Entropy, Gibbs Energy and Spontaneity: Quick Revision

A brief note on Entropy, Gibbs Energy and Spontaneity. Read it before a test. Try each question at the end before you open its answer.

Entropy

• A spontaneous change happens on its own. That says nothing about how fast.
• A fall in enthalpy helps, but some endothermic changes are spontaneous too.
• Entropy, $S$, measures how spread out matter and energy are. Gas $>$ liquid $>$ solid.
• $\Delta S > 0$ for melting, boiling, dissolving and reactions that make more moles of gas.
• $\Delta S = q_{rev}/T$; for a change of state, $\Delta S = \Delta H / T$.
• Second law: $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0$ for a spontaneous change, with $\Delta S_{surr} = -\Delta H/T$.
• Third law: a perfect crystal has $S = 0$ at $0\ \text{K}$. $\Delta_r S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})$.

Gibbs energy

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

Gibbs energy and K

• $\Delta G^\circ = -RT\ln K = -2.303\,RT\log K$.
• $\Delta G^\circ < 0$: $K > 1$. $\Delta G^\circ = 0$: $K = 1$. $\Delta G^\circ > 0$: $K < 1$.
• At $298\ \text{K}$, $2.303\,RT \approx 5.71\ \text{kJ/mol}$.

A worked example

A change has $\Delta H = +40\ \text{kJ/mol}$ and $\Delta S = +120\ \text{J/(K mol)}$. Is it spontaneous at $298\ \text{K}$, and where does it turn?

1. Units. $\Delta S = 0.120\ \text{kJ/(K mol)}$.
2. At 298 K. $\Delta G = 40 - 298 \times 0.120 = +4.24\ \text{kJ/mol}$. Not spontaneous.
3. The turning point. $T = \dfrac{40\,000}{120} = 333\ \text{K}$.
4. Meaning. Both signs are positive, so it is spontaneous above $333\ \text{K}$.

Practice with new numbers

A program to try

The program tries each of the four sign cases at a cold and a hot temperature.

Mistakes to avoid

• Only exothermic changes are spontaneous. Melting ice takes in heat and still happens.
• Mixing kJ and J. Divide $\Delta S$ by $1000$ before combining it with $\Delta H$ in kJ/mol.
• Celsius in the formulas. Every $T$ must be in kelvin.
• $\Delta G$ and $\Delta G^\circ$ confused. At equilibrium $\Delta G = 0$; $\Delta G^\circ$ fixes $K$.
• A negative $\Delta G$ means fast. It says the change can happen, not how quickly.

Check yourself

1. Predict the sign of $\Delta S$ when ammonia gas dissolves in water.

2. Why is a reaction with $\Delta H > 0$ and $\Delta S < 0$ never spontaneous?

3. A reaction has $K = 1000$ at $298\ \text{K}$. Is $\Delta G^\circ$ positive or negative, and about how large?

4. What is $\Delta S_{total}$ for a system at equilibrium?