The quantum casino: simulations        # ∆G and temperature

Graphs of ∆G against temperature show how reaction feasibility changes. As a rule of thumb, if ∆G is more negative than –60 kJ mol-1, the reaction is considered to be complete, and if more positive than +60 kJ mol-1, it is considered not to ‘go’ at all (ie the reverse reaction is complete. If ∆G is in the grey area between these values, the reaction is considered to be reversible.

Graphs of ∆G against temperature for the formation of metal oxides from their elements are called Ellingham diagrams. You can plot Ellingham diagrams for a number of industrially important reactions. All the equations have been written (and the data calculated) so that they involve 1 mole of oxygen molecules and are written as oxidations (ie element + oxygen oxide). For most of the reactions, as written, ∆G is negative over the whole temperature range, so the formation of the metal oxide is feasible and the reverse reaction (extracting the metal from its oxide is not.

The interesting situation is when two elements ‘compete’ for the oxygen. At any temperature, the reaction represented by the lower line will have the more negative value of ∆G. If the two reactions are ‘competing’, the lower one will go forwards and drive the other in reverse.

### Add diagram or choose from the library below

∆H / kJ mol-1 ∆S / kJ K  mol-1 diagram

## ### Reaction library

Reversible reactions ∆H / kJ mol-1 ∆S / kJ K  mol-1 diagram
N2(g) + 3H2(g) 2NH3(g) –100 –0.198 add | remove
N2O4(g) 2NO(g)

+58

H2(g) + CO2(g) HO2(g) + CO(g) +40 +0.0392 add | remove
2SO2(g) + O2(g) 2SO3(g) –197 –0.187 add | remove
H2(g) + I 2(g) 2HI(g) +27 –0.239 add | remove
CaO + CO2 CaCO3 –178 –0.161 add | remove
Formation of oxides ∆H / kJ mol-1 ∆S / kJ K  mol-1 diagram
2C + O2 2CO –221 +0.179 add | remove
C + O2 CO2 –393.5 +0.0056 add | remove
Si + O2 SiO2 –910.9 –0.182 (T < 1683 K)
–0.208 (T > 1683 K)
2Mg + O2 2MgO –1203.4 –0.2166 (T < 922 K)
–0.2201 (922 < T < 1380 K)
–0.448 (T > 1380 K)
2Zn + O2 2ZnO –696.6 –0.201 (T < 693 K)
–0.219 (693 < T < 1180 K)
–0.439 (1180 < T < 2248 K)
4Ag + O2 2Ag2O –62 –0.1328 (T < 503 K) add | remove
4/3 Fe + O2 2/3 Fe2O3 –549.5 –0.183 (T < 1808 K)
–0.193 (1808 < T < 1830 K)
2Fe + O2 2FeO –543.8 –0.143 (T < 1642 K)
A key point in the graphs is the crossover point of two reaction lines. Here ∆G for each reaction has the same numerical value. Either side of the crossover point, the reaction represented by the lower line (the one with the more negative value of ∆G) will go forwards while that represented by the upper line will go in reverse. It is therefore possible to predict the temperature above which, say, carbon will reduce any particular metal oxide. Since the 2C + O2 2CO line slopes downwards with temperature and the lines for the oxidations of metals slope upwards, this temperature will always be above the temperature of the crossover point. 