Chemistry · General chemistry II · Worked example
Find ΔG° from enthalpies and entropies
Use the data in the table to find ΔH°, ΔS° and ΔG° at 298.15 K for 2H₂(g) + O₂(g) → 2H₂O(l).
Collect the data
Elements in their standard states have ΔHf° = 0, but their standard entropies are not zero.
| Species | ΔHf° (kJ/mol) | S° (J/(mol·K)) |
|---|---|---|
| H₂(g) | 0 | 130.68 |
| O₂(g) | 0 | 205.15 |
| H₂O(l) | −285.83 | 69.91 |
ΔH° from enthalpies of formation
Products minus reactants, each times its coefficient: 2(−285.83) − 0.
ΔS° from standard entropies
Three moles of gas become two moles of liquid, so the entropy falls.
Combine at 298.15 K
Convert ΔS° to −0.32669 kJ/K so the units match ΔH°.
Interpret
ΔG° is negative, so the reaction is spontaneous as written at 298 K even though its entropy falls: the large, negative ΔH° outweighs the entropy term. It still needs a spark to start.
Result
ΔH° = −571.66 kJ, ΔS° = −326.69 J/K and ΔG° = −474.26 kJ at 298.15 K.
Your turn
With the same data, find ΔG° at 298.15 K for H₂(g) + ½O₂(g) → H₂O(l).
Show the answer and explanation
−237.13 kJ.
ΔH° = −285.83 kJ and ΔS° = 69.91 − 130.68 − ½(205.15) = −163.345 J/K, so ΔG° = −285.83 − (298.15)(−0.163345) = −237.13 kJ, half the value for two moles of water.
Keep exploring
In Hess law & thermodynamics, reverse the reaction to 2H₂O(l) → 2H₂(g) + O₂(g). Every sign flips: ΔG° = +474.26 kJ, so splitting water is not spontaneous.
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