Gibbs energy

One quantity instead of two

Working out the entropy of the surroundings every time is tiring. Chemists would like a test that uses the system alone. The American scientist J. Willard Gibbs found one.

He defined the Gibbs energy:

• $G = H - TS$

Like $H$ and $S$, $G$ is a state function. At constant temperature, its change is

• $\Delta G = \Delta H - T\Delta S$

Why delta G works

Start from the second law. With $\Delta S_{surr} = -\Delta H / T$,

• $\Delta S_{total} = \Delta S - \dfrac{\Delta H}{T}$

Multiply both sides by $-T$:

• $-T\Delta S_{total} = \Delta H - T\Delta S = \Delta G$

So $\Delta G$ is just $-T\Delta S_{total}$. When the total entropy rises, $\Delta G$ is negative. At constant temperature and pressure:

• $\Delta G < 0$: the change is spontaneous.
• $\Delta G = 0$: the system is at equilibrium.
• $\Delta G > 0$: the change is not spontaneous; the reverse change is.

$\Delta G$ also measures the useful work a change can do, other than the work of expansion. A negative $\Delta G$ means energy is free to do work, hence the old name "free energy".

The four cases

$\Delta G = \Delta H - T\Delta S$ has two parts. A negative $\Delta H$ helps, and so does a positive $\Delta S$, through $-T\Delta S$.

In the last two rows the two parts pull opposite ways. Temperature decides the winner, because the entropy part grows with $T$.

The turning temperature

In the last two cases, $\Delta G$ changes sign where it is zero:

• $T = \dfrac{\Delta H}{\Delta S}$

Take the decomposition of limestone, $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$. Here $\Delta H^\circ = +178.3\ \text{kJ/mol}$ and $\Delta S^\circ = +160.6\ \text{J/(K mol)}$.

• At $298\ \text{K}$: $\Delta G^\circ = 178.3 - 298 \times 0.1606 = +130.4\ \text{kJ/mol}$. Limestone is safe on a hillside.
• The turning point: $T = \dfrac{178\,300\ \text{J/mol}}{160.6\ \text{J/(K mol)}} = 1110\ \text{K}$.

Above about $1110\ \text{K}$ the decomposition is spontaneous. That is why lime kilns run at more than $1100\ \text{K}$.

Water freezing is the third case. $\Delta H$ and $\Delta S$ are both negative, so it is spontaneous only below $273\ \text{K}$.

Practise with new numbers

Try it in Python

The program finds $\Delta G$ for the decomposition of limestone at several temperatures, and the temperature where it changes sign.

Common mistakes

• Adding kJ and J. Divide $\Delta S$ by $1000$ before you multiply by $T$ and subtract from $\Delta H$ in kJ/mol.
• Thinking a negative $\Delta G$ means a fast reaction. It says the change can happen, not how quickly.
• Forgetting the minus sign in $-T\Delta S$. A positive $\Delta S$ makes $\Delta G$ more negative.
• Using $T = \Delta H/\Delta S$ for every reaction. It gives a real turning point only when $\Delta H$ and $\Delta S$ have the same sign.

Remember these

• $G = H - TS$ and $\Delta G = \Delta H - T\Delta S$ at constant temperature.
• $\Delta G < 0$ spontaneous; $\Delta G = 0$ equilibrium; $\Delta G > 0$ not spontaneous.
• $\Delta H < 0$ and $\Delta S > 0$: always spontaneous. $\Delta H > 0$ and $\Delta S < 0$: never.
• When the signs match, $\Delta G$ changes sign at $T = \Delta H/\Delta S$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A reaction has $\Delta H < 0$ and $\Delta S > 0$. At what temperatures is it spontaneous?

2. Why does the sign of $T\Delta S$ matter more at high temperature?

3. A change has $\Delta H = +12\ \text{kJ/mol}$ and $\Delta S = +40\ \text{J/(K mol)}$. Above what temperature is it spontaneous?

4. What does $\Delta G = 0$ tell you about a system?