Ellingham Diagram: Reading Metal Extraction Temperatures
An Ellingham diagram is a graph of ΔG° against temperature for the formation of oxides, always written per mole of O2. It answers a practical question: at what temperature can carbon (or another metal) reduce a given metal oxide? Questions on it appear in metallurgy sections of IIT-JAM, GATE and CSIR-NET style papers, mostly as "read the graph" or "find the temperature" problems. This page shows how to build and read the lines from ΔH° and ΔS°, with the arithmetic done step by step.
Where the straight lines come from
For any reaction, ΔG° = ΔH° − TΔS°. If ΔH° and ΔS° are taken as constant, this is the equation of a straight line in T. The intercept is ΔH° and the slope is −ΔS°. Writing every oxidation as "metal + O2 → oxide" (one mole of O2 each time) lets all the lines sit on one graph and be compared.
2M + O2 → 2MO: gas is used up, ΔS° is negative, so the line slopes upwards
2C + O2 → 2CO: gas increases from 1 mol to 2 mol, ΔS° is positive, so the line slopes downwards
Reduction of MO by C is possible when the carbon line lies below the metal line
What decides the slope
- Most metals: one mole of gaseous O2 disappears, so the system becomes more ordered. ΔS° is negative, about −200 J/K per mole of O2, and the line rises.
- C → CO2: one mole of gas is used and one mole formed, so ΔS° is close to zero and the line is nearly flat.
- 2C → 2CO: the gas moles double, ΔS° is large and positive, and the line falls steeply. This is why carbon becomes a better reducing agent at higher temperature.
- Kinks: when the metal or oxide melts or boils, ΔS° changes and the slope changes. The simple calculations below ignore this, so they are estimates.
Worked examples
The data are typical data-book values at 298 K: ΔHf° (kJ/mol) CO −110.5, CO2 −393.5, ZnO −350.5. S° (J/mol K) C(graphite) 5.74, O2 205.1, CO 197.7, CO2 213.8, Zn 41.6, ZnO 43.6. Check your own textbook, because small differences change the final temperature slightly.
Step 1: ΔH° = 2 × (−110.5) = −221.0 kJ.
Step 2: ΔS° = 2(197.7) − [2(5.74) + 205.1] = 395.4 − 216.58 = +178.8 J/K = 0.1788 kJ/K.
Step 3: ΔG° = −221.0 − 0.1788 T (kJ per mole of O2).
Step 4: At 1000 K: ΔG° = −221.0 − 178.8 = −399.8 kJ.
Answer: ΔG° = −221.0 − 0.1788T. The negative slope confirms the line falls with temperature.
Step 1: For C + O2 → CO2: ΔH° = −393.5 kJ and ΔS° = 213.8 − (5.74 + 205.1) = +2.96 J/K. So ΔG° = −393.5 − 0.00296 T.
Step 2: Set the two lines equal: −221.0 − 0.1788T = −393.5 − 0.00296T.
Step 3: 393.5 − 221.0 = (0.1788 − 0.00296) T, so 172.5 = 0.17584 T, giving T = 981 K.
Step 4: Check at 981 K: CO line = −221.0 − 175.4 = −396.4 kJ. CO2 line = −393.5 − 2.9 = −396.4 kJ ✓.
Answer: About 981 K. Below this temperature C → CO2 is the lower line and CO2 is the more stable product. Above it, CO is.
Step 1: ΔH° = 2 × (−350.5) = −701.0 kJ.
Step 2: ΔS° = 2(43.6) − [2(41.6) + 205.1] = 87.2 − 288.3 = −201.1 J/K. So ΔG° = −701.0 + 0.2011 T.
Step 3: Set equal to the CO line: −221.0 − 0.1788T = −701.0 + 0.2011T.
Step 4: 701.0 − 221.0 = (0.2011 + 0.1788) T, so 480.0 = 0.3799 T, giving T = 1263 K.
Step 5: Check the net reaction 2ZnO + 2C → 2Zn + 2CO. Its ΔG° = (CO line) − (ZnO line). At 1000 K: −399.8 − (−499.9) = +100.1 kJ, not spontaneous. At 1500 K: −489.2 − (−399.4) = −89.9 kJ, spontaneous.
Answer: About 1263 K on this constant-ΔH°, constant-ΔS° estimate. Real furnace temperatures also depend on zinc's boiling, which changes the slope, so treat the number as a first estimate.
Reading the diagram without numbers
| What you see | What it means |
|---|---|
| A line lower on the graph | That oxide is more stable (more negative ΔG°) |
| Metal line above the carbon line at a given T | Carbon can reduce that metal oxide at that T |
| Two lines cross | At that T, the two oxides have equal stability; the reduction is just at the borderline (ΔG° = 0) |
| A line far below the carbon line at every T | Carbon cannot reduce that oxide in a practical range; another method (for example electrolysis, as for the most reactive metals) is needed |
| Steeper rise for a metal line | More negative ΔS°; the oxide loses stability faster as T increases |
- Writing the equation per mole of metal. Ellingham lines are per mole of O2. If you write 4/3 Al + O2 → 2/3 Al2O3, scale ΔH° and ΔS° by the same factor.
- Mixing up ΔH and ΔG at crossings. At the crossing temperature the net ΔG° is zero, not ΔH°.
- Using ΔG° = ΔH° + TΔS°. The sign is minus. A wrong sign flips every slope.
- Mixing kJ and J. ΔS° is in J/K and ΔH° in kJ. Convert before multiplying.
- Trusting a constant-ΔS° line across a phase change. Melting or boiling gives a kink. State that your answer is approximate.
- Reading the diagram as a rate chart. It tells you what is thermodynamically possible, not how fast it happens.
Exam relevance
| Question type | Method |
|---|---|
| "Which oxide can be reduced by C at 1000 K?" | Compare ΔG° values of the lines at that T |
| "Find the minimum temperature of reduction" | Set the carbon line equal to the metal line, solve for T |
| "Why does the CO line slope down?" | ΔS° is positive because gas moles increase |
| "Why can Al reduce Cr2O3?" | The Al line lies below the Cr line |
Check the current official syllabus and notification for your exam to see whether metallurgy and the Ellingham diagram are part of it. Do not assume a fixed number of questions.
Use the calculator to check ΔG° = ΔH° − TΔS° at any temperature once you have worked the line by hand.
Open the Gibbs Free Energy Calculator →