Pressure/Volume Changes on Equilibrium — Predict effects of pressure/volume changes on gaseous equilibria - General Chemistry 2

SECTION 1

Historical Context & Motivation

The quest to understand how chemical systems respond to external perturbations has deep roots in nineteenth-century thermodynamics and industrial chemistry. As chemists and engineers sought to optimize large-scale chemical processes—particularly the synthesis of ammonia, sulfuric acid, and other industrially vital compounds—they recognized that simply mixing reactants was insufficient; the conditions under which a reaction proceeds profoundly influence the yield of products. The interplay between pressure, volume, and chemical equilibrium became one of the central practical and theoretical problems of physical chemistry, bridging the gap between Gibbs's abstract thermodynamic formalism and the reality of high-pressure reactors.

1803 • Dalton's Law of Partial Pressures

John Dalton published his law stating that the total pressure of a gas mixture equals the sum of the partial pressures of each component. This foundational concept later proved essential for analyzing how individual gaseous species contribute to equilibrium behavior under changing total pressure.

1864 • Guldberg and Waage's Law of Mass Action

Norwegian chemists Cato Guldberg and Peter Waage formulated the law of mass action, establishing that the rate of a reaction is proportional to the concentrations of the reactants raised to appropriate powers. This principle laid the mathematical groundwork for equilibrium expressions that relate concentrations—and by extension, partial pressures—to the position of equilibrium.

1884 • Le Châtelier's Principle

Henri-Louis Le Châtelier articulated his general principle: when a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system shifts to partially counteract the imposed change. This qualitative rule became the primary tool for predicting the direction of equilibrium shifts under pressure and volume perturbations.

1909 • The Haber Process

Fritz Haber demonstrated the high-pressure synthesis of ammonia from nitrogen and hydrogen. This process provided dramatic industrial validation of Le Châtelier's principle: operating at elevated pressures (150–200 atm) shifted the equilibrium toward ammonia because the product side has fewer moles of gas. The Haber process remains the most celebrated example of pressure manipulation in chemical equilibrium.

1920s • Thermodynamic Formalization

With the maturation of chemical thermodynamics, the relationship between the equilibrium constant expressed in terms of pressure (Kₚ) and the standard Gibbs energy change (ΔG°) was rigorously established. This formalism clarified why Kₚ itself is independent of total pressure, while the equilibrium composition is not when the reaction changes the total number of moles of gas.

The central question this lesson addresses is deceptively simple: if you compress or expand a gaseous equilibrium mixture, which direction does the equilibrium shift, and why? Answering this question requires connecting the ideal gas law, the equilibrium expression, and Le Châtelier's principle into a coherent predictive framework. Understanding these connections is essential not only for general chemistry but also for fields ranging from atmospheric science to biochemical gas exchange.

SECTION 2

Core Principles & Definitions

Before predicting how pressure and volume changes affect gaseous equilibria, it is essential to establish a clear conceptual vocabulary. The behavior of gases at equilibrium is governed by the interplay of several foundational ideas: the relationship between moles and partial pressures, the form of the equilibrium expression, and the mechanism by which Le Châtelier's principle operates at the molecular level. The following core principles form the basis for all quantitative and qualitative predictions about pressure-induced equilibrium shifts.

Partial Pressure & Mole Fraction

The partial pressure of a gas in a mixture is pᵢ = χᵢ × P_total, where χᵢ is the mole fraction. When total pressure increases at constant temperature and composition, every partial pressure increases proportionally.

Equilibrium Constant Kₚ

For a gaseous reaction, Kₚ is defined in terms of partial pressures of products over reactants, each raised to its stoichiometric coefficient. Crucially, Kₚ depends only on temperature—not on total pressure or volume.

Reaction Quotient Qₚ

The reaction quotient Qₚ has the same mathematical form as Kₚ but is evaluated with current (non-equilibrium) partial pressures. When Qₚ < Kₚ, the reaction shifts forward; when Qₚ > Kₚ, it shifts in reverse.

Δn(gas) — Change in Moles of Gas

The quantity Δn = (moles of gaseous products) − (moles of gaseous reactants) determines sensitivity to pressure. When Δn ≠ 0, changes in total pressure alter the equilibrium position; when Δn = 0, pressure changes have no effect.

Le Châtelier's Principle Applied to Pressure

An increase in pressure (via volume decrease) favors the side with fewer moles of gas, because shifting toward fewer gas molecules partially relieves the imposed pressure increase. Conversely, decreasing pressure (expanding volume) favors the side with more moles of gas.

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KEY TAKEAWAY

Think of a gaseous equilibrium as a crowded room with two exits—one leading to a hallway that holds 4 people (the side with more moles of gas) and another that holds 2 people (the side with fewer moles). If you suddenly shrink the room (increase pressure by decreasing volume), people will preferentially flow toward the hallway that accommodates fewer occupants, because the system seeks to reduce the total number of gas-phase particles pressing against the walls. This is precisely what Le Châtelier's principle predicts: the equilibrium shifts toward the side with fewer moles of gas to partially offset the pressure increase.

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Important Distinction

Adding an inert gas at constant volume increases total pressure but does not change the partial pressures of the reacting species (since their moles and the volume are unchanged). Therefore, adding an inert gas at constant volume has no effect on the equilibrium position. However, adding an inert gas at constant total pressure requires the volume to expand, which dilutes the reacting gases and can shift the equilibrium toward the side with more moles of gas.

SECTION 3

Visual Explanation — Equilibrium Shifts Under Compression

The following diagram illustrates the classic equilibrium N₂O₄(g) ⇌ 2 NO₂(g) in three states: at the initial equilibrium, immediately after a volume decrease (compression), and at the new equilibrium after the system has re-equilibrated. This reaction is particularly instructive because the forward direction produces more moles of gas (Δn = +1), so compressing the system shifts the equilibrium to the left, favoring the colorless N₂O₄ over the brown NO₂.

The diagram shows three states for the N₂O₄ ⇌ 2 NO₂ equilibrium. The purple circles represent N₂O₄ molecules and the orange circles represent NO₂ molecules. After compression (middle panel), all concentrations increase, making Q > K. The system then shifts left to re-establish equilibrium (right panel), producing more N₂O₄ and consuming NO₂.

The visual above makes a critical point: compression does not selectively increase the pressure of one species. Rather, when the volume decreases, the partial pressures of all gaseous species increase simultaneously. However, because the equilibrium expression involves products and reactants raised to different powers (reflecting different stoichiometric coefficients), the reaction quotient Q is affected asymmetrically. For the N₂O₄ ⇌ 2 NO₂ system, the numerator of Q (which contains p_NO₂ squared) increases faster than the denominator (which contains p_N₂O₄ to the first power). This makes Q > K, and the system shifts in the reverse direction (toward fewer moles of gas) to restore equilibrium.

SECTION 4

Mathematical Framework

The qualitative predictions of Le Châtelier's principle can be made rigorous through the mathematical relationship between the equilibrium constant Kₚ, the reaction quotient Qₚ, and the change in moles of gas Δn. This section develops the key equations and demonstrates why the direction of the shift depends entirely on the sign of Δn.

EQUILIBRIUM CONSTANT IN TERMS OF PARTIAL PRESSURES

Kₚ = ∏(pᵢ)^νᵢ (products over reactants)

For a general reaction aA(g) + bB(g) ⇌ cC(g) + dD(g): Kₚ = (p_C^c × p_D^d) / (p_A^a × p_B^b). Here pᵢ is the partial pressure of species i, and the stoichiometric coefficients serve as exponents.

RELATIONSHIP BETWEEN Kₚ AND Kc

Kₚ = Kc × (RT)^Δn

Where Δn = (c + d) − (a + b) is the change in total moles of gas, R = 0.08206 L·atm·mol⁻¹·K⁻¹, and T is the absolute temperature in Kelvin. This equation shows that Kₚ and Kc are equal only when Δn = 0.

EFFECT OF VOLUME CHANGE ON Qₚ

If V → V/α (compression by factor α), then pᵢ → α × pᵢ for each species

Substituting into the Qₚ expression: Qₚ(new) = α^Δn × Kₚ. When Δn > 0, compression (α > 1) makes Qₚ > Kₚ, shifting the reaction in reverse (toward fewer moles). When Δn < 0, Qₚ < Kₚ, shifting the reaction forward. When Δn = 0, Qₚ = Kₚ and there is no shift.

Derivation of the Compression Factor

Consider a system initially at equilibrium so that Qₚ = Kₚ. If the volume is halved (α = 2), each partial pressure doubles because pᵢ = nᵢRT/V. The new reaction quotient becomes:

GENERAL COMPRESSION RESULT

Qₚ(new) = (2p_C)^c × (2p_D)^d / [(2p_A)^a × (2p_B)^b] = 2^(c+d−a−b) × Kₚ = 2^Δn × Kₚ

Since each partial pressure is multiplied by 2, and the exponents differ between numerator and denominator, the factor 2^Δn emerges naturally. For general compression by factor α, replace 2 with α. This is the mathematical origin of Le Châtelier's principle for pressure changes: the direction of the shift is determined entirely by the sign of Δn.

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Summary of Predictions

For a volume decrease (pressure increase): if Δn > 0, shift toward reactants; if Δn < 0, shift toward products; if Δn = 0, no shift. For a volume increase (pressure decrease): the predictions reverse.

SECTION 5

Classifying Pressure/Volume Perturbations

Not all pressure changes affect equilibrium identically. It is a common misconception that any increase in total pressure will shift the equilibrium. In fact, the method by which pressure is altered determines whether and how the equilibrium responds. The following classification organizes the three principal ways total pressure can change, along with their distinct effects on the equilibrium position.

This classification diagram compares three methods of changing total pressure. Method 1 (volume change) and Method 3 (inert gas at constant pressure) both alter partial pressures of reacting species and can shift equilibrium. Method 2 (inert gas at constant volume) does not change partial pressures and has no effect on the equilibrium position.

Reaction	Δn	Effect of Compression	Effect of Expansion
N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g)	−2	Shift → products (NH₃)	Shift → reactants
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)	+1	Shift → reactants (PCl₅)	Shift → products
H₂(g) + I₂(g) ⇌ 2 HI(g)	0	No shift	No shift
2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g)	−1	Shift → products (SO₃)	Shift → reactants
CaCO₃(s) ⇌ CaO(s) + CO₂(g)	+1 (gas only)	Shift → reactants	Shift → products

Summary of pressure effects for common equilibria. Only gaseous species are counted in Δn.

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Solids and Liquids

When calculating Δn, count only gaseous species. Pure solids and pure liquids do not appear in the equilibrium expression and their concentrations are not affected by pressure or volume changes. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), the only gaseous species is CO₂ on the product side, so Δn = +1.

SECTION 6

Worked Example — The Haber Process Under Compression

The industrial synthesis of ammonia provides an ideal context for a quantitative worked example. Consider the equilibrium N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), with Kₚ = 6.0 × 10⁵ at 298 K (though industrially the reaction is run at higher temperatures where Kₚ is smaller). Suppose the system is at equilibrium and the volume is suddenly halved at constant temperature. We wish to determine the direction of the shift and qualitatively describe how partial pressures change.

Predicting the Shift for N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) After Volume Halved

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Step 1 — Calculate Δn

Count the moles of gaseous products minus moles of gaseous reactants. Products: 2 mol NH₃. Reactants: 1 mol N₂ + 3 mol H₂ = 4 mol total. Therefore, Δn = 2 − 4 = −2.

Δn = −2
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Step 2 — Determine the Compression Factor α

The volume is halved, so V_new = V/2. Since pᵢ = nᵢRT/V, halving V doubles all partial pressures. The compression factor is α = V_old/V_new = 2.

α = 2
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Step 3 — Calculate the New Reaction Quotient

Using the general result Qₚ(new) = α^Δn × Kₚ: Qₚ(new) = 2⁻² × Kₚ = (1/4) × Kₚ = 0.25 Kₚ. Since Qₚ = 0.25 Kₚ < Kₚ, the system is no longer at equilibrium.

Qₚ = 0.25 Kₚ < Kₚ
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Step 4 — Predict the Direction of Shift

Since Qₚ < Kₚ, the reaction must proceed in the forward direction (toward products) to increase Qₚ back to Kₚ. This means more NH₃ is formed at the expense of N₂ and H₂. This result is consistent with Le Châtelier's principle: compression favors the side with fewer moles of gas, and the product side (2 mol) has fewer gas-phase molecules than the reactant side (4 mol).

Equilibrium shifts → forward (toward NH₃)
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Step 5 — Qualitative Description of Final State

At the new equilibrium, the partial pressure of NH₃ will be higher than at the initial equilibrium, while the partial pressures of N₂ and H₂ will be higher than their original values (because the total pressure has increased) but lower than they were immediately after compression (because the forward shift consumes some of each). The total pressure at the new equilibrium will be lower than the pressure immediately after compression but higher than the original pressure, because the shift reduces the total number of gas-phase moles.

p(NH₃) ↑, p(N₂) and p(H₂) partially relieved, P_total slightly decreased from compressed value

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WHY α^Δn IS THE KEY

The expression Qₚ(new) = α^Δn × Kₚ provides a universal shortcut. You only need two pieces of information—the compression factor α and the sign of Δn—to determine the direction of the shift without substituting actual pressure values. When Δn is negative and α > 1 (compression), α^Δn < 1, so Qₚ < Kₚ, and the forward direction is favored. This is exactly the logic the Haber process exploits industrially.

SECTION 7

Strengths, Limitations & Common Pitfalls

Le Châtelier's principle is a powerful qualitative tool, but it has important limitations and is frequently misapplied in the context of pressure and volume changes. Understanding both the strengths and the pitfalls of this approach helps develop deeper chemical intuition and prepares you for more advanced thermodynamic treatments.

Strengths	Limitations / Pitfalls
Quick qualitative predictions without calculations—just compare Δn to zero and know the shift direction.	Does not tell you the magnitude of the shift. Only a full equilibrium calculation (ICE table with the new conditions) can quantify how much the composition changes.
Works universally for ideal gas mixtures at any temperature, provided Kₚ is known to be constant at that temperature.	Fails for non-ideal (real) gas mixtures at very high pressures, where fugacities replace partial pressures and Kₚ may appear to change.
The α^Δn framework connects Le Châtelier's qualitative prediction to the rigorous Qₚ vs. Kₚ comparison.	Students often forget that Kₚ itself does NOT change with pressure—only the equilibrium composition changes. Confusing Kₚ shifts with composition shifts is a major exam error.
Applicable to heterogeneous equilibria as long as only gaseous species are counted in Δn.	Misapplication when adding inert gas at constant volume: students frequently predict a shift, but no shift occurs because partial pressures are unchanged.
Integrates seamlessly with the ideal gas law, making it easy to connect PV = nRT reasoning to equilibrium shifts.	Le Châtelier's principle does not apply to reactions in condensed phases (liquids/solids only) because their volumes are nearly incompressible; pressure effects on those equilibria require different thermodynamic treatment (ΔV of reaction).

Strengths and limitations of the Le Châtelier pressure-volume analysis for gaseous equilibria

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AVOIDING THE #1 EXAM MISTAKE

The most common error is stating that "increasing pressure always shifts the equilibrium toward fewer moles of gas." This is only true if the pressure increase is achieved by decreasing the volume. If pressure increases because an inert gas is added at constant volume, no shift occurs. Always ask: how was the pressure changed? The mechanism of the pressure change determines whether partial pressures of the reacting species are actually altered.

SECTION 8

Connection to Advanced Thermodynamics

The qualitative Le Châtelier analysis presented in this lesson is grounded in the ideal gas approximation. In more advanced courses—physical chemistry and chemical engineering thermodynamics—the treatment extends to real gases using fugacities, activity coefficients, and the full Gibbs energy landscape. Understanding how the basic framework connects to these advanced treatments provides valuable perspective and motivates the study of non-ideal systems.

Feature	General Chemistry Treatment	Advanced Thermodynamic Treatment
Equilibrium expression	Kₚ in terms of partial pressures (ideal gas)	K in terms of fugacities (fᵢ = φᵢpᵢ), where φᵢ is the fugacity coefficient
Pressure dependence of K	Kₚ is independent of pressure at constant T	K (thermodynamic) is truly P-independent; apparent Kₚ may vary if fugacity corrections are significant
Shift prediction	Le Châtelier's principle / Qₚ vs Kₚ comparison	(∂ ln K / ∂P)_T = −ΔV°/RT for condensed-phase equilibria; Gibbs-Helmholtz type relations
Quantitative composition	ICE table with ideal gas assumption	Minimization of total Gibbs energy G(T, P, {nᵢ}) using numerical methods
Applicable pressure range	Low to moderate pressures (~1–10 atm)	All pressures, including supercritical conditions

Comparison of general chemistry and advanced thermodynamic approaches to pressure-equilibrium relationships

The transition from the ideal gas Kₚ framework to fugacity-based equilibrium becomes essential when dealing with reactions at very high pressures (such as the industrial Haber process at 150–300 atm) or with gases near their critical point. At these conditions, intermolecular forces and molecular volumes cause significant deviations from ideal behavior. The fugacity coefficient φᵢ quantifies this deviation: when φᵢ = 1, the gas is ideal, and when φᵢ ≠ 1, corrections are needed. Despite these refinements, the fundamental logic remains the same: compare the reaction quotient (now in terms of fugacities) to the equilibrium constant, and the system shifts to make them equal. Le Châtelier's principle still provides the correct qualitative prediction; the advanced treatment merely refines the quantitative details.

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Looking Ahead

In physical chemistry, you will encounter the van't Hoff equation for temperature dependence of K, and the pressure-dependent Gibbs energy for condensed-phase reactions: (∂G/∂P)_T = V. These tools generalize the pressure-equilibrium relationship to systems where Le Châtelier's simple mole-counting argument does not directly apply.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

For the reaction 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), predict the effect on the equilibrium position if the container is compressed to one-third of its original volume at constant temperature. Explain your reasoning in terms of both Le Châtelier's principle and the reaction quotient.

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The change in moles of gas is Δn = 2 − (2 + 1) = −1. Compression to one-third volume triples all partial pressures (α = 3). The new reaction quotient is Qₚ = 3⁻¹ × Kₚ = Kₚ/3 < Kₚ. Since Qₚ < Kₚ, the reaction shifts forward toward products (more SO₃). By Le Châtelier's principle, the system counteracts the pressure increase by shifting toward the side with fewer moles of gas (2 product moles vs. 3 reactant moles), which is consistent with the Qₚ analysis.

Question 2: BASIC CALCULATION

Consider the dissociation PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). At equilibrium in a 10.0 L vessel at 250 °C, the partial pressures are p(PCl₅) = 1.20 atm, p(PCl₃) = 0.60 atm, and p(Cl₂) = 0.60 atm. (a) Calculate Kₚ. (b) If the volume is reduced to 5.0 L, calculate the new Qₚ immediately after compression and predict the direction of the shift.

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(a) Kₚ = p(PCl₃) × p(Cl₂) / p(PCl₅) = (0.60)(0.60) / (1.20) = 0.30 atm. (b) Halving the volume doubles all partial pressures: p(PCl₅) = 2.40 atm, p(PCl₃) = 1.20 atm, p(Cl₂) = 1.20 atm. The new Qₚ = (1.20)(1.20) / (2.40) = 0.60 atm. Since Qₚ = 0.60 > Kₚ = 0.30, the reaction shifts in reverse (toward PCl₅). This is expected because Δn = +1, so compression favors the side with fewer moles of gas, which is the reactant side. Equivalently, Qₚ = α^Δn × Kₚ = 2¹ × 0.30 = 0.60, confirming our calculation.

Question 3: INTERMEDIATE

A mixture of N₂, H₂, and NH₃ is at equilibrium in a flexible container (constant external pressure) at 500 K. Argon gas is slowly injected into the container at constant total pressure. (a) Does the equilibrium shift? If so, in which direction? (b) What would happen instead if the argon were injected into a rigid container at constant volume?

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(a) Adding argon at constant total pressure forces the container to expand (since the flexible walls maintain constant P). The expansion dilutes all reacting gases, effectively lowering their partial pressures. For N₂ + 3 H₂ ⇌ 2 NH₃, Δn = −2. Dilution is equivalent to a volume increase, which favors the side with more moles of gas. The equilibrium shifts in reverse (toward N₂ and H₂). (b) At constant volume, adding argon increases total pressure but does not change the partial pressures of N₂, H₂, or NH₃ (since pᵢ = nᵢRT/V and neither nᵢ nor V changes). Therefore, Qₚ remains equal to Kₚ and there is no shift in equilibrium.

Question 4: APPLIED

In the Contact process for manufacturing sulfuric acid, one key step is 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g) (Δn = −1). Industrial plants typically operate at about 1–2 atm rather than at very high pressures, despite the fact that higher pressure would increase SO₃ yield. Suggest a reason—drawing on both equilibrium principles and practical engineering considerations—for why moderate pressure is chosen.

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While Le Châtelier's principle confirms that higher pressure would shift this equilibrium toward SO₃ (since Δn = −1), the equilibrium already lies far to the right at the operating temperature (~450 °C) with the V₂O₅ catalyst, giving conversions above 95%. The marginal gain in yield from further pressure increase is small. High-pressure equipment (thick-walled reactors, high-pressure pumps, reinforced piping) is extremely expensive and introduces safety hazards. Unlike the Haber process (Δn = −2), where the pressure effect is much more dramatic, the Contact process gains little thermodynamic advantage from very high pressure, so the cost-benefit analysis favors moderate pressure. This illustrates that industrial process design integrates equilibrium thermodynamics with kinetics, catalyst performance, and economic constraints.

Question 5: CRITICAL THINKING

Consider a hypothetical gaseous equilibrium A(g) ⇌ B(g) + C(g) with Kₚ = 4.0 atm at 400 K. The system starts at equilibrium with p(A) = 1.0 atm, p(B) = 2.0 atm, and p(C) = 2.0 atm. The volume is suddenly doubled. (a) Calculate Qₚ immediately after expansion. (b) Set up (but do not need to fully solve) the ICE-style expression that would allow you to find the new equilibrium partial pressures. (c) Without solving, explain whether the new equilibrium pressure of B will be greater than, less than, or equal to 2.0 atm.

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(a) Doubling the volume halves all partial pressures: p(A) = 0.50 atm, p(B) = 1.0 atm, p(C) = 1.0 atm. Qₚ = (1.0)(1.0) / (0.50) = 2.0 atm. Since Qₚ = 2.0 < Kₚ = 4.0, the reaction shifts forward. Alternatively, α = 1/2 for an expansion, so Qₚ = (1/2)⁺¹ × 4.0 = 2.0, consistent. (b) Let x = change in partial pressure of A consumed. Then: p(A) = 0.50 − x, p(B) = 1.0 + x, p(C) = 1.0 + x. Substituting into Kₚ: (1.0 + x)² / (0.50 − x) = 4.0. This is a quadratic in x that can be solved for the equilibrium state. (c) Since x > 0, p(B) at the new equilibrium = 1.0 + x > 1.0 atm but we must compare to the original 2.0 atm. The forward shift increases B from its post-expansion value of 1.0, but cannot restore it fully to 2.0—Le Châtelier's principle only partially counteracts the perturbation. Therefore, p(B) at the new equilibrium will be less than 2.0 atm (somewhere between 1.0 and 2.0 atm). The exact value requires solving the quadratic.

SUMMARY

Lesson Summary

Summary

Gaseous equilibria respond to pressure and volume changes in a predictable manner governed by Le Châtelier's principle and the reaction quotient Qₚ. The key variable is Δn, the change in total moles of gas (products minus reactants). When the volume is decreased (pressure increased by compression), the system shifts toward the side with fewer moles of gas; when the volume is increased, the shift favors more moles of gas. When Δn = 0, pressure and volume changes have no effect on the equilibrium position.

Mathematically, compression by a factor α changes the reaction quotient to Qₚ = α^Δn × Kₚ, providing a rigorous basis for the qualitative Le Châtelier prediction. Crucially, Kₚ itself never changes with pressure—only the equilibrium composition adjusts. Students must also remember that adding an inert gas at constant volume does not shift the equilibrium because partial pressures of the reacting species remain unchanged, whereas adding an inert gas at constant total pressure effectively dilutes the mixture and can cause a shift. These principles underpin major industrial processes such as the Haber process and the Contact process, where pressure optimization directly translates to economic yield.