Gibbs free energy — predicting spontaneous reactions — Chemistry

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Gibbs free energy at Pre-University · coming soon

Two entropy changes, every time

The entropy lecture answered a hard question: why does the cold pack work? Ammonium nitrate dissolves and takes in heat, yet it happens by itself. The answer needed two calculations. First, the entropy change of the system, +108.7 J/K/mol. Second, the entropy change of the surroundings, $- Δ H / T$ , about −86.2 J/K/mol. Only their total, +22.5, gave the verdict.

That works. But it means doing two calculations and keeping track of the surroundings every time. The entropy lecture ended with a promise: the next lecture would combine the two entropy changes into one quantity, the Gibbs free energy. Here it is. It looks only at the system, it uses ΔH and ΔS that you can already find from data, and it gives the same verdict in one line.

From the total entropy to ΔG

Start from the second law, as the entropy lecture wrote it. A change happens by itself only if

Δ S_{total} = Δ S_{system} - \frac{Δ H}{T} > 0

Now multiply every term by $- T$ . Two things happen. The fraction disappears, because $T$ cancels. And the inequality turns round, because multiplying by a negative number reverses it:

- T ΔS total = ΔH - T ΔS system < 0

The right-hand side is the new quantity. It is called the change in Gibbs free energy, $Δ G$ :

Gibbs free energy

$Δ G = Δ H - T Δ S$

$Δ H$ and $Δ S$ belong to the system (the reaction). $T$ is the temperature in kelvin.

A change can happen by itself (it is feasible) when $Δ G$ is negative.

Look at what just happened. $Δ G$ is simply $- T$ times the total entropy change. Temperature in kelvin is always positive. So whenever the total entropy rises, $Δ G$ is negative, and whenever the total entropy falls, $Δ G$ is positive. It is the second law in a different form, and it needs only the system's own values.

That is it. That is the whole idea. A negative $Δ G$ means that the universe's entropy increases, so the change can happen by itself.

Test it on the cold pack. $Δ H = + 25.7$ kJ/mol and $Δ S = + 108.7$ J/K/mol. Units first: ΔH is in kilojoules, so change ΔS to kilojoules too, $+ 0.1087$ kJ/K/mol. Then

ΔG = 25.7 - 298 \times 0.1087 \approx -6.7 kJ/mol

Negative, so the salt dissolves by itself. And $- 298 \times 22.5$ J/K/mol $\approx - 6.7$ kJ/mol, so it is exactly the entropy lecture's answer in a new form.

Three words need care.

Feasible, or spontaneous, means "can happen by itself, without a continuous supply of energy from outside". It does not mean fast. It does not even mean that it will happen today. Speed is a separate question.
ΔG = 0 means neither direction is favoured. Ice and water at 0 °C are like this: they stay together, with melting and freezing balanced.
When the data are standard values, the result is written ΔG^⦵. It describes the reaction under standard conditions: 100 kPa for gases and 1 mol/dm³ for solutions. A positive ΔG^⦵ does not mean that no product ever forms. It means that, under those conditions, the reactants are favoured. The equilibrium lectures make this precise.

Four kinds of reaction

The equation has two parts that compete: $Δ H$ and $- T Δ S$ . Think about the signs. You do not need to learn a table. You can read each case from the equation.

ΔH negative, ΔS positive. Both parts are negative, so ΔG is negative at every temperature. Always feasible.
ΔH positive, ΔS negative. Both parts are positive, so ΔG is positive at every temperature. Never feasible by itself.
ΔH negative, ΔS negative. ΔH helps, but $- T Δ S$ is positive and grows as $T$ rises. At low temperature ΔH wins. Feasible at low temperature only.
ΔH positive, ΔS positive. ΔH works against the change, but $- T Δ S$ is negative and grows as $T$ rises. At high temperature it wins. Feasible at high temperature only.

A picture makes this clear. Because $Δ G = Δ H - T Δ S$ has the form of a straight line, a graph of ΔG against $T$ is a straight line. It starts at ΔH when $T = 0$ , and its gradient is $- Δ S$ .

ΔG against temperature for four invented reactions, one of each kind. Each line starts at its ΔH and slopes with gradient −ΔS. Below the axis (shaded) the reaction is feasible. The two dashed lines cross the axis, so their verdict changes with temperature. The solid lines never cross it.

Stop and think

Water freezing gives out heat (ΔH negative) and lowers the water's entropy (ΔS negative). Which kind of reaction is it, and does that match what you see in a freezer?

Show the answer

It is "feasible at low temperature only". ΔH helps, but $- T Δ S$ is positive and grows with temperature. Below 0 °C the ΔH term wins, so water freezes. Above 0 °C the $- T Δ S$ term wins, so it does not. The everyday fact that water freezes only when it is cold enough is this equation at work.

Now sort some real reactions. Use the signs, not memory.

Sorter — which kind of reaction?

Decide the sign of ΔS for each reaction from the moles of gas or the change of state. Then put the card in the right box. Read the reason after each one.

Eight reactions sorted by the signs of ΔH and ΔS.

Always feasible (ΔH −, ΔS +): 2H₂O₂(l) → 2H₂O(l) + O₂(g), ΔH = −196 kJ/mol; C(diamond) → C(graphite), ΔH = −1.9 kJ/mol, ΔS = +3.3 J/K/mol.
Never feasible (ΔH +, ΔS −): 3O₂(g) → 2O₃(g), ΔH = +285 kJ/mol.
Feasible at low temperature only (ΔH −, ΔS −): N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −91.8 kJ/mol; H₂O(l) → H₂O(s), ΔH = −6.01 kJ/mol.
Feasible at high temperature only (ΔH +, ΔS +): CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ/mol; H₂O(s) → H₂O(l), ΔH = +6.01 kJ/mol; NH₄NO₃(s) dissolving, ΔH = +25.7 kJ/mol.

The temperature where the verdict changes

For the two "it depends" kinds, there is a temperature where the line crosses the axis. There, ΔG = 0, so

Δ H - T Δ S = 0, so T = \frac{Δ H}{Δ S}

Keep the units matched: ΔH in kJ/mol and ΔS in kJ/K/mol, so that $T$ comes out in kelvin. This temperature is where the two parts balance. On one side the reaction is feasible, and on the other side it is not. The values of ΔH and ΔS change a little with temperature, so treat the answer as a good estimate, not an exact value.

Explore the line in the graph builder.

Graph builder — ΔG against temperature

Choose a reaction, or set ΔH and ΔS yourself. The line shows ΔG at every temperature. Then move the temperature and read ΔG. Watch where the line crosses into the shaded region.

Reactions:

ΔH = kJ/mol

ΔS = J/K/mol

Temperature T = K

ΔG against temperature for two reactions. Each line crosses the axis where ΔG = 0, at $T = Δ H /Δ S$ .

CaCO₃ → CaO + CO₂: ΔG ≈ +130.3 kJ/mol at 298 K, 0 at about 1111 K, ≈ −25.9 kJ/mol at 1273 K. Feasible only when hot.
N₂ + 3H₂ → 2NH₃: ΔG ≈ −32.8 kJ/mol at 298 K, 0 at about 463 K, ≈ +51.4 kJ/mol at 723 K. Feasible only when cool.

What should you have seen? Each line starts at its ΔH on the left and slopes with gradient −ΔS. For limestone the line starts high and falls, crossing into the shaded region near 1111 K. For ammonia it starts low and rises, leaving the shaded region near 463 K. For hydrogen peroxide the line stays below the axis, and for ozone it stays above, whatever the temperature.

Using ΔG

Worked Example 1 — Making ammonia: feasible, but at what temperature?

N₂(g) + 3H₂(g) → 2NH₃(g). Use ΔH^⦵ = −91.8 kJ/mol and ΔS^⦵ = −198.1 J/K/mol, as in the entropy lecture. Find ΔG^⦵ at 298 K and at 723 K (450 °C), and the temperature where it changes sign.

Convert ΔS: $- 198.1$ J/K/mol $= - 0.1981$ kJ/K/mol.

At 298 K: $Δ G = - 91.8 - 298 \times (- 0.1981) = - 91.8 + 59.0 \approx - 32.8$ kJ/mol. Feasible.

At 723 K: $Δ G = - 91.8 + 723 \times 0.1981 \approx + 51.4$ kJ/mol. Not feasible under standard conditions.

Crossover: $T = 91.8 \div 0.1981 \approx 463$ K, about 190 °C.

So why do ammonia plants run at about 400–450 °C? Because at room temperature the reaction is far too slow: the N≡N bond is very strong. A higher temperature and an iron catalyst make it fast. The price is that, at 450 °C, the reactants are favoured. So plants also use a high pressure, about 200 atmospheres, and only about 15% of the gases turn into ammonia on each pass. The ammonia is removed and the unreacted gases are recycled (Chemguide). Feasible and fast pull in opposite directions here, and industry has to compromise.

Check the sign: ΔH and ΔS are both negative, so this is "feasible at low temperature only". 298 K is below 463 K, and 723 K is above it. ✓

Worked Example 2 — Heating limestone

CaCO₃(s) → CaO(s) + CO₂(g). ΔH^⦵ = +178 kJ/mol and ΔS^⦵ = +160.2 J/K/mol (from the entropy lecture). Above what temperature is the decomposition feasible?

Both are positive, so it is feasible at high temperature only.

$T = \frac{Δ H}{Δ S} = \frac{178}{0.1602} \approx 1111$ K, which is about 838 °C.

At 298 K, $Δ G = 178 - 298 \times 0.1602 \approx + 130$ kJ/mol, strongly positive. Limestone is stable in buildings and cliffs. At 1273 K (1000 °C), $Δ G \approx - 25.9$ kJ/mol, so it decomposes. This is why lime kilns must be very hot. Strictly, the 1111 K applies when the carbon dioxide is at standard pressure, 100 kPa, so treat it as an estimate.

Check: $1111 \times 0.1602 \approx 178$ . ✓

Worked Example 3 — Ice again, and why kelvin matters

For melting ice, ΔH = +6.01 kJ/mol and ΔS = +22.0 J/K/mol. Find ΔG at 25 °C and at −10 °C.

At 25 °C, $T = 298$ K: $Δ G = 6.01 - 298 \times 0.0220 \approx - 0.55$ kJ/mol. Ice melts.

At −10 °C, $T = 263$ K: $Δ G = 6.01 - 263 \times 0.0220 \approx + 0.22$ kJ/mol. Ice does not melt.

Crossover: $6.01 \div 0.0220 \approx 273$ K, the melting point. The entropy lecture found the same temperature from the total entropy change: at 298 K it gave +1.8 J/K/mol, and $- 298 \times 1.8 \div 1000 \approx - 0.54$ kJ/mol. Same verdict, one step shorter.

Now see what happens if you forget to convert to kelvin. With $T = 25$ : $Δ G = 6.01 - 25 \times 0.0220 = + 5.46$ kJ/mol. That says ice does not melt at 25 °C, which is plainly wrong.

Worked Example 4 — Feasible is not fast

For C(diamond) → C(graphite), ΔH^⦵ = −1.9 kJ/mol and ΔS^⦵ = 5.7 − 2.4 = +3.3 J/K/mol. Is the change feasible at 298 K?

$Δ G = - 1.9 - 298 \times 0.0033 \approx - 2.9$ kJ/mol. Negative, and with ΔH negative and ΔS positive, it is negative at every temperature.

So diamond "should" turn into graphite. Yet diamonds last for millions of years. The reason is speed. To rearrange, carbon atoms must break strong covalent bonds in the diamond network. At room temperature that almost never happens. Thermodynamics says whether a change can happen. Kinetics, the next unit, says how fast.

Three classic mistakes

Trap 1 — Using °C instead of K

$T$ in $Δ G = Δ H - T Δ S$ must be in kelvin: $T (K) = T (° C) + 273$ . Celsius can give the wrong sign, as the ice example showed. At 0 °C it would even make $T Δ S$ vanish.

Trap 2 — Mixing kJ and J

ΔH is usually in kJ/mol, but ΔS is in J/K/mol. Divide ΔS by 1000 before you use it. Forgetting makes the $T Δ S$ term 1000 times too big, and the answer is nonsense.

Trap 3 — "Negative ΔG, so it happens straight away"

A negative ΔG means the change can happen by itself. It says nothing about how fast. Diamond to graphite, and hydrogen with oxygen in a balloon, are feasible but do not happen at any noticeable rate at room temperature. And a positive ΔG^⦵ means the reactants are favoured under standard conditions, not that nothing at all can form.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

For H₂(g) + ½O₂(g) → H₂O(l), ΔH^⦵ = −285.8 kJ/mol and ΔS^⦵ = −163.3 J/K/mol. (a) Find ΔG^⦵ at 298 K. (b) Which kind of reaction is this? (c) Estimate the temperature above which it is no longer feasible.

Hint 1
Convert ΔS to kJ/K/mol first.

Hint 2
$Δ G = - 285.8 - 298 \times (- 0.1633)$ . For (c), $T = Δ H /Δ S$ .

Full answer
(a) $Δ G = - 285.8 + 48.7 \approx - 237.1$ kJ/mol. (b) ΔH and ΔS are both negative, so it is feasible at low temperature only. (c) $T = 285.8 \div 0.1633 \approx 1750$ K. Check: data books list the standard Gibbs energy of formation of liquid water as −237.1 kJ/mol, the same value. ✓
Use the limestone data from Worked Example 2. (a) Find ΔG^⦵ at 1273 K. (b) A student heats a lump of limestone with a small flame that brings it to only about 1000 K (an invented value). Will it decompose much? Explain.

Hint 1
$Δ G = 178 - 1273 \times 0.1602$ .

Hint 2
Compare 1000 K with the crossover temperature, 1111 K.

Full answer
(a) $Δ G \approx 178 - 203.9 = - 25.9$ kJ/mol, feasible. (b) At 1000 K, $Δ G \approx 178 - 160.2 = + 17.8$ kJ/mol, still positive, so under standard conditions the reactants are favoured. Little decomposition happens. In practice a small amount can still decompose, because the carbon dioxide escapes into the air at a pressure far below 100 kPa. Check (a): $1273 \times 0.1602 \approx 203.9$ . ✓
Ozone forms from oxygen: 3O₂(g) → 2O₃(g). ΔH^⦵ = +285.4 kJ/mol. S^⦵ values: O₂ 205.2 and O₃ 238.9 J/K/mol (OpenStax data table). (a) Find ΔS^⦵ and ΔG^⦵ at 298 K. (b) Ozone is made high in the atmosphere every day. How can that be, if the reaction is never feasible?

Hint 1
ΔS^⦵ = 2 × 238.9 − 3 × 205.2.

Hint 2
"Feasible" means "can happen by itself". What supplies energy high in the atmosphere?

Full answer
(a) ΔS^⦵ = 477.8 − 615.6 = −137.8 J/K/mol. $Δ G = 285.4 + 298 \times 0.1378 \approx + 326.5$ kJ/mol. With ΔH positive and ΔS negative, it is positive at every temperature. (b) It does not happen by itself. Ultraviolet light from the Sun supplies the energy, splitting O₂ molecules into atoms, which then join other O₂ molecules. A change with positive ΔG can be driven by a continuous supply of energy from outside. Check: $298 \times 0.1378 \approx 41.1$ , and $285.4 + 41.1 = 326.5$ . ✓
Living cells drive "uphill" reactions by joining them to the breakdown of ATP. One step in a cell has ΔG = +14 kJ/mol (invented value), so it cannot happen on its own. Breaking down ATP has ΔG ≈ −30.5 kJ/mol under standard conditions, and about −50 kJ/mol under the conditions inside a cell (biochemistry textbook values). (a) Find ΔG for the joined process in each case. (b) Why must the two changes be joined in one process, rather than happening separately?

Hint 1
When two changes happen together as one process, their ΔG values add, as enthalpies do in Hess's law.

Hint 2
+14 + (−30.5), and +14 + (−50).

Full answer
(a) Standard: $14 - 30.5 = - 16.5$ kJ/mol. In the cell: $14 - 50 = - 36$ kJ/mol. Both are negative, so the joined process is feasible. (b) If ATP broke down on its own, its energy would simply spread out as heat, and the uphill step would still have a positive ΔG. Enzymes link the two so that they happen as one process. Only then does the combined ΔG decide. Check: $- 36 < - 16.5 < 0$ , so the cell's conditions make the joined process even more favourable. ✓

The bigger idea

The American scientist Josiah Willard Gibbs set out these ideas in a long work, On the Equilibrium of Heterogeneous Substances, published in two parts in 1876 and 1878. It appeared in a small journal and was very hard to read, so it spread slowly. It became widely known in Europe after Wilhelm Ostwald translated it into German in the early 1890s. By 1901, Gibbs had received the Copley Medal of the Royal Society, one of the highest honours in science.

His idea now reaches far beyond test tubes. Every step in a living cell is judged by ΔG, and cells use ATP to pay for the uphill steps. This lecture closes the Energetics unit. Energetics answers one question: can a reaction happen by itself? The next unit, Kinetics, answers the other: how fast will it happen? A chemist needs both answers.

Key concepts in this lesson

ΔG = ΔH − TΔS, with T in kelvin and ΔS converted to kJ/K/mol, and it equals −T times the total entropy change.
A change is feasible, meaning it can happen by itself, when ΔG is negative, and feasible does not mean fast.
The signs of ΔH and ΔS decide the four kinds: always feasible (ΔH −, ΔS +), never feasible (ΔH +, ΔS −), feasible at low temperature (both −) and feasible at high temperature (both +).
For the two "it depends" kinds, the verdict changes where ΔG = 0, at T = ΔH ÷ ΔS.
A standard value, ΔG⦵, describes standard conditions, so a positive ΔG⦵ means the reactants are favoured there, not that nothing can form.