Effect of temperature and pressure on equilibrium conversion

Why it matters

Ammonia and methanol plants run at 100 to 300 bar; steam reformers run near 900 °C; the water-gas shift is split into a hot and a cold stage. Each of these choices comes from how temperature and pressure move the equilibrium conversion. Quantifying that effect, rather than quoting Le Chatelier, tells you how much conversion a change buys and what it costs.

Key ideas

Temperature changes K. K depends only on T. The van't Hoff equation, d ln K/dT = ΔH°/(RT²), follows from the Gibbs-Helmholtz equation applied to ΔG° = −RT ln K. For an exothermic reaction (ΔH° < 0) K falls as T rises; for an endothermic reaction K rises. The steeper |ΔH°|, the stronger the effect.

Integrating van't Hoff.

• Shortcut: if ΔH° is taken as constant, ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁), so ln K is linear in 1/T. Adequate over modest temperature ranges.
• Rigorous: ΔH° itself changes with T through ΔCp° = Σνᵢ Cp,ᵢ°: ΔH°(T) = ΔH°₀ + ∫ΔCp° dT. Integrating van't Hoff with this ΔH°(T) (or using ΔG°/RT expressions in your textbook) is needed over large ranges, such as from 298 K data to a reactor at 800 K.

Pressure changes composition, not K. For ideal gases Πyᵢ^νᵢ = K(P/P°)^(−ν). If ν < 0 (fewer gas moles in products, as in NH₃ or CH₃OH synthesis), raising P raises the product mole fractions; if ν > 0 (steam reforming, dehydrogenation), lowering P helps; if ν = 0 (water-gas shift), pressure has no effect on ideal-gas equilibrium. At high pressure the φ̂ᵢ terms also matter and are computed from an equation of state.

Inerts and feed ratio. Adding an inert at fixed total P dilutes the reacting species, which acts like lowering the pressure: it helps reactions with ν > 0 (steam as diluent in ethylbenzene dehydrogenation) and hurts those with ν < 0. Excess of a cheap reactant raises the equilibrium conversion of the expensive one.

The exothermic dilemma. For exothermic reversible reactions, low T gives high equilibrium conversion but slow rates; high T gives fast rates but low conversion. Industrial reactors compromise: an optimum temperature progression, staged adiabatic beds with interstage cooling (SO₂ oxidation, ammonia), and a catalyst active enough to work at lower T.

Adiabatic reactors. In an adiabatic bed the temperature changes with conversion along an energy-balance line. The bed's final conversion lies where this line meets the equilibrium-conversion curve, which is why exothermic reactions are run in several beds with cooling in between.

Formulas

d ln K/dT = ΔH°/(R·T²) (van't Hoff) ln(K₂/K₁) = −(ΔH°/R)·(1/T₂ − 1/T₁) (ΔH° constant) ΔH° = Σ νᵢ·ΔH°f,ᵢ ΔCp° = Σ νᵢ·Cp,ᵢ°, ΔH°(T) = ΔH°(T₀) + ∫ΔCp° dT ΔG° = ΔH° − T·ΔS°, ln K = −ΔG°/(R·T) Π yᵢ^νᵢ = K·(P/P°)^(−ν) (ideal gases) 16·ε²·(2 − ε)² / [27·(1 − ε)⁴] = K·(P/P°)² (N₂ + 3H₂ ⇌ 2NH₃, stoichiometric feed, ideal gas)

Symbols: K equilibrium constant (dimensionless); T (K); ΔH° standard heat of reaction (J/mol), negative for exothermic; ΔH°f standard heat of formation (J/mol); ΔS° standard entropy change (J/mol·K); ΔCp° heat-capacity change of reaction (J/mol·K); νᵢ stoichiometric numbers, ν = Σνᵢ; yᵢ mole fractions; ε extent of reaction per mole of N₂ fed; P (bar); P° = 1 bar; R = 8.314 J/(mol·K).

Worked examples

Example 1 (standard): shortcut van't Hoff for the water-gas shift. Given: CO + H₂O ⇌ CO₂ + H₂ has K = 1.03 × 10⁵ (ln K = 11.546) at 298.15 K; ΔH°f (kJ/mol): CO −110.53, H₂O(g) −241.82, CO₂ −393.51. Estimate K at 800 K assuming constant ΔH°.

1. ΔH° = Σνᵢ ΔH°f,ᵢ = −393.51 − (−110.53 − 241.82) = −41.16 kJ/mol (exothermic).
2. ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁) = (41 160/8.314)(1/800 − 1/298.15) = 4950.7 × (−0.0021040) = −10.416.
3. ln K₂ = 11.546 − 10.416 = 1.130, so K₂ = 3.10.
4. K drops by a factor of about 33 000 between 298 K and 800 K, as expected for an exothermic reaction. Over such a wide range the shortcut is only an estimate; rigorous integration with ΔCp° data gives a noticeably different value. Answer: K(800 K) ≈ 3.1 (shortcut estimate)

Example 2 (GATE level): pressure effect on ammonia synthesis. Given: N₂ + 3H₂ ⇌ 2NH₃ with K = 8.3 × 10⁻⁵ at the reactor temperature (for the reaction as written); stoichiometric feed (1 mol N₂ : 3 mol H₂); ideal gases. Find the N₂ conversion at 1 bar and at 100 bar.

1. Moles: N₂ = 1 − ε, H₂ = 3(1 − ε), NH₃ = 2ε, total = 2(2 − ε). ν = −2.
2. Mole fractions: y_NH₃ = ε/(2 − ε), y_N₂ = (1 − ε)/[2(2 − ε)], y_H₂ = 3(1 − ε)/[2(2 − ε)].
3. Πyᵢ^νᵢ = K(P/P°)² simplifies to 16ε²(2 − ε)²/[27(1 − ε)⁴] = K·P².
4. Taking square roots: ε(2 − ε)/(1 − ε)² = (√27/4)·√K·P = 1.2990 × 0.009110 × P = 0.011835·P.
5. With u = 1 − ε, ε(2 − ε) = 1 − u², so 1/u² = 1 + 0.011835·P.
6. At 1 bar: u = 1/√1.011835 = 0.99413, ε = 0.0059. At 100 bar: u = 1/√2.1835 = 0.67675, ε = 0.323.
7. Ammonia mole fraction at 100 bar: y_NH₃ = 0.323/(2 − 0.323) = 0.193. Answer: conversion 0.6% at 1 bar, 32.3% at 100 bar (y_NH₃ ≈ 0.19)

Common mistakes

• Saying pressure changes K; it changes the equilibrium composition through (P/P°)^(−ν).
• Sign errors in van't Hoff: for exothermic reactions ln K must fall as T rises; check the sign of your answer.
• Using °C in 1/T, or kJ with R in J.
• Applying the constant-ΔH° shortcut over hundreds of kelvin and treating the result as exact.
• Assuming inerts never matter; at constant P they shift equilibrium whenever ν ≠ 0.
• Confusing a catalyst's effect (faster approach) with a shift in equilibrium (none).

For GATE CH

• NAT problems: K at a new temperature from van't Hoff with given ΔH°; ΔH° from two K values; equilibrium conversion at different P, with inerts or excess reactant.
• Conceptual MCQs on the direction of shifts with T, P, inerts and feed ratio, and on why exothermic reactors are staged with interstage cooling.
• Occasionally the full ΔCp° integration with given heat-capacity data.
• Practise reducing ammonia-type and dissociation-type equilibrium expressions to a single solvable equation.

Quick check

1. Write the van't Hoff equation.
2. For an endothermic reaction, does K rise or fall with temperature?
3. For the water-gas shift (ν = 0), how does pressure affect ideal-gas equilibrium conversion?
4. Does adding nitrogen as an inert at constant total pressure help or hurt ethylbenzene → styrene + H₂?
5. K = 10 at 300 K and ΔH° = −40 kJ/mol (constant). Is K at 400 K about 0.18, 1.8 or 18?

Answers: 1. d ln K/dT = ΔH°/(RT²). 2. Rises. 3. No effect. 4. Helps (ν > 0, dilution acts like lower pressure). 5. About 0.18.