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). The reaction cannot be run cleanly in a bomb calorimeter, so its enthalpy is routed through three steps that can: burning carbon monoxide, −283.0 kJ/mol; burning hydrogen to steam, −483.6 kJ/mol; and the reversed combustion of methanol, +676.5 kJ/mol. The reaction's entropy change is −219.0 J/(mol·K). Find the enthalpy of the target reaction, its free energy change in a 500 K reactor, and the equilibrium constant that implies.

Step 1 of 3 · solve for Enthalpy change of the target reaction

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.

Rearranged for ΔHrxn
ΔHrxn=ΔH1+ΔH2+ΔH3\Delta H_{\text{rxn}} = \Delta H_1 + \Delta H_2 + \Delta H_3
Your values, in your units
ΔHrxn=(283 kJ/mol)+(483.6 kJ/mol)+(676.5 kJ/mol)\Delta H_{\text{rxn}} = \left( -283\ \text{kJ/mol} \right) + \left( -483.6\ \text{kJ/mol} \right) + \left( 676.5\ \text{kJ/mol} \right)
Answer
ΔHrxn=90.1\Delta H_{\text{rxn}} = -90.1

Carried onward at full precision, not this rounded figure.

Open the Hess's Law (Three-Step Sum) solver →

Step 2 of 3 · solve for Gibbs free energy change

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.

Rearranged for ΔG
ΔG=ΔHTΔS\Delta G = \Delta H - T\,\Delta S
-90.1 kJ/molcarried from step 1
Your values, in your units
ΔG=(90,100 J/mol)(500 K)×(219 J/(mol⋅K))\Delta G = \left( -90{,}100\ \text{J/mol} \right) - \left( 500\ \text{K} \right) \times \left( -219\ \text{J/(mol·K)} \right)
Converted to base units
ΔG=(90.1 kJ/mol)(226.85 C)×(219 J/(mol⋅K))\Delta G = \left( -90.1\ \text{kJ/mol} \right) - \left( 226.85\ ^{\circ}\text{C} \right) \times \left( -219\ \text{J/(mol·K)} \right)
Answer
ΔG=19.4\Delta G = 19.4

Carried onward at full precision, not this rounded figure.

Open the Gibbs Free Energy Change (ΔG = ΔH − TΔS) solver →

Step 3 of 3 · solve for Equilibrium constant

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.

Rearranged for K
K=exp ⁣(ΔGRT)K = \exp\!\left(-\frac{\Delta G^{\circ}}{RT}\right)
19.4 kJ/molcarried from step 2
Your values, in your units
K=exp ⁣((19,400 J/mol)R(500 K))K = \exp\!\left(-\frac{\left( 19{,}400\ \text{J/mol} \right)}{R \, \left( 500\ \text{K} \right)}\right)
Converted to base units
K=exp ⁣((19.4 kJ/mol)R(226.85 C))K = \exp\!\left(-\frac{\left( 19.4\ \text{kJ/mol} \right)}{R \, \left( 226.85\ ^{\circ}\text{C} \right)}\right)
Answer
K=0.0094K = 0.0094

Carried onward at full precision, not this rounded figure.

Open the Gibbs Free Energy and the Equilibrium Constant solver →

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.