Enthalpy says one thing. Entropy says another. When they disagree, who wins? Gibbs free energy is the referee: one number that takes both sides into account and tells you straight out whether a reaction can happen on its own.
📚 What you need to know
ΔG° = ΔH° − TΔS° — this one is in the data booklet, section 1.
A negative ΔG means the reaction can happen on its own. Positive means it cannot.
ΔH is in kJ mol−1, ΔS is in J K−1 mol−1, so divide ΔS by 1000 before you use it.
T must be in kelvin: add 273 to a Celsius temperature.
Two ways to find ΔG°: from ΔH° and ΔS°, or by adding up ΔG°f values like a Hess sum.
Free energy is the energy actually available to do something useful, once entropy has taken its cut.
Because T multiplies the entropy term, temperature can change the answer completely.
Two forces, one number
Picture a tug of war. On one side, enthalpy: reactions like giving out heat, so a negative ΔH pulls towards “yes, go”. On the other side, entropy: things like spreading out, so a positive ΔS also pulls towards “yes, go”.
Sometimes both pull the same way and the reaction is a certainty. Sometimes they fight, and then the winner depends on the temperature. Gibbs free energy adds the two pulls together into a single score.
The Gibbs equation
ΔG° = ΔH°reaction − TΔS°system
Notice the minus sign in front of the entropy term. A positive ΔS makes ΔG more negative, which is why spreading out helps a reaction go.
Read the equation out loud as a sentence: “the energy I can actually use equals the heat I release, minus what entropy takes as its fee.” That fee is TΔS, and it gets more expensive as things heat up.
The units trap
This is the single biggest source of lost marks in the whole topic, and it is not chemistry — it is arithmetic. ΔH comes in kilojoules. ΔS comes in joules. You cannot subtract one from the other until they match.
Do the division on its own line, before it goes anywhere near the Gibbs equation. Trying to do it inside the brackets is how the 1000 gets lost.
🧩 Route 1: from ΔH° and ΔS°
Write down ΔH° in kJ mol−1 exactly as given.
Divide ΔS° by 1000 to get kJ K−1 mol−1. Write the converted value down.
Check T is in kelvin. If the question says 25 °C, that is 298 K.
Substitute into ΔG° = ΔH° − TΔS°, keeping the signs in brackets.
Answer in kJ mol−1, then say what the sign means.
Signs inside brackets. If ΔS° is negative, −TΔS° becomes a plus. Write it as −92 − (298 × −0.199) and let the brackets do the thinking, rather than trying to work out the sign in your head.
Route 2: adding up ΔG°f values
Standard free energies of formation behave exactly like enthalpies of formation. If the question hands you a table of ΔG°f values, you never need the Gibbs equation at all.
The Hess-style route
ΔG° = ΣΔG°f(products) − ΣΔG°f(reactants)
Here elements in their standard states are zero, just like with enthalpy: ΔG°f for O2(g) is 0. That is the one place where entropy and free energy behave differently, and it catches people out both ways.
Symbol
What it is
Units
Watch out for
ΔG°
Free energy change: your verdict on the reaction
kJ mol−1
Negative means it can go
ΔH°
Enthalpy change: heat taken in or given out
kJ mol−1
Already in kJ, leave it alone
ΔS°
Entropy change: how much things spread out
J K−1 mol−1
Must be divided by 1000
T
Temperature of the reaction
K
Add 273 if given in °C
ΔG°f
Free energy of formation of one substance
kJ mol−1
Zero for elements, unlike S°
Worked examples
WORKED EXAMPLE
For N2(g) + 3H2(g) → 2NH3(g), ΔH° = −92 kJ mol−1 and ΔS° = −199 J K−1 mol−1. Calculate ΔG° at 298 K.
Step 1: convert the entropy value−199 ÷ 1000 = −0.199 kJ K⁻¹ mol⁻¹Step 2: substitute, keeping the signs in bracketsΔG° = −92 − (298 × −0.199)Step 3: the bracket first298 × −0.199 = −59.3, and −92 − (−59.3) = −92 + 59.3ΔG° = −32.7 kJ mol⁻¹, so it can happen at 298 KBoth terms fought: enthalpy said yes, entropy said no. Enthalpy won, but only by 33 kJ — hold that thought for the next page.
WORKED EXAMPLE
For MgCO3(s) → MgO(s) + CO2(g), ΔH° = +100 kJ mol−1 and ΔS° = +175 J K−1 mol−1. Calculate ΔG° at 298 K and comment.
Step 1: convert+175 ÷ 1000 = +0.175 kJ K⁻¹ mol⁻¹Step 2: substituteΔG° = +100 − (298 × 0.175)Step 3: work it out298 × 0.175 = 52.15, so 100 − 52.15 = +47.85ΔG° = +47.9 kJ mol⁻¹, so it will not go at room temperatureAnd that is right: you have to heat a carbonate in a Bunsen flame to decompose it. Entropy is helping, just not enough yet.
WORKED EXAMPLE
Use ΔG°f values to find ΔG° for CH4(g) + 2O2(g) → CO2(g) + 2H2O(l). Values in kJ mol−1: CH4 −51, O2 0, CO2 −394, H2O(l) −237.
Step 1: products, with balancing numbers−394 + (2 × −237) = −394 − 474 = −868Step 2: reactants (oxygen is an element, so zero)−51 + (2 × 0) = −51Step 3: products − reactants−868 − (−51) = −868 + 51ΔG° = −817 kJ mol⁻¹Hugely negative, which is why methane burns the instant you give it a spark. No temperature or entropy data needed for this route.
💡 Exam tip
Show the ÷ 1000 as its own line. One mark is often reserved for the conversion alone.
Use brackets round every negative number you substitute. It removes the sign guesswork.
Check which route the question wants. A table of ΔG°f values means route 2; a temperature in the question means route 1.
Give three significant figures unless told otherwise, and always attach kJ mol−1.
Finish with a sentence. “ΔG° is negative, so the reaction is spontaneous at this temperature” is usually the last mark.
Do not round the converted entropy. Keep −0.199, not −0.2, or your answer drifts.
⚠ Common mix-up
Forgetting to divide ΔS by 1000. You get an answer thousands of kJ out and a sign that may be wrong too.
Using °C instead of K. Using 25 instead of 298 shrinks the entropy term to nothing.
Losing the minus on −TΔS. A negative ΔS makes that term positive, and skipping it flips your conclusion.
Setting elements to zero for ΔS°. Zero is only for ΔG°f and ΔH°f of elements.
Saying a reaction is fast because ΔG is negative. Free energy says nothing about rate. Diamond turning into graphite has a negative ΔG and takes millions of years.
Reading ΔG as a kind of energy you can measure with a thermometer. It is a calculated balance, not heat.
You can now get a number for ΔG at one temperature. The really useful question is what happens when you change that temperature — and which reactions can be switched on with a Bunsen burner. Up next: Spontaneous Reactions.
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