← SciSim / Chemistry
Chemical Equilibrium & Le Chatelier's Principle
Physical Chemistry #03
⚗️ Section 1 — Interactive Simulation / Equilibrium Dynamics
N₂+3H₂⇌2NH₃
N₂O₄⇌2NO₂
H₂+I₂⇌2HI
Add Reactant
Change T/P
[Products]
—
mol/L
[Reactants]
—
mol/L
Kc
—
varies
Qc (current)
—
—
Shift Direction
—
—
ΔG (kJ/mol)
—
kJ/mol
% Conversion
—
%
Presets
Haber
450°C
NO₂
25°C
HI
450°C
Haber
High P
CO+3H₂
→CH₄
N₂O₄
55°C
Controls
Conditions
Temperature723K
Pressure200atm
[Reactant]₀1.000mol/L
ΔH° reaction-92kJ/mol
Add to system0.00mol/L
Speed
Sim Speed1.0×
Display
Particle Labels
Shift Arrow
Collision Sparks
💡 Section 2 — The Idea, Step by Step / From a Balanced Tug-of-War to Kc

Start simple: a busy revolving door

Picture a revolving door between a warm lobby and a cold street. People keep pushing through in both directions all day. After a while the number of people in the lobby stops changing — not because anyone stopped moving, but because just as many walk out as walk in each minute. That steady, two-way balance is exactly what chemists mean by equilibrium. The reaction never stops; the forward and reverse changes simply run at the same rate, so the amounts you can measure hold still.

Name the pieces and the rule

A reversible reaction is written with a double arrow, $\text{A} \rightleftharpoons \text{B}$. The forward step turns reactant into product; the reverse step turns it back. When their rates match, the mixture sits at a fixed ratio. We capture that ratio with the equilibrium constant, $K_c = \dfrac{[\text{products}]}{[\text{reactants}]}$, where each concentration is raised to its coefficient in the balanced equation.

One worked number: suppose $\text{A} \rightleftharpoons \text{B}$ has $K_c = 4$ and you start with $1$ mol/L of A. The amount converted is $x = \dfrac{K_c C_0}{1+K_c} = \dfrac{4(1)}{5} = 0.8$ mol/L. So at equilibrium you hold $0.8$ M product and $0.2$ M reactant — and indeed $0.8/0.2 = 4 = K_c$. A bigger $K_c$ simply parks the balance further toward products.

The precise picture and the sliders

For a general reaction the full form is $K_c = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$. Its twin, the reaction quotient $Q_c$, uses the same formula but with the concentrations you have right now. Comparing them gives direction: $Q < K$ means too few products, so the reaction runs forward; $Q > K$ means too many, so it runs back; $Q = K$ is equilibrium. That single comparison is Le Chatelier's principle — poke the system and it shifts to partly undo the poke. Two pokes deserve care: changing temperature changes $K$ itself (van't Hoff, $\frac{d\ln K}{dT} = \frac{\Delta H^\circ}{RT^2}$, so heating an exothermic reaction lowers $K$), while changing pressure shifts a gas reaction toward the side with fewer moles when $\Delta n \neq 0$. In the sim above, the Temperature and ΔH° sliders set $K_c$, the Pressure slider acts when $\Delta n \neq 0$, and the Add to system slider pushes $Q$ off $K$ so you can watch the shift.

Try this in the sim above

First, drag Add to system positive and watch the Shift Direction card flip to "→ Forward" as $Q$ drops below $K$. Second, with the exothermic Haber preset, raise Temperature and watch $K_c$ (and % conversion) fall — the counter-intuitive cost of running hot. Third, switch to the $\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2$ mode and raise Pressure: with more gas moles on the right, the balance retreats toward the single-molecule side.

📐 Section 3 — Equation Derivation / Equilibrium Constant & Van't Hoff
Law of Mass Action & Van't Hoff Equation
$$ K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} \quad \Delta G^\circ = -RT\ln K_c $$ $$ \frac{d\ln K}{dT} = \frac{\Delta H^\circ}{RT^2} \implies \ln\frac{K_2}{K_1} = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) $$

Symbol Definitions

SymbolMeaningUnit
$K_c$Equilibrium constant in terms of concentrationsdimensionless (or mol^Δn L^-Δn)
$[\text{X}]$Equilibrium molar concentration of species Xmol L⁻¹
$a,b,c,d$Stoichiometric coefficients in balanced equationdimensionless
$\Delta G^\circ$Standard Gibbs free energy change at 298 K, 1 barkJ mol⁻¹
$R$Gas constant = 8.314 J mol⁻¹ K⁻¹J mol⁻¹ K⁻¹
$T$Absolute temperatureK
$Q_c$Reaction quotient (same form as $K_c$ but not at equilibrium)same as $K_c$
$\Delta H^\circ$Standard enthalpy change of reactionkJ mol⁻¹

Step-by-Step Derivation

Step 1 — Thermodynamic Definition of K
From chemical thermodynamics, at constant T and P the reaction proceeds until $\Delta G = 0$ (minimum free energy). The free energy of mixing gives: $$ \Delta G = \Delta G^\circ + RT\ln Q $$ At equilibrium, $\Delta G = 0$ and $Q = K$, therefore: $$ 0 = \Delta G^\circ + RT\ln K \implies \Delta G^\circ = -RT\ln K $$
Step 2 — Law of Mass Action from Activities
For ideal solutions, the chemical potential $\mu_i = \mu_i^\circ + RT\ln(c_i/c^\circ)$ where $c^\circ = 1$ mol/L. Summing over products minus reactants: $$ \Delta G = \sum_\text{prod}\mu_i - \sum_\text{react}\mu_j $$ At equilibrium this gives: $K_c = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$
Step 3 — Gibbs-Helmholtz to Van't Hoff
From $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ and $\Delta G^\circ = -RT\ln K$: $$ \ln K = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R} $$ Differentiate with respect to T: $$ \frac{d\ln K}{dT} = \frac{\Delta H^\circ}{RT^2} \quad \text{(van't Hoff equation)} $$
Step 4 — Integrated van't Hoff
Assuming $\Delta H^\circ$ constant over temperature range (valid over small intervals): $$ \int_{K_1}^{K_2}d\ln K = \frac{\Delta H^\circ}{R}\int_{T_1}^{T_2}\frac{dT}{T^2} $$ $$ \ln\frac{K_2}{K_1} = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) $$
Step 5 — Le Chatelier's Principle from Q vs K
If $Q < K$: reaction proceeds forward (products form) until $Q = K$. If $Q > K$: reaction proceeds reverse until $Q = K$. Adding reactant increases denominator of Q → Q < K → forward shift. Increasing T for exothermic (ΔH < 0): K decreases (van't Hoff, negative dln K/dT) → less product at equilibrium.
Step 6 — ICE Table Solution
For $\text{A} \rightleftharpoons \text{B}$ with $K_c$ and $C_0$ initial [A]: $K_c = x/(C_0-x) \implies x = K_c C_0/(1+K_c)$, product $= x$, reactant $= C_0-x$. % conversion $= 100x/C_0 = 100K_c/(1+K_c)$.

Simulation Variable Mapping

Temperature slider
Computes Kc via van't Hoff from reference Kc at 723 K. ΔH slider sets the sign and magnitude of response.
Pressure slider
For reactions with Δn ≠ 0 (e.g. Haber: Δn = −2), increasing P shifts equilibrium toward fewer moles.
Add slider
Injects or removes reactant, computing new Q and showing the direction of Le Chatelier shift.
Particle colors
Blue = reactants (A, B), green = products (C, D). Number ratio reflects equilibrium concentrations.

Worked Example — Haber Process: N₂ + 3H₂ ⇌ 2NH₃

$K_c \approx 1.0\times10^{10}$ at 298 K; $\Delta H^\circ = -92.4$ kJ/mol. Find $K_c$ at 723 K (450°C, industrial). (Ammonia's $K_c$ at room temperature is enormous; the exothermic reaction's $K$ collapses as it is heated.)

$\ln(K_{723}/K_{298}) = -\frac{-92400}{8.314}\left(\frac{1}{723}-\frac{1}{298}\right) = 11114 \times (-0.001973) = -21.9$

$K_{723} = 1.0\times10^{10} \times e^{-21.9} = 1.0\times10^{10} \times 3.0\times10^{-10} = \mathbf{3.0}$

Starting with [N₂]₀ = 1, [H₂]₀ = 3 M: ICE gives [NH₃] = 0.93 M (32% conversion at 450°C, 200 atm).

References: Atkins & de Paula — Physical Chemistry, 11th Ed., Ch. 6.1–6.2: "The equilibrium constant." | Zumdahl — Chemical Principles, 8th Ed., Ch. 13: "Chemical Equilibrium." | HyperPhysics — "Chemical Equilibrium" (hyperphysics.phy-astr.gsu.edu)
❓ Section 4 — FAQ
🧪 ConceptualWhat does the equilibrium constant K actually measure?▼
K measures the ratio of product concentrations to reactant concentrations at equilibrium, with each raised to its stoichiometric coefficient. It is a thermodynamic quantity that depends only on temperature — not on initial concentrations, volume, or pressure. A large K (e.g., K = 10⁶) means products are strongly favored at equilibrium; small K (e.g., K = 10⁻⁶) means reactants dominate. K is related to the standard Gibbs free energy by ΔG° = −RT ln K — this connects equilibrium position to thermodynamic stability.
✦ Key Takeaway: K tells you where the reaction "wants" to go at a given temperature. K > 1 = products favored; K < 1 = reactants favored.
🌍 Real LifeHow does the Haber process use Le Chatelier's principle industrially?▼
N₂ + 3H₂ ⇌ 2NH₃ (ΔH = −92 kJ/mol) has several competing Le Chatelier factors. Lower temperature shifts equilibrium toward NH₃ (exothermic), but the reaction is too slow at low T. Higher pressure (the reaction produces 4 mol → 2 mol, so high P favors products) increases yield. The industrial compromise: 400–500°C (fast enough kinetics), 150–300 atm (yield acceptable), iron catalyst (speeds up equilibrium attainment). Continuous removal of NH₃ keeps Q < K, pulling the equilibrium forward perpetually. This process feeds ~50% of the world's population through synthetic fertilizers.
✦ Key Takeaway: Haber process is Le Chatelier in action — every operating parameter is a trade-off between equilibrium yield and reaction rate.
🔬 SimulationWhat do Q and K tell us about the direction of reaction in the simulation?▼
Q (reaction quotient) has the same mathematical form as K but uses the current concentrations, not equilibrium ones. If Q < K, the numerator (products) is too small relative to equilibrium — the reaction proceeds forward to increase product concentration until Q = K. If Q > K, there are too many products — the reaction runs in reverse. When you use the "Add" slider to inject reactants, Q drops below K momentarily, and you see the simulation shift right. The Shift Direction card shows "→ Forward" or "← Reverse" in real time.
✦ Key Takeaway: Compare Q to K. Q < K → forward. Q > K → reverse. Q = K → equilibrium (no net change).
💡 Non-ObviousWhy doesn't adding a catalyst change the equilibrium position?▼
A catalyst speeds up both the forward and reverse reactions equally, because it lowers the activation energy of the transition state without changing the thermodynamic stability of reactants or products. Since K = kforward/kreverse (rate constant ratio), and a catalyst multiplies both k values by the same factor, K is unchanged. The equilibrium concentrations — and therefore the % yield — are identical with and without a catalyst. What the catalyst does is get you there faster. Le Chatelier's principle predicts shifts due to changes in concentration, temperature, or pressure — not to catalysts.
✦ Key Takeaway: Catalyst → faster equilibrium, same position. K is thermodynamic (temperature-only), not kinetic.
🧮 MathematicalHow do I solve an equilibrium problem when the ICE table gives a cubic equation?▼
For reactions like N₂ + 3H₂ ⇌ 2NH₃, the ICE table gives: Kc = (2x)²/((1−x)(3−3x)³) which is indeed a quartic in x. For large Kc values, use the approximation that conversion is high and solve iteratively. For small Kc, use the 5% approximation (x ≪ C₀). In general, numerical methods work best: start with x₀ = 0.5C₀, then Newton-Raphson: x_{n+1} = x_n − f(x_n)/f'(x_n). The simulation uses this numerical approach. For the special case where Δn = 0 (like H₂ + I₂ ⇌ 2HI), the equation simplifies to a quadratic.
✦ Key Takeaway: For Δn ≠ 0 equilibria, use numerical iteration (or the 5% rule if Kc is small). The simulation uses Newton-Raphson internally.
🌌 Deep / AdvancedHow does Le Chatelier's principle connect to the Second Law of Thermodynamics?▼
Le Chatelier's principle is a macroscopic consequence of entropy maximization. When a perturbation is applied (say, adding heat), the system responds to maximize entropy of the universe — the system absorbs the perturbation in a way that dissipates the imposed change. Endothermic shifts at high T increase the system's entropy by absorbing heat from surroundings. Shifts toward fewer moles at high pressure reduce the work done against the surroundings. Mathematically, this follows from dG = −SdT + VdP at constant composition: the system minimizes G by adjusting composition. Le Chatelier's "opposition" is really entropy seeking its maximum in disguise.
✦ Key Takeaway: Le Chatelier's principle is the 2nd Law in chemical disguise — every shift maximizes entropy of the universe.
Section 4 References: LibreTexts Chemistry — "Chemical Equilibrium" (https://chem.libretexts.org) | Khan Academy — "Equilibrium" (khanacademy.org/science/ap-chemistry-beta) | MIT OCW 5.111, Lecture 26-28 | Chemguide.co.uk — "Equilibria"
⚠️ Section 5 — Common Misconceptions
❌
Misconception: "At equilibrium, the forward and reverse reactions stop completely."
✅
Equilibrium is dynamic: both forward and reverse reactions continue at equal rates. Macroscopic concentrations appear constant only because the rates balance. If you use radioactive isotope tracers in a reaction at equilibrium, you can show that atoms are still being exchanged between reactants and products constantly — just at equal rates. The concept of "stopping" confuses macroscopic constancy with molecular stasis. This dynamic nature is why catalysts can still accelerate reaching equilibrium without changing its position.
📖 Atkins & de Paula — Physical Chemistry, 11th Ed., Ch. 6.1. | Driver et al. — Making Sense of Secondary Science, Routledge, 1994.
❌
Misconception: "A large K means the reaction goes to completion."
✅
Even with K = 10⁶, both reactants and products are present at equilibrium — just in an extreme ratio. "Completion" strictly applies only when K → ∞, which is approached but never truly reached thermodynamically. Practically, a reaction is considered complete when < 0.01% reactant remains, requiring K > 10⁸. The % conversion depends on both K and initial concentrations: for K = 100 with equimolar reactants, conversion is 91%; for K = 10⁶, conversion is 99.9%. Near-complete reactions are common in strong acid/base neutralization and precipitation reactions, but even these have finite K.
📖 Zumdahl — Chemical Principles, 8th Ed., Ch. 13.4: "Solving equilibrium problems."
❌
Misconception: "Adding an inert gas always shifts equilibrium to the side with more moles."
✅
If the container volume is fixed (rigid container), adding an inert gas increases total pressure but does NOT change the partial pressures or concentrations of reactive species. Therefore Q is unchanged and equilibrium does not shift. If the container is flexible (constant total pressure), the addition of inert gas causes expansion, which dilutes all species, lowering partial pressures — this shifts equilibrium toward more moles (Le Chatelier). The key distinction is constant volume (no shift) vs constant pressure (dilution shift).
📖 Silberberg — Chemistry, 9th Ed., Ch. 17.6: "The effect of pressure on equilibrium."
❌
Misconception: "Kc changes when you change concentration, pressure, or add a catalyst."
✅
Kc depends ONLY on temperature. Adding reactants changes Q temporarily (Q < K), and the system shifts to restore Q = K — but K itself remains the same. Changing pressure redistributes concentrations but Kc (in terms of concentrations) is unchanged (though Kp may differ if Δn ≠ 0). Catalysts accelerate both forward and reverse rates equally, so K = kf/kr is unchanged. Only temperature changes Kc, via the van't Hoff equation. This is because K is a ratio of thermodynamic activities determined by ΔG°, which is temperature-dependent.
📖 Atkins & de Paula — Physical Chemistry, 11th Ed., Ch. 6.2. | Levine — Physical Chemistry, 6th Ed., Ch. 11.
❌
Misconception: "Le Chatelier says the system fully opposes any change — equilibrium is restored to exactly the original state."
✅
Le Chatelier says the system shifts to partially oppose the change — not completely. If you add reactant, the equilibrium shifts right but the final equilibrium has MORE reactant than the original, just less than immediately after addition. The system never returns to exactly the original state. The new equilibrium concentrations satisfy the same K but at different absolute values. Think of it as damage control, not full restoration: the system minimizes the impact without eliminating it entirely.
📖 Zumdahl — Chemical Principles, 8th Ed., Ch. 13.6: "Le Chatelier's principle." | J. Chem. Educ., various misconception surveys.
Section 5 Education Research References: Tyson, L. et al. — "A view into the classroom: Using interview data to understand student thinking about equilibrium." J. Chem. Educ. 1999, 76, 554. | Nakhleh, M.B. — J. Chem. Educ. 1992, 69, 191. | Taber, K.S. — Chemical Misconceptions, RSC, 2002.