E. Chemical Systems and Equilibrium · Chemical EquilibriumExtension: Entropy, Gibbs Free Energy and K
E3.2 E3.1 A1.12 A1.13About 70 minutes
Learning goals- I can describe entropy as a measure of how spread out energy and matter are, and predict the sign of ΔS.
- I can use ΔG = ΔH − TΔS to decide whether a process is thermodynamically favourable at a given temperature.
- I can relate the sign and size of ΔG° to K using ΔG° = −RT ln K.
- I can explain why equilibrium positions change with temperature, connecting to Le Châtelier's principle.
Words to know
entropy (S)
A measure of how spread out (dispersed) the energy and particles of a system are; units J/(mol·K).
thermodynamically favourable (spontaneous)
A process that, once started, can continue without outside energy being continually supplied. It says nothing about how fast the process is.
Gibbs free energy change (ΔG)
ΔG = ΔH − TΔS. At constant temperature and pressure, a process is favourable when ΔG < 0.
standard free energy change (ΔG°)
ΔG for a reaction with all substances in their standard states (1 mol/L for solutions, 1 bar for gases); related to K by ΔG° = −RT ln K.
natural logarithm (ln)
The logarithm with base e ≈ 2.718; ln K and K = e^x are inverse operations (use the ln and eˣ keys on a scientific calculator).
Why does a reaction stop where it does?
So far, K has been a number to measure and use. But why is K for ammonia formation large at room temperature and small in a hot reactor? Why does ice melt above 0 °C but water freeze below it? The answer comes from thermodynamics, which balances two tendencies: the drive to release energy (lower enthalpy) and the drive to spread out (higher entropy). This lesson is an extension: many SCH4U classes use it to explain the position of equilibrium, and it is central to first-year university chemistry. The energy terms (ΔH) build on Energy Changes and Rates of Reaction.
Entropy: the tendency to spread out
Entropy (S) measures how spread out a system's energy and particles are. Nature tends toward greater total entropy: a drop of food colouring spreads through water; perfume spreads through a room; it never happens in reverse by itself.
Predicting the sign of ΔS for a change:
- Solid → liquid → gas: S increases (gases have by far the most entropy).
- More moles of gas among the products: ΔS > 0. Fewer moles of gas: ΔS < 0.
- Dissolving a solid in a liquid usually increases S.
- Heating a substance increases its S.
Examples: H₂O(s) → H₂O(l): ΔS > 0. N₂(g) + 3 H₂(g) → 2 NH₃(g): 4 mol of gas become 2, so ΔS < 0. CaCO₃(s) → CaO(s) + CO₂(g): a gas forms, so ΔS > 0.
Gibbs free energy: putting it together
At constant temperature and pressure, the Gibbs free energy change combines both tendencies:
ΔG = ΔH − TΔS (T in kelvins; convert ΔS from J/K to kJ/K)
- ΔG < 0: the process is thermodynamically favourable (it can proceed forward on its own).
- ΔG > 0: not favourable in the forward direction (the reverse is favourable).
- ΔG = 0: the system is at equilibrium.
The signs of ΔH and ΔS decide how temperature matters:
| ΔH | ΔS | ΔG = ΔH − TΔS | Example |
|---|
| − (exothermic) | + (more spread out) | negative at all T: always favourable | burning fuels |
| + (endothermic) | − (less spread out) | positive at all T: never favourable | 6 CO₂ + 6 H₂O → glucose + 6 O₂ without light |
| − (exothermic) | − | favourable at LOW T | water freezing; ammonia synthesis |
| + (endothermic) | + | favourable at HIGH T | ice melting; limestone decomposing |
Worked exampleExample 1: The melting point is an equilibrium
For H₂O(s) → H₂O(l), ΔH = +6.01 kJ/mol and ΔS = +22.0 J/(mol·K). Find ΔG at −10 °C and +10 °C, and find the temperature where ice and water are at equilibrium.
- Convert: ΔS = 0.0220 kJ/(mol·K).
- At −10 °C (263 K): ΔG = 6.01 − 263(0.0220) = 6.01 − 5.79 = +0.22 kJ/mol > 0: melting is not favourable (water freezes).
- At +10 °C (283 K): ΔG = 6.01 − 283(0.0220) = 6.01 − 6.23 = −0.22 kJ/mol < 0: melting is favourable.
- At equilibrium ΔG = 0, so T = ΔH ÷ ΔS = 6.01 ÷ 0.0220 = 273 K.
Ice melts above 273 K (0 °C) and water freezes below it; at 273 K, ice and water are in dynamic equilibrium (ΔG = 0). This is the phase equilibrium from lesson 1, explained by thermodynamics.
ΔG° and the equilibrium constant
The standard free energy change, ΔG°, is directly linked to K:
ΔG° = −RT ln K or K = e^(−ΔG°/RT)
with R = 8.314 J/(mol·K) and ΔG° in J (not kJ).
- ΔG° < 0 → ln K > 0 → K > 1 (products favoured).
- ΔG° = 0 → K = 1.
- ΔG° > 0 → K < 1 (reactants favoured).
Because K depends on e raised to a power, a small change in ΔG° makes a big change in K: every 5.7 kJ makes K about 10 times larger or smaller at 25 °C.
A note on units
The K in ΔG° = −RT ln K is the thermodynamic equilibrium constant. It uses concentrations in mol/L for solutions and partial pressures in bar for gases (1 bar = 100 kPa). So for gas reactions it is Kp in bar, which differs from the Kp in kPa used earlier in this unit when Δn ≠ 0. For solution reactions it matches the K you have been using.
Worked exampleExample 2: Ammonia synthesis, at room temperature and in a hot reactor
For N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), ΔH° = −92.2 kJ and ΔS° = −198.1 J/K. (a) Find ΔG° and K at 298 K. (b) Above what temperature does K become less than 1? (Assume ΔH° and ΔS° stay roughly constant.)
- (a) ΔG° = ΔH° − TΔS° = −92.2 − 298(−0.1981) = −92.2 + 59.0 = −33.2 kJ = −3.32 × 10⁴ J.
- ln K = −ΔG° ÷ RT = 33 200 ÷ (8.314 × 298) = 13.40, so K = e^13.40 = 6.6 × 10⁵.
- (b) K < 1 when ΔG° > 0, that is, when TΔS° outweighs ΔH°.
- ΔG° = 0 at T = ΔH° ÷ ΔS° = (−92 200 J) ÷ (−198.1 J/K) = 465 K (about 192 °C).
(a) ΔG° = −33.2 kJ, K ≈ 6.6 × 10⁵ (in bar units): products strongly favoured at 25 °C. (b) Above about 465 K, K < 1. This is the thermodynamic reason an exothermic reaction's K falls as temperature rises, matching Le Châtelier's principle.
ΔG° for N₂ + 3 H₂ ⇌ 2 NH₃ rises with temperature because ΔS° is negative. Below about 465 K, K > 1; above it, K < 1. (The line assumes ΔH° and ΔS° stay constant.)Worked exampleExample 3: When does limestone decompose?
For CaCO₃(s) ⇌ CaO(s) + CO₂(g), ΔH° = +178 kJ and ΔS° = +160 J/K. Estimate the temperature above which K > 1.
- Both ΔH° and ΔS° are positive: favourable only at high temperature.
- T = ΔH° ÷ ΔS° = 178 000 J ÷ 160 J/K = 1.11 × 10³ K, about 840 °C.
Above about 1110 K (about 840 °C), K > 1. This is why lime kilns, which make quicklime (CaO) for cement and steel, are run at high temperatures, and why limestone is stable at everyday temperatures.
Just for funYour bedroom does not get messy because of entropy alone. But entropy does explain why it never tidies itself.
Hands-on activityWatching entropy and enthalpy compete
You need: two clear glasses, cold tap water, hot tap water (not boiling; a grown-up should handle hot water), food colouring, a few ice cubes, a thermometer if you have one.
- Fill one glass with cold water and one with hot water. Let them settle for a minute. Add one drop of food colouring to the centre of each without stirring. Watch for 2 minutes.
- Put two ice cubes in a glass of room-temperature water. Record the water temperature over 10 minutes.
- For each observation, say whether ΔS is positive or negative for the process, and whether ΔH is positive, negative or close to zero.
What to look for: the colour spreads by itself in both glasses (ΔS > 0, ΔH ≈ 0), and faster in hot water because particles move faster; that is a rate effect, not a change in whether it happens. The ice melts (ΔH > 0, ΔS > 0) because at room temperature the TΔS term is bigger than ΔH, so ΔG < 0. The water temperature falls toward 0 °C because melting absorbs heat.
Your turn
1Predict the sign of ΔS: (a) H₂O(l) → H₂O(g); (b) 2 SO₂(g) + O₂(g) → 2 SO₃(g); (c) NaCl(s) → Na⁺(aq) + Cl⁻(aq).
(a) positive (liquid to gas); (b) negative (3 mol gas to 2); (c) positive (a solid dissolving into ions).
2A process has ΔH = −50 kJ and ΔS = +100 J/K. Is it favourable at 298 K?
ΔG = −50 − 298(0.100) = −50 − 29.8 = −79.8 kJ < 0. Yes (and at every temperature).
3For H₂O(l) → H₂O(g), ΔH = +40.7 kJ/mol and ΔS = +109 J/(mol·K). Find the temperature at which liquid and vapour are at equilibrium at standard pressure.
T = 40.7 ÷ 0.109 = 373 K (100 °C), the normal boiling point.
4A reaction has ΔG° = −22.8 kJ at 700 K. Find K.
ln K = 22 800 ÷ (8.314 × 700) = 3.918, so K = e^3.918 = 50.3.
5A reaction has K = 1.0 × 10⁻⁵ at 298 K. Is ΔG° positive or negative? Calculate it.
K < 1, so ΔG° > 0. ΔG° = −(8.314)(298) ln(1.0 × 10⁻⁵) = −(2478)(−11.51) = +2.85 × 10⁴ J = +28.5 kJ.
6Explain why a reaction with ΔG° < 0 can still be too slow to notice.
ΔG° tells you whether products are favoured at equilibrium, not how fast the reaction gets there. A high activation energy can make it extremely slow (diamond turning into graphite is a famous example).
More practice: warm-up
1Which state of a substance has the highest entropy?
2Write the Gibbs equation.
3What is ΔG for a system at equilibrium?
4If ΔG° is negative, is K greater or less than 1?
5Convert ΔS = −198 J/K to kJ/K.
More practice: on your own
1A reaction has ΔH = +30.0 kJ and ΔS = +80.0 J/K. Above what temperature is it favourable?
T = 30.0 ÷ 0.0800 = 375 K. It is favourable above 375 K (102 °C).
2For 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), ΔH° = −198 kJ and ΔS° = −188 J/K. Find ΔG° at 298 K and decide whether K is larger or smaller than 1.
ΔG° = −198 − 298(−0.188) = −198 + 56.0 = −142 kJ. K is much greater than 1 at 298 K.
3Use the same data to estimate the temperature at which K = 1.
T = ΔH° ÷ ΔS° = 198 000 ÷ 188 = 1053 K ≈ 1.05 × 10³ K (about 780 °C). Above this, reactants are favoured.
4Find ΔG° at 298 K for a reaction with K = 4.0.
ΔG° = −(8.314)(298) ln 4.0 = −(2477.6)(1.386) = −3.43 × 10³ J = −3.4 kJ.
5Show that a ΔG° change of 5.7 kJ changes K by a factor of about 10 at 298 K.
K changes by e^(5700/(8.314 × 298)) = e^2.30 = 10.0.
Challenge
1Use ΔG° = ΔH° − TΔS° and ΔG° = −RT ln K to show that for an exothermic reaction, K decreases as T increases (assume ΔH° and ΔS° are constant).
Combining: ln K = −ΔH° ÷ RT + ΔS° ÷ R. For an exothermic reaction, ΔH° < 0, so −ΔH°/RT is positive and gets smaller as T increases. So ln K, and K, decrease as T rises, which is exactly what Le Châtelier's principle predicts.
2For the Haber reaction (ΔH° = −92.2 kJ, ΔS° = −198.1 J/K), estimate K at 700 K (about 427 °C) using the constant ΔH°/ΔS° assumption.
ΔG° = −92.2 − 700(−0.1981) = −92.2 + 138.7 = +46.5 kJ. ln K = −46 500 ÷ (8.314 × 700) = −7.99, so K = e^(−7.99) ≈ 3.4 × 10⁻⁴ (in bar units). Products are not favoured, which is why industry uses very high pressure to push the equilibrium toward NH₃.
3Photosynthesis, 6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂, has ΔG° of about +2.9 × 10³ kJ. How can plants make glucose if ΔG° is so positive?
A process with ΔG > 0 can happen when it is coupled to an energy source. Plants capture light energy, which drives the reaction; the light energy more than makes up the positive ΔG. Without light, the reaction does not proceed.
Quiz
1. Which change has a negative ΔS?
2. A process with ΔH > 0 and ΔS > 0 is favourable:
3. For ice melting, ΔH = +6.01 kJ/mol and ΔS = +22.0 J/(mol·K). At 298 K, ΔG is about:
4. If ΔG° = 0 for a reaction, then K equals:
5. A reaction has K = 1.0 × 10⁸ at 298 K. Its ΔG° is:
6. At equilibrium, which statement is true?
Big idea (extension to E3.1 and E3.2): entropy measures spreading out; ΔG = ΔH − TΔS decides whether a process is favourable; ΔG° = −RT ln K links thermodynamics to the equilibrium constant. This explains why K changes with temperature, and why phase changes happen at definite temperatures.
How to help: this lesson goes a little beyond the Ontario expectations, as many Grade 12 courses do, and is excellent preparation for university chemistry. If your teen finds the logarithms tricky, practise the calculator keys: ln and eˣ undo each other.
Common difficulties:
- Mixing kJ and J (ΔH is usually in kJ, ΔS in J/K, and R in J/(mol·K)).
- Using °C instead of K.
- Thinking ΔG < 0 means fast. Thermodynamics says "whether", kinetics says "how fast".
Safety: an adult should handle hot water in the activity.