Methanol synthesis: K from tabulated data

UKChO Round 1 · UKChO Round 2 · IChO · unreviewed

CO + 2H₂ ⇌ CH₃OH with standard ΔfH° and S° tabulated; find K at a given temperature.

Worked example

For CO(g) + 2H₂(g) ⇌ CH₃OH(g), use ΔfH°(CO) = −110.5 kJ mol⁻¹, ΔfH°(CH₃OH) = −201.0 kJ mol⁻¹, S°(CO) = 197.7 J K⁻¹ mol⁻¹, S°(H₂) = 130.7 J K⁻¹ mol⁻¹ and S°(CH₃OH) = 239.9 J K⁻¹ mol⁻¹. Assuming ΔrH° and ΔrS° do not vary with temperature, calculate K at 590 K.

1. Weighted sum of 1·ΔfH°(CH₃OH) − 1·ΔfH°(CO) → ΔrH°
= −90.50 kJ mol⁻¹
2. Weighted sum of 1·S°(CH₃OH) − 1·S°(CO) − 2·S°(H₂) → ΔrS°
= −219.2 J K⁻¹ mol⁻¹
3. Gibbs energy from ΔH and ΔS: solve for ΔG
$\Delta G = \Delta H - T\Delta S$
= 38.83 kJ mol⁻¹
4. Standard Gibbs energy and K: solve for K
$K = \exp\!\left(-\dfrac{\Delta G^\circ}{RT}\right)$
= 3.652 × 10⁻⁴

Answer: K = 3.65 × 10⁻⁴

Method

ΔrH° = ΔfH°(CH₃OH) − ΔfH°(CO) (H₂ is an element, ΔfH° = 0). ΔrS° = S°(CH₃OH) − S°(CO) − 2S°(H₂). Then ΔrG° = ΔrH° − TΔrS° and K = exp(−ΔrG°/RT). K falls steeply with T because the reaction is exothermic.

Common mistakes

There are two moles of H₂: ΔrS° = S°(CH₃OH) − S°(CO) − 2S°(H₂).
That uses ΔG° ≈ ΔH° and ignores the large negative ΔS° (three moles of gas become one).

Formulas used

Gibbs energy from ΔH and ΔS
$\Delta G = \Delta H - T\Delta S$
Standard Gibbs energy and K
$\Delta G^\circ = -RT\ln K$

Practise this with fresh numbers

Methanol synthesis: K from tabulated data

Worked example

Method

Common mistakes

Formulas used

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