Why does limestone only break down in a hot kiln, while a lump of magnesium is desperate to burn at any temperature? Same equation, same two terms — the difference is which signs ΔH and ΔS have, and how big T is. Get this page straight and you can predict the behaviour of a reaction you have never met.
📚 What you need to know
Spontaneous means ΔG ≤ 0 — the reaction can happen on its own, with no help.
Spontaneous says nothing about speed. It can be spontaneous and still take a million years.
There are only four sign combinations of ΔH and ΔS, and each one behaves in a known way.
Exothermic + entropy up = always spontaneous. Endothermic + entropy down = never.
The other two cases depend on temperature, because T multiplies the entropy term.
At the switch-over point ΔG = 0, so T = ΔH° / ΔS°.
This is exactly why metal extraction needs a furnace: heating buys you a bigger entropy term.
What “spontaneous” actually means
In everyday English, spontaneous suggests sudden. In chemistry it means something narrower and less exciting: the reaction is allowed to go by itself, without anything pushing it. That is all.
Petrol and air sitting in a tank have a hugely negative ΔG, yet they sit there for months doing nothing, because the activation energy is too high. Spontaneous tells you where the reaction wants to end up. Rate tells you how long it will take to get there. Two completely separate questions.
If a question asks “will this reaction occur?” they want ΔG. If it asks “will it occur quickly?” they want activation energy and collisions. Answer the one you were asked, and never use ΔG to explain a rate.
The test
ΔG° negative → spontaneous ΔG° positive → not spontaneous ΔG° = 0 → at equilibrium, no push either way
The four cases
Look at the equation again: ΔG° = ΔH° − TΔS°. You are adding two terms. The first term is negative when the reaction is exothermic. The second term, −TΔS°, is negative when ΔS° is positive. So there are four ways this can play out.
The two boxes on the diagonal are the ones worth memorising: exothermic with more spreading out always goes, endothermic with less spreading out never does.
ΔH
ΔS
ΔG
Spontaneous?
Because
Negative (exothermic)
Positive
Always negative
Always
Both terms push the same way
Negative (exothermic)
Negative
Negative at low T
Only when cool
TΔS beats ΔH once it is hot
Positive (endothermic)
Positive
Negative at high T
Only when hot
TΔS must grow big enough to beat ΔH
Positive (endothermic)
Negative
Always positive
Never
Both terms push against it
Temperature is the switch
Here is the key insight, and it is hiding in plain sight. ΔH° and ΔS° barely change with temperature. What changes is T, and it is multiplying the entropy term. So if you plot ΔG against T, you get a straight line: the intercept is ΔH° and the gradient is −ΔS°.
The two lines that cross the dashed zero line are the temperature-dependent cases. Where they cross is the temperature you are asked to calculate.
Reading the graph. A line that starts below zero and rises (amber) is an exothermic reaction that entropy is fighting — heat it too much and it switches off. A line that starts above zero and falls (blue) is an endothermic reaction that entropy is helping — heat it enough and it switches on. That second one is how we get metals out of their ores.
Finding the switch-over temperature
At the exact temperature where a reaction changes from “no” to “yes”, ΔG° is zero. Put zero into the Gibbs equation and rearrange:
At the switch-over point
0 = ΔH° − TΔS° so T = ΔH° ÷ ΔS°
🧩 How to answer “at what temperature does it become spontaneous?”
Check the signs first. If ΔH and ΔS have the same sign, there is a switch-over temperature. If not, say so — there isn’t one.
Convert ΔS° to kJ by dividing by 1000, or your temperature will be 1000 times too small.
Divide:T = ΔH° ÷ ΔS°. Both signs cancel, so T comes out positive.
Say which side of it works. Endothermic with ΔS positive: spontaneous above that temperature. Exothermic with ΔS negative: spontaneous below it.
Give the answer in kelvin, and convert to °C only if asked.
Worked examples
WORKED EXAMPLE
For 2Mg(s) + O2(g) → 2MgO(s), ΔH° = −1204 kJ mol−1 and ΔS° = −217 J K−1 mol−1. Is it spontaneous at 298 K?
Step 1: convert the entropy value−217 ÷ 1000 = −0.217 kJ K⁻¹ mol⁻¹Step 2: substituteΔG° = −1204 − (298 × −0.217) = −1204 + 64.7Step 3: read the signΔG° = −1139 kJ mol⁻¹Strongly negative, so yes, spontaneousBoth signs are negative, so in theory there is a switch-off temperature: 1204 ÷ 0.217 = 5548 K. Magnesium oxide has long since melted by then, so in practice this one always goes.
WORKED EXAMPLE
MgCO3(s) → MgO(s) + CO2(g) has ΔH° = +100 kJ mol−1 and ΔS° = +175 J K−1 mol−1. Above what temperature does it become spontaneous?
Step 1: check it can switch
Both are positive, so there is a temperature where ΔG° turns negative.
Step 2: set ΔG° to zero and rearrangeT = ΔH° ÷ ΔS°Step 3: convert the entropy, then divideT = 100 ÷ 0.175 = 571 KSpontaneous above about 571 K (roughly 300 °C)Endothermic and entropy-favoured, so it goes ABOVE this temperature. Below it, the reverse reaction is the spontaneous one.
WORKED EXAMPLE
2NO2(g) → N2O4(g) has ΔH° = −57 kJ mol−1 and ΔS° = −176 J K−1 mol−1. Find ΔG° at 298 K, and the temperature above which it stops being spontaneous.
Step 1: ΔG° at 298 K−57 − (298 × −0.176) = −57 + 52.4 = −4.6 kJ mol⁻¹Step 2: only just negative, so the switch is close byT = 57 ÷ 0.176 = 324 KStep 3: which side works?
Exothermic with ΔS° negative, so it is spontaneous below 324 K.
ΔG° = −4.6 kJ mol⁻¹; stops being spontaneous above 324 K (51 °C)Warm this gas mixture in your hand and it goes brown as N₂O₄ splits back into NO₂. You can literally see the sign of ΔG flip.
💡 Exam tip
Check the two signs before calculating anything. Half of these questions can be answered from the signs alone.
“Feasible” and “spontaneous” mean the same thing in IB mark schemes. So does “thermodynamically favoured”.
Say above or below. A temperature with no direction attached rarely gets full marks.
Divide by 1000 before dividing ΔH by ΔS. Skip it and you get 0.571 K instead of 571 K.
Quote a real example if asked to explain. Iron extraction in the blast furnace only works hot, because it is endothermic with a positive ΔS.
Never say a positive ΔG means “nothing happens”. It means the reverse reaction is the spontaneous one.
⚠ Common mix-up
Thinking spontaneous means fast. These are separate ideas with separate equations behind them.
Assuming heating always helps. It only helps when ΔS° is positive. For an exothermic reaction with negative ΔS°, heating is exactly what you must not do.
Getting above and below the wrong way round. Tie it to the sign of ΔH: endothermic reactions need heat, so they go above the switch temperature.
Using °C in the Gibbs equation. Kelvin only, always.
Reporting a negative temperature. If ΔH and ΔS have opposite signs there is no switch-over point, and the answer is a sentence, not a number.
Forgetting that ΔG = 0 is a real state. It is not “nothing happening” — it is equilibrium, which is where the next page begins.
You have met ΔG = 0 twice now: at the switch-over temperature, and as the balance point. That is not a coincidence — it is equilibrium, and it links free energy straight to K. Up next: Gibbs Free Energy and the Equilibrium Constant.
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