Chemistry · General chemistry I · Worked example
Use enthalpies of formation to find ΔH°rxn
Find ΔH° for MgO(s) + H₂(g) → Mg(s) + H₂O(l). Use ΔHf° = −602 kJ/mol for MgO(s), from the Hess’s law example, and −285.83 kJ/mol for H₂O(l).
List the formation enthalpies
H₂(g) and Mg(s) are elements in their standard states, so their ΔHf° is zero. Every coefficient is 1.
| Substance | Side | ΔHf° (kJ/mol) |
|---|---|---|
| MgO(s) | reactant | −602 |
| H₂(g) | reactant | 0 |
| Mg(s) | product | 0 |
| H₂O(l) | product | −285.83 |
Products minus reactants
The products sum to −285.83 kJ and the reactants to −602 kJ. Subtracting gives 316.17 kJ, which is +316 kJ to the ones place of −602.
Check with Hess’s law
The reaction is the formation of MgO reversed (+602 kJ) plus the formation of water (−285.83 kJ): 602 − 285.83 = 316 kJ, the same answer.
Interpret the sign
ΔH° = +316 kJ: the reaction is strongly endothermic. Enthalpy alone does not decide whether a reaction happens; the Gibbs free energy, which also counts entropy, does.
Result
ΔH° = +316 kJ for MgO(s) + H₂(g) → Mg(s) + H₂O(l).
Your turn
Find ΔH° for Mg(s) + H₂O(l) → MgO(s) + H₂(g) from the same formation enthalpies.
Show the answer and explanation
−316 kJ.
Products minus reactants: [−602 + 0] − [0 + (−285.83)] = −316.17, which is −316 kJ. It is the reverse of the worked reaction, so its ΔH° has the opposite sign.
Keep exploring
In Hess law & thermodynamics, give H₂(g) a formation enthalpy of −100 kJ/mol by mistake. ΔH° jumps to +416 kJ, which is why an element in its standard state must stay at zero.
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