Le Chatelier's principle — pressure, temperature, concentration, catalyst effects

WHY does this happen? At equilibrium, forward and reverse rates are equal. When you disturb the system, you temporarily break this balance. The system responds by favoring whichever direction re-establishes equilibrium under the new conditions.

Core Definition

Key term: Stress = any external change that disturbs the ratio $Q$ away from $K$.

Derivation from First Principles

Starting point: For a reaction $aA + bB \rightleftharpoons cC + dD$, equilibrium constant is:

$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

The reaction quotient $Q$ has the same form but uses current concentrations (not equilibrium ones):

$Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}\bigg|_{\text{current}}$

The shift rule:

• If $Q < K$: Too much reactant → shift forward (produce more products)
• If $Q > K$: Too much product → shift backward (produce more reactants)
• If $Q = K$: At equilibrium, no net shift

WHY? Because $\Delta G = \Delta G^\circ + RT\ln Q$. When $Q \neq K$, $\Delta G \neq 0$, so the reaction is spontaneous in whichever direction brings $Q$ toward $K$.

Effect 1: Concentration Changes

Derivation: For $N_2 + 3H_2 \rightleftharpoons 2NH_3$, if you add $N_2$:

$Q = \frac{[NH_3]^2}{[N_2][H_2]^3}$

Increasing $[N_2]$ makes the denominator larger, so $Q$ decreases. Since $Q < K$ now, the system shifts forward to consume the added $N_2$ and increase $[NH_3]$ until $Q = K$ again.

Stress: Add $0.2\,M$ of $SO_2$.

Step 1 — Calculate Q: $Q = \frac{[SO_3]^2}{[SO_2]^2[O_2]} = \frac{(0.8)^2}{(0.7)^2(0.3)} = \frac{0.64}{0.147} = 4.35$

Why this step? New $[SO_2] = 0.5+ 0.2 = 0.7\,M$. We need to see if $Q$ changed relative to $K$.

Step 2 — If $K = 5.0$: Since $Q = 4.35 < K = 5.0$, the system shifts forward.

Why? Not enough product relative to reactants under the new conditions.

Step 3 — Prediction: $[SO_3]$ will increase, $[SO_2]$ and $[O_2]$ will decrease until $Q = K$.

Effect 2: Pressure and Volume Changes

WHY? For gas-phase reactions, $K_p$ relates to partial pressures. By ideal gas law, $P \propto n/V$ at constant $T$. Changing $V$ changes all concentrations proportionally, but the equilibrium expression has different powers of each species.

Derivation: For $N_2 + 3H_2 \rightleftharpoons 2NH_3$ (4 moles reactant gas → 2 moles product gas):

$K_p = \frac{P_{NH_3}^2}{P_{N_2} \cdot P_{H_2}^3}$

If you compress (decrease $V$), all partial pressures increase. But the denominator has higher total exponent ($1 + 3 = 4$ vs numerator's $2$), so $Q$ decreases more than $K$. Thus $Q < K$ → shift forward.

Count moles:

• Reactants: $1 + 3 = 4$ moles gas
• Products: $2$ moles gas
• Compression favors products (fewer moles)

Stress: Compress to $1\,L$ (volume halved, pressure doubled).

Step 1 — Count moles:

• Reactant: 1 mole $N_2O_4$
• Product: 2 moles $NO_2$
• More moles on product side

Why this step? The side with more moles is more sensitive to pressure changes.

Step 2 — Predict shift: Pressure increase favors fewer moles → shift backward (toward $N_2O_4$).

Why? Forming $N_2O_4$ reduces total gas molecules, partially relieving the pressure stress.

Key insight: If equal moles on both sides, pressure change causes no shift.

Effect 3: Temperature Changes

WHY is temperature special? Unlike concentration or pressure stresses, changing temperature actually changes the value of $K$ itself (via the van't Hoff equation). This is the only stress that modifies $K$.

Derivation from van 't Hoff:

$\frac{d\ln K}{dT} = \frac{\Delta H^\circ}{RT^2}$

For exothermic ($\Delta H^\circ < 0$): As $T$ increases, $\frac{d\ln K}{dT} < 0$, so $K$ decreases. The system must shift backward to match the new, smaller $K$.

For endothermic ($\Delta H^\circ > 0$): As $T$ increases, $K so the system shifts forward.

Stress: Increase temperature from $400°C$ to $500°C$.

Step 1 — Identify reaction type: $\Delta H < 0$ → exothermic → heat is a "product"

Why this step? We need to know which direction absorbs or releases heat.

Step 2 — Apply heat as product: Increasing $T$ is like "adding product" → shift backward (toward reactants)

Why? The system tries to consume the added heat by favoring the endothermic (reverse) direction.

Step 3 — Predict: $[N_2]$ and $[H_2]$ increase, $[NH_3]$ decreases. Also, $K$ decreases (verified by van 't Hoff).

Practical note: This is why Haber process runs at moderate temperatures despite wanting high yield—high $T$ speeds reaction but shifts equilibrium unfavorably.

Effect 4: Catalyst

WHY? A catalyst lowers $E_a$ for both directions by the same amount. Since $K = \frac{k_f}{k_r}$, and both rate constants increase by the same factor, their ratio (and thus $K$) is unchanged.

$K = \frac{k_f}{k_r} = \frac{A_f e^{-E_{a,f}/RT}}{A_r e^{-E_{a,r}/RT}} = \text{constant}$

Catalyst changes: $E_{a,f}$ and $E_{a,r}$ both decrease by $\Delta E_a$

New ratio: $\frac{A_f e^{-(E_{a,f}-\Delta E_a)/RT}}{A_r e^{-(E_{a,r}-\Delta E_a)/RT}} = \frac{A_f e^{-E_{a,f}/RT}}{A_r e^{-E_{a,r}/RT}} \cdot \frac{e^{\Delta E_a/RT}}{e^{\Delta E_a/RT}} = K$ (same!)

Stress: Add $MnO_2$ catalyst.

Step 1 — What happens to rates?

• Forward rate: increases (say, 1000×)
• Reverse rate: increases (same factor, 1000×)

Why this step? Catalyst provides alternate pathway with lower $E_a$.

Step 2 — What happens to equilibrium?

• Equilibrium position: unchanged
• Time to reach equilibrium: much faster

Why? $K = \frac{k_f}{k_r}$ stays constant when both change proportionally.

Practical: Catalysts don't increase yield, but they make reactions industrially viable by reducing time from years to seconds.

Summary Table of Stress Effects

Why it feels right: Catalysts do speed reactions, and we see product form faster.

Steel-man: The intuition that "faster forward = more product" would be correct if the reverse reaction didn't exist. In a one-way reaction, catalyst does increase product per unit time.

The fix: In an equilibrium, reverse reaction also speeds up equally. Both rates scale by the same factor, so the ratio (which determines equilibrium position) is unchanged. Catalyst changes kinetics (time), not thermodynamics (position).

Test it: If catalyst shifted equilibrium, we could extract infinite energy from equilibrium systems—violates thermodynamics!

Why it feels right: In many industrial reactions (Haber, contact process), high pressure does favor products. Students overgeneralize from these examples.

Steel-man: For reactions where products have fewer moles (like $N_2 + 3H_2 \rightleftharpoons 2NH_3$), this is actually correct. The mistake is assuming it's always true.

The fix: Pressure favors the side with fewer total moles of gas. For $N_2O_4 \rightleftharpoons 2NO_2$ (1 mole → 2 moles), pressure favors reactants, not products. Always count moles first.

Now, what if someone pushes down on your friend's side? The seesaw isn't balanced anymore. But here's the cool part: the seesaw naturally tries to balance itself again by having your side go down a bit and your friend's side come up.

Le Chatelier's principle is just like that seesaw. When you "push" on a chemical reaction (by adding more chemicals, squeezing it, or heating it up), the reaction automatically pushes back the other way to get balanced again.

• Add more reactant? Reaction makes more product to use it up.
• Heat it up? If the reaction normally gives off heat, it slows down to cool things down.
• Squeeze it? If the reaction makes fewer gas molecules, it shifts that way to relieve the squeeze.

The reaction is always trying to get back to balance, just like a seesaw!

Pressure mnemonic: "Fewer Favored when Flattened" (compression favors fewer moles)

Temperature mnemonic: "Endo Eats heat, Exo Expels heat" → adding heat shifts toward endothermic direction

Connections

• Chemical Equilibrium — Le Chatelier explains how equilibrium responds to disturbances
• Equilibrium Constant — $K$ is the target; $Q$ tells us which way to shift
• Reaction Quotient Q — comparing $Q$ vs $K$ predicts shift direction
• Haber Process — industrial application using pressure and temperature to maximize yield
• van 't Hoff Equation — quantifies how $K$ changes with temperature
• Activation Energy — why catalysts speed both directions equally
• Gibs Free Energy — $\Delta G = RT\ln(Q/K)$ shows why systems shift when $Q \neq K$

#flashcards/chemistry

What is Le Chatelier's principle? :: If a system at equilibrium is subjected to a stress (change in concentration, pressure, volume, or temperature), the equilibrium shifts to counteract that stress and partially restore equilibrium.

Concept Map

Hinglish (regional understanding)

Hinglish (regional understanding)