.15 K, and the temperature above which the reaction is no longer spontaneous.

$\Delta S° = 2(192.8) - [191.6 + 3(130.7)] = 385.6 - 583.7 = -198.1\ \text{J/K} = -0.1981\ \text{kJ/K}$ (negative: 4 mol of gas become 2)
$\Delta G°$ :
$\small\begin{aligned} &\Delta G° = -91.8\ \text{kJ} \\[4pt] &\quad - (298.15\ \text{K}) \\[4pt] &\quad\quad \times (-0.1981\ \text{kJ/K}) \\[4pt] &\quad = -91.8\ \text{kJ} + 59.06\ \text{kJ} \\[4pt] &\quad = \mathbf{-32.7\ kJ} \end{aligned}$
Spontaneous at 298 K. Switching temperature: $T = \dfrac{\Delta H°}{\Delta S°} = \dfrac{-91.8\ \text{kJ}}{-0.1981\ \text{kJ/K}} =$ 463 K. Above this, $\Delta G°$ is positive.

Worked example: Why limestone must be heated

Question: For $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$ , find $\Delta H°$ , $\Delta S°$ and $\Delta G°$ at 298.15 K, and the minimum temperature for decomposition. (Data in the note above.)

$\Delta H° = [-634.9 + (-393.5)] - (-1207.6) = +179.2\ \text{kJ}$
$\Delta S° = [38.1 + 213.8] - 91.7 = +160.2\ \text{J/K} = +0.1602\ \text{kJ/K}$
$\Delta G°_{298} = 179.2\ \text{kJ} - (298.15\ \text{K})(0.1602\ \text{kJ/K}) = 179.2\ \text{kJ} - 47.76\ \text{kJ} =$ +131.4 kJ: not spontaneous at room temperature.
$T = \dfrac{179.2\ \text{kJ}}{0.1602\ \text{kJ/K}} =$ 1119 K (about 846 °C). Lime kilns run hotter than this.

Worked example: From ΔG° to K

Question: Estimate $K$ at 298.15 K for ammonia synthesis ( $\Delta G°$ = −32.7 kJ = −3.27 × 10⁴ J).

$\ln K = -\dfrac{\Delta G°}{RT} = -\dfrac{-3.27 \times 10^{4}\ \text{J/mol}}{(8.314\ \text{J mol}^{-1}\,\text{K}^{-1})(298.15\ \text{K})} = 13.2$
$K = e^{13.2} \approx$ 5 × 10⁵: products are strongly favoured at 298 K (even though the reaction is too slow without a catalyst).

Common mistake

Common mistake: Mixing J and kJ

$\Delta S°$ is usually in J/K but $\Delta H°$ in kJ. Convert $\Delta S°$ to kJ/K (divide by 1000) before calculating $\Delta H - T\Delta S$ , or the entropy term will be 1000 times too large.

Common mistake: Using °C in ΔG = ΔH − TΔS

$T$ must be in kelvin. At 25 °C, use 298.15 K; using 25 would make the entropy term about twelve times too small.

Common mistake: Thinking spontaneous means fast

A negative $\Delta G$ says a reaction can happen, not how fast. Speed is kinetics (activation energy); a mixture of hydrogen and oxygen is spontaneous but does nothing until a spark starts it.

Notation note

The degree sign (°) means standard conditions: 1 bar and, unless stated, 298.15 K.
Entropy units: J mol⁻¹ K⁻¹ for $S°$ , J/K for $\Delta S°$ of a reaction as written.
Some books use “feasible” or “thermodynamically favourable” instead of “spontaneous”.

Remember this

Entropy: gases high, solids low; ΔS is positive when moles of gas increase.
$\Delta S° = \sum nS°(\text{products}) - \sum nS°(\text{reactants})$ , in J/K; convert to kJ/K.
$\Delta G = \Delta H - T\Delta S$ (T in K). Negative ΔG: spontaneous. Switching temperature $T = \Delta H/\Delta S$ .
$\Delta G° = -RT\ln K$ and $\Delta G° = -nFE°$ .

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/gibbs-free-energy/