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Chemical Equilibrium & Le Chatelier's Principle

Dynamic balance between forward and reverse reactions

A reaction that never finishes — on purpose. The forward reaction runs, the reverse reaction runs, and when they run at exactly the same speed the mixture looks frozen but is anything but: molecules keep trading places millions of times per second. Now push on this balance — dump in more reactant — and the system pushes back. Partially. It never fully wins. That stubborn, incomplete pushback is Le Chatelier's principle, and it is the reason industry can make ammonia at all.

What you'll be able to do

  • Describe equilibrium as a dynamic state: forward and reverse rates equal, molecular exchange ongoing
  • Use Q versus K to predict the direction of shift after a concentration disturbance
  • Apply Le Chatelier's principle with correct partial-counteraction reasoning to concentration, pressure, and temperature stresses

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)

Make a prediction

Your reversible reaction sits quietly at equilibrium. You suddenly inject extra reactant A. After the system re-settles, how does the amount of product compare with before the injection?

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Your prediction

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Answer: More product — the system shifts forward, but only partially consumes the extra A

Le Chatelier's principle: the system counteracts the disturbance — partially, never completely. Injecting A drops Q below K, so the forward reaction surges until Q climbs back to K: extra product forms, AND the new equilibrium still holds more A than before the injection (it ate some of the spike, not all of it). A constant K does not freeze concentrations — it fixes their ratio at equilibrium. And the dilution answer confuses rate with position: equilibrium is about where the ratio settles, not how fast it gets there.

Quiz (0/4)

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), if at equilibrium [N₂]=0.5, [H₂]=0.3, [NH₃]=0.2 M, calculate Kc.

The Haber process is exothermic. Why is it industrially run at high temperature (400-500°C) despite Le Chatelier's prediction?

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), predict the effect of doubling the total pressure.

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?

You can now

  • Describe equilibrium as a dynamic state: forward and reverse rates equal, molecular exchange ongoing
  • Use Q versus K to predict the direction of shift after a concentration disturbance
  • Apply Le Chatelier's principle with correct partial-counteraction reasoning to concentration, pressure, and temperature stresses

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.