OpenSpecEngineering

Finds a reaction’s equilibrium constant at a new temperature from its value at a known one, and the most a simple reversible reaction can convert at each temperature. Process engineers and chemists use it to choose a reactor temperature: heat speeds a reaction up, but for an exothermic reaction it also lowers the conversion that equilibrium allows.

Equations

ln(K₂/K₁) = −(ΔH°/R)·(1/T₂ − 1/T₁)

X_e = K/(1 + K)

The first is the integrated van’t Hoff equation [1]. K₁ is the equilibrium constant at the reference temperature T₁, and K₂ its value at the target temperature T₂, both absolute. ΔH° is the standard heat of reaction in kJ/mol: negative for an exothermic reaction, positive for an endothermic one. R is the gas constant, 8.314 J/(mol·K). The second gives the equilibrium conversion X_e of a reversible reaction A ⇌ B that starts from pure A, where K = C_B/C_A at equilibrium [2].

Limits and common mistakes

• ΔH° is taken as constant between T₁ and T₂. Over a wide range, or when the heat capacities of the products and reactants differ, correct for ΔC_p [1].
• X_e = K/(1 + K) holds only for A ⇌ B with one mole on each side and no B in the feed. Other stoichiometries, inerts or pressure effects need their own equilibrium expression.
• Watch the sign of ΔH°. Entering an exothermic heat of reaction as positive makes K rise with temperature instead of fall.
• ΔH° is per mole of reaction as written, in kJ, not J.
• X_e is a ceiling. A real reactor reaches less, set by its size and the reaction rate.

Worked example

An isomerization A ⇌ B has K = 20 at 25 °C and ΔH° = −30 kJ/mol. The reaction is slow at room temperature, so the team wants to run it at 120 °C. How much conversion would that give up?

1. Enter Equilibrium Constant at T1 (K₁) 20 and Reference Temperature (T₁) 25 °C.
2. Enter Target Temperature (T₂) 120 °C and Standard Heat of Reaction (ΔH°, kJ/mol) −30.
3. Press Solve. The Solution shows: K2 at T2: 1.07408
   Equilibrium conversion Xe at T1 (A ⇌ B): 0.952381
   Equilibrium conversion Xe at T2 (A ⇌ B): 0.517858
   Behaviour: Exothermic — K falls as T rises

The ceiling drops from 95 % to about 52 %. Check by hand: 1/393.15 − 1/298.15 = −8.105 × 10⁻⁴ K⁻¹, so ln(K₂/K₁) = (30 000/8.314) × (−8.105 × 10⁻⁴) = −2.924. So K₂ = 20 × e^−2.924 = 1.074, and X_e = 1.074/2.074 = 0.518. A lower temperature, or cooling in stages, would keep more of the conversion.