Hess's law to free energy to the equilibrium constant
SCH4U Grade 12 Chemistry · Energy Changes and Rates of Reaction
Methanol is made industrially from synthesis gas: CO(g) + 2H₂(g) → CH₃OH(g). Load that reaction into a bomb calorimeter and nothing useful happens — the synthesis will not run cleanly in the steel bomb, so there is no single clean temperature rise to measure. The thermochemistry bench therefore routes the enthalpy through three combustions the calorimeter handles well, each one a weighed sample, an oxygen fill, a fired ignition wire, and a thermometer trace: burning carbon monoxide gives −283.0 kJ/mol, burning hydrogen to steam gives −483.6 kJ/mol, and the combustion of methanol, reversed on paper so its arrow points the right way, contributes +676.5 kJ/mol. Tabulated third-law entropies put the reaction's entropy change at −219.0 J/(mol·K). The plant runs its converter hot, so the question is asked at operating temperature: find the enthalpy of the target reaction, its free energy change in a 500 K reactor, and the equilibrium constant that implies.
Every number in this problem is editable — change any value below and the whole chain recalculates.
- ΔH₁ = -283 kJ/mol — Burning carbon monoxide
- ΔH₂ = -483.6 kJ/mol — Burning hydrogen to steam
- ΔH₃ = 676.5 kJ/mol — Combustion of methanol, reversed
- ΔS = -219 J/(mol·K) — Entropy change of the target reaction
- T = 500 K — Reactor temperature
- (a)the enthalpy of the target reaction
- (b)its free energy change at 500 K
- (c)the equilibrium constant that implies
Enthalpy is a state function, so any route that starts and ends where the target reaction does gives its ΔH. The third value is positive because that step is a combustion run backwards, and reversing a step flips its sign.
Carried onward at full precision, not this rounded figure.
Enthalpy alone cannot say whether the reaction runs. Three moles of gas collapse into one, so the entropy term is strongly negative — and at 500 K it is large enough to overturn the favourable enthalpy.
Carried onward at full precision, not this rounded figure.
The exchange rate between thermodynamics and equilibrium. A positive ΔG° at this temperature must give K < 1, and it does — barely one part in a hundred.
Carried onward at full precision, not this rounded figure.
Open the Gibbs Free Energy and the Equilibrium Constant solver →
Therefore the synthesis-gas reaction is exothermic at ΔH = −90.1 kJ/mol, the collapse of three moles of gas into one drives ΔG to +19.4 kJ/mol at 500 K, and the equilibrium constant lands near 0.0094 — the number that sends methanol plants hunting for pressure.
Why this order
The order is forced by what each equation can see. Hess's law knows only about enthalpy and says nothing about whether anything happens; the Gibbs equation adds the entropy and the temperature and answers the spontaneity question; ΔG° = −RT ln K then converts that answer into a number an equilibrium table can be compared against. Skipping the middle step is the classic error — a strongly exothermic reaction feels like it must go, and here it does not, because turning three moles of gas into one is an entropic catastrophe that 500 K amplifies into +19.4 kJ/mol.
Watch the prefixes. Enthalpies are tabulated in kJ/mol and entropies in J/(mol·K), a factor of a thousand apart, and a student who subtracts 500 × 219 from 90.1 without converting concludes the reaction is impossibly endothermic. This site keeps molar energy canonically in J/mol and gives entropy its own unit type so the mismatch shows on screen rather than hiding in a placeholder. The chemistry is real, too: K ≈ 0.0094 at 500 K is why methanol plants run at 50 to 100 atmospheres. Le Chatelier's principle says squeezing three moles of gas into one rewards pressure, and the pressure is what makes an unfavourable equilibrium into a profitable one.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.