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Pro 🔒~18 min

Chemical Equilibrium & Le Chatelier's Principle

Dynamic balance between forward and reverse reactions

How it works

Chemical equilibrium is a dynamic state where the forward and reverse reaction rates are equal, so concentrations remain constant (but not necessarily equal). The equilibrium constant Kc is calculated from equilibrium concentrations: Kc = [products]^m / [reactants]^n. Large K (>1) favors products; small K (<1) favors reactants. Le Chatelier's Principle: if a system at equilibrium is disturbed, it shifts to partially counteract the disturbance. Adding a reactant → shifts right (more products). Removing a product → shifts right. Increasing pressure (for gases) → shifts toward fewer moles of gas. Increasing temperature → shifts in the endothermic direction (changes K). Catalysts do NOT change the equilibrium position — only reach it faster.

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Step-by-step

  1. Press Play to watch the reaction approach equilibrium — concentration lines converge.
  2. Once at equilibrium, apply stresses using the controls.
  3. Adding [A] (reactant) → watch the system shift right.
  4. Increasing temperature on an exothermic reaction → K decreases, equilibrium shifts left.
  5. The Q vs K indicator shows which way the reaction is going.

Key formulas

  • Kc=[C]c[D]d[A]a[B]b(at equilibrium)K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \quad \text{(at equilibrium)}Equilibrium constant Kc (products over reactants, raised to stoichiometric powers)
  • Q<Kc→forward;Q>Kc→reverseQ < K_c \to \text{forward}; \quad Q > K_c \to \text{reverse}Reaction quotient Q determines which direction equilibrium shifts
  • Kp=Kc(RT)ΔngK_p = K_c(RT)^{\Delta n_g}Converting between Kc and Kp (Δng = change in moles of gas)

Frequently asked questions

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), if at equilibrium [N₂]=0.5, [H₂]=0.3, [NH₃]=0.2 M, calculate Kc.
Kc = [NH₃]²/([N₂][H₂]³) = (0.2)²/(0.5×0.3³) = 0.04/(0.5×0.027) = 0.04/0.0135 ≈ 2.96.
The Haber process is exothermic. Why is it industrially run at high temperature (400-500°C) despite Le Chatelier's prediction?
High T decreases K (less NH₃ at equilibrium) but dramatically increases reaction rate. The industrial compromise: high T for acceptable rate, high pressure and catalyst to improve yield. Remove NH₃ continuously to shift equilibrium right.
For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), predict the effect of doubling the total pressure.
Left side: 2+1=3 mol gas. Right side: 2 mol gas. Increasing pressure shifts toward fewer moles of gas → shifts right → more SO₃.
Kc = 0.025 at 700°C for the reaction CO(g)+H₂O(g)⇌CO₂(g)+H₂(g). If Q = 0.1 > Kc, what will the reaction do?
Q > Kc means too many products relative to equilibrium → reaction shifts LEFT (toward reactants) to decrease Q until Q = Kc.