Chemical Equilibrium & Le Chatelier Lab

See when volume shifts equilibrium (only when Δn ≠ 0) and watch temperature alter K while concentrations shift Q.

Worked example — Chemical Equilibrium & Le Chatelier Lab

Equilibrium Constant K	50.5000
Initial Quotient Q₀	0.0000
Extent of Reaction ξ (mol)	0.7804
Reactant 1 Conversion	78.0%
Reactant 2 Conversion	78.0%
Mole Change Δn	+0

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Squeezing only shifts a reaction when mole counts change 🖖

A volume squeeze doubles every concentration in a gas reaction. For H₂ + I₂ ⇌ 2HI, two reactants become two products (Δn = 0), so the doublings in the numerator and denominator cancel out exactly — leaving Q unchanged and conversion flat at 78.0% at every volume. For N₂ + 3H₂ ⇌ 2NH₃ (Δn = -2), four gas molecules compress into two, so on compression the denominator grows as 1/V⁴ while the numerator grows only as 1/V², driving the reaction strongly toward ammonia.

Only temperature moves K; everything else moves Q 🖖

Adding reagents, removing products or changing the container volume changes the reaction quotient Q without touching the equilibrium constant K. Adding an inert gas at constant volume changes neither, since every concentration is unmoved. Temperature is the sole control that alters K itself, dictated by the van 't Hoff equation and the sign of ΔH°. An exothermic reaction (like Haber ammonia synthesis) sees K collapse as temperature rises, while endothermic N₂O₄ dissociation sees K climb.

Adding reactant converts more of the other and less of itself 🖖

Pouring excess reactant B into a mixture of A and B drives A's conversion toward 100%, but the fraction of B consumed drops. In industrial chemistry, 'driving a reaction to completion' always means completion with respect to the more expensive reagent by flooding the reactor with the cheaper one.

Problems solved in full

Why squeezing does nothing to hydrogen iodide and everything to ammonia 6 steps
Halve the volume on H₂ + I₂ ⇌ 2HI and the conversion does not move. Halve it on N₂ + 3H₂ ⇌ 2NH₃ and it climbs. Write out K for both and show where the volume goes — then say how far compression can take you.
1. Work in extent. If ξ moles of reaction have run from 1 mol of each reactant, H₂ and I₂ are each 1 − ξ and HI is 2ξ, all divided by V to make concentrations.
2. Put them into K. The numerator (2ξ/V)² carries V⁻², and the denominator (1−ξ)/V × (1−ξ)/V carries V⁻² as well, so the volumes cancel and K = 4ξ²/(1−ξ)². At ξ = 0.7804 that is 50.5, which is the K the panel started with.
  
  $K = \frac{[ HI ] ^{2}}{[ H _{2} ] [ I _{2} ]} = \frac{( 2 ξ / V ) ^{2}}{( \frac{1 - ξ}{V} ) ^{2}} = \frac{4 ξ ^{2}}{( 1 - ξ ) ^{2}}$
3. V does not appear, so no value of it can change ξ. The panel prints an extent of 0.7804 at 2 L, at 1 L, at 0.5 L and at 0.25 L — the same four digits four times, which is not the tool being lazy.
  
  $ξ = 0.78037 \Rightarrow \frac{4 ( 0.78037 ) ^{2}}{( 0.21963 ) ^{2}} = 50.5 (V absent)$
4. Now ammonia, where the counts differ. N₂ is 1 − ξ, H₂ is 3 − 3ξ, NH₃ is 2ξ. The numerator carries V⁻² and the denominator V⁻⁴, so this time K = 4ξ²V²/(27(1−ξ)⁴) and a V² survives.
  
  $K = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}} = \frac{( 2 ξ / V ) ^{2}}{\frac{1 - ξ}{V} ( \frac{3 - 3 ξ}{V} ) ^{3}} = \frac{4 ξ ^{2} V ^{2}}{27 ( 1 - ξ ) ^{4}}$
5. That surviving V² is the whole effect. K is fixed, so shrinking V forces ξ up to compensate: the extent reads 0.3675 at 2 L, 0.4858 at 1 L, 0.5969 at 0.5 L and 0.6929 at 0.25 L, which the panel shows as a conversion climbing from 36.7% to 69.3%. Put any of them back into the expression and K comes out at 0.500, to the four digits the extent carries.
  
  $V = 1 \to 0.5 L : ξ = 0.48578 \to 0.59693$
6. The exponent is not arbitrary — it is −Δn, the change in gas moles. HI has Δn = 0 and gets V⁰. Ammonia goes from four molecules to two, Δn = −2, and gets V². Any reaction that makes fewer gas molecules than it consumes rewards compression, in exactly that power.
  
  $exponent of V = - Δ n_{gas} : HI : 0, N H_{3} : - (2 - 4) = 2$
Answer

The volume exponent is −Δn: zero for HI, two for ammonia, and that single number decides whether a pressure vessel is worth building. Which is why the Haber process runs at a couple of hundred atmospheres and the HI equilibrium is studied in ordinary glassware — one of them pays for steel and the other cannot. But follow the compression down and it disappoints: 79.23% at 0.1 L, 92.89% at 0.01 L, 97.69% at 0.001 L. Every factor of ten in volume buys less than the one before, because ξ enters as ξ² against (1−ξ)⁴ and the fourth power wins as ξ approaches 1. You cannot squeeze your way to completion, only towards it, and the real plant stops long before the arithmetic does — at some point the vessel wall costs more than the ammonia. Le Chatelier tells you the direction. Only the expression tells you how much, and how fast the how-much runs out.
How many degrees a flask of N₂O₄ is worth per doubling of its volume 6 steps
Load Temperature sweep: N₂O₄ ⇌ 2NO₂: one mole in a 1 L flask at 25 °C, K = 0.0058, extent 0.0374, Δn of +1. Find the temperature that does exactly what doubling the flask does — then decide, for the ammonia reaction, which of the two knobs a plant can actually reach for.
1. Write K out in the extent, one mole in, ξ moles reacted. Two NO₂ appear for every N₂O₄ that goes, so the numerator is squared and the denominator is not, and one factor of V survives the division. Move it across: the left side is now pure ξ and everything the outside world can do sits on the right as a single product.
  
  $K = \frac{[ N O _{2} ] ^{2}}{[ N _{2} O _{4} ]} = \frac{( 2 ξ / V ) ^{2}}{( 1 - ξ ) / V} = \frac{4 ξ ^{2}}{V ( 1 - ξ )} ⟹ \frac{4 ξ ^{2}}{1 - ξ} = K V$
2. That is the general shape, and the exponent is Δn — the change in the number of gas molecules. Hydrogen iodide makes two from two, so V drops out and the right side is K alone. Ammonia makes two from four, so V arrives squared and underneath. The extent never sees temperature and volume separately; it sees one number.
  
  $f (ξ) = K V^{Δ n_{gas}} : HI : \frac{4 ξ ^{2}}{( 1 - ξ ) ^{2}} = K, N H_{3} : \frac{4 ξ ^{2}}{27 ( 1 - ξ ) ^{4}} = \frac{K}{V ^{2}}$
3. Which means the two controls are interchangeable, and the rate is readable straight off. Here Δn is +1, so doubling the flask doubles the product, and the extent goes from 0.0374 to 0.0524.
  
  $Δ n = + 1 : V \to 2 V \equiv K \to 2 K; ξ : 0.0374 \to 0.0524$
4. Now buy the same doubling with heat instead. N₂O₄ dissociation is endothermic at +57.2 kJ per mole, so warming raises K, and van 't Hoff says by how much. Doubling K takes 9.2 degrees. Set the flask to 34.2 °C at its original 1 litre and the extent card reads 0.0524 — the same four digits.
  
  $ln \frac{K _{2}}{K _{1}} = - \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}}) = ln 2 ⟹ T_{2} = 307.38 K = 34.2^{\circ} C$
5. The rate is not fixed, though, and the reason is in the equation: the sensitivity of ln K to temperature goes as 1/T², so it fades as the flask warms. The first doubling costs 9.2 degrees, the next one 9.8, and the linearised estimate at 25 °C is 8.96.
  
  $Δ T \approx \frac{R T ^{2} ln 2}{Δ H ^{\circ}} = \frac{8.314 \times 298.1 5 ^{2} \times 0.6931}{57 , 200} = 8.96 K; next doubling 9.82 K$
6. Run the same conversion on ammonia, where Δn is −2 and the sign of everything flips. Halving the reactor is worth four times the product, and because the reaction is exothermic that is bought by cooling: from 400 °C down to 620.9 K, which is 347.8 °C. The extent is 0.5969 either way, to sixteen digits.
  
  $N H_{3}, Δ n = - 2 : V \to \frac{V}{2} \equiv K \to 4 K \equiv 673.15 \to 620.90 K; ξ = 0.5969$
Answer

9.2 degrees per doubling of the flask, at 25 °C, and the rate rises as the flask warms. Temperature and volume are not two effects to be reasoned about separately; they are one number, K·V^Δn, and Δn is the exchange rate between them.

The ammonia line is the one that built a chemical industry. Squeezing the reactor to half its size does exactly what cooling it by 52 degrees would do — and cooling it is the one thing a Haber plant cannot afford, because at 348 °C the catalyst is too slow to reach that equilibrium in any useful time. Compression buys the yield that temperature is forbidden to buy. So the pressure vessel is not there because pressure is a better lever than heat; it is there because it is the only lever left once the rate has claimed the other one.

And Δn = 0 is the case with no exchange rate at all. Hydrogen iodide gets nothing from volume however hard you press it, which leaves temperature as the sole control, and a reaction with a ΔH° of only −9.6 kJ per mole barely responds to that either. Some equilibria simply cannot be moved, and the two exponents tell you which before you build anything.

$Δ T \approx \frac{R T ^{2}}{Δ H ^{\circ}} Δ n ln \frac{V _{2}}{V _{1}}; Δ n = 0 \Rightarrow no temperature is equivalent$

References (2)

The standard formation enthalpies and entropies behind every reaction enthalpy the tool uses NIST Chemistry WebBook, SRD 69: gas-phase Δ_f H° and S° for H₂, I₂, HI, N₂, NH₃, NO₂, N₂O₄, CO, CO₂ and H₂O.
The equilibrium constants themselves, and the temperature each was measured at Atkins, P. and de Paula, J. (2018). Atkins’ Physical Chemistry (11th ed.). Oxford University Press — chemical equilibrium and the van ’t Hoff equation.

By the 42Math team · Updated 2026-08-23

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Volume sweep: H₂ + I₂ (Δn = 0) - Halving volume doubles all concentrations equally; Q₀ remains 0 and conversion sits at a flat 78.0% everywhere.
Volume sweep: N₂ + 3H₂ (Δn = -2) - At 1 L volume, N₂ + 3H₂ reaches 48.6% conversion; squeezing to higher pressure drives conversion hard because Δn = -2.
Temperature sweep: N₂O₄ ⇌ 2NO₂ - At 298 K, N₂O₄ has K = 0.0058 and 3.7% conversion; heating the endothermic reaction drives the forward shift.
Excess reactant: H₂ + I₂ - Setting I₂ to 3.0 mol against 1.0 mol of H₂, a threefold excess, drives H₂ conversion to 96.4% while I₂ conversion drops to 32.1%.