Reaction Quotient and Le Chatelier's Principle

Use Q to predict reaction direction and apply Le Chatelier's principle to predict equilibrium shifts.

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Reaction Quotient and Le Chatelier's Principle

Reaction Quotient (Q)

Q: Same form as K, but uses any concentrations (not just equilibrium)

For reaction: aA + bB ⇌ cC + dD

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

Difference from K:

K: only at equilibrium
Q: at any moment
Same expression, different values

Comparing Q and K

Predicting direction:

| Comparison | Direction | What happens | |------------|-----------|--------------| | Q < K | Forward → | Make more products | | Q = K | Equilibrium | No net change | | Q > K | Reverse ← | Make more reactants |

Memory aid:

Q < K: Need more products (go forward)
Q > K: Too many products (go backward)

Example:

For N₂ + 3H₂ ⇌ 2NH₃, K = 0.50

If [N₂] = 1.0, [H₂] = 1.0, [NH₃] = 0.50:

$Q = \frac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3} = \frac{(0.50)^2}{(1.0)(1.0)^3} = 0.25$

Q (0.25) < K (0.50) → Forward reaction proceeds

Le Chatelier's Principle

"When stress applied to equilibrium, system shifts to relieve stress"

Types of stress:

Concentration changes
Pressure/volume changes
Temperature changes

Concentration Changes

Add reactant:

Shift right (toward products)
Consume added reactant
Make more products

Add product:

Shift left (toward reactants)
Consume added product
Make more reactants

Remove reactant:

Shift left (toward reactants)
Replace removed reactant

Remove product:

Shift right (toward products)
Replace removed product

Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

| Change | Shift | Why | |--------|-------|-----| | Add N₂ | Right → | Consume excess N₂ | | Remove H₂ | Left ← | Replace removed H₂ | | Add NH₃ | Left ← | Consume excess NH₃ |

Note: Adding/removing solid or pure liquid has NO effect

Pressure/Volume Changes

For reactions with gases:

Decrease Volume (Increase Pressure):

Shifts toward side with fewer moles of gas
Relieves pressure by reducing gas molecules

Increase Volume (Decrease Pressure):

Shifts toward side with more moles of gas
Fills space with more gas molecules

Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Count moles gas:

Left: 1 + 3 = 4 moles
Right: 2 moles

Increase pressure: Shift right (fewer moles) Decrease pressure: Shift left (more moles)

Special case: Δn = 0 (equal moles)

Pressure change: no shift
Example: H₂ + I₂ ⇌ 2HI (2 moles each side)

Adding Inert Gas:

At constant volume: No shift (partial pressures unchanged) At constant pressure: Volume increases, shift to more moles

Temperature Changes

Temperature is different - actually changes K value!

Exothermic Reaction (ΔH < 0):

Think of heat as product:

A + B ⇌ C + D + heat

Increase T: Shift left (consume heat, favor reactants)

K decreases

Decrease T: Shift right (produce heat, favor products)

K increases

Endothermic Reaction (ΔH > 0):

Think of heat as reactant:

A + B + heat ⇌ C + D

Increase T: Shift right (use heat, favor products)

K increases

Decrease T: Shift left (less heat available, favor reactants)

K decreases

Summary:

| Reaction Type | Increase T | K value | |---------------|------------|---------| | Exothermic (ΔH < 0) | Shift left | Decreases | | Endothermic (ΔH > 0) | Shift right | Increases |

Catalyst Effect

Catalyst:

Speeds up both forward AND reverse reactions equally
NO shift in equilibrium position
NO change in K
Reaches equilibrium faster

Summary of Le Chatelier

Effect on equilibrium position:

| Stress | Equilibrium Shift | K Changes? | |--------|-------------------|------------| | Add/remove substance | Yes | No | | Change pressure/volume | Yes (if Δn ≠ 0) | No | | Change temperature | Yes | YES | | Add catalyst | No shift | No |

Only temperature changes K value!

📚 Practice Problems

1Problem 1easy

❓ Question:

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), K_c = 4.0 at 1000 K. At a certain time, [SO₂] = 0.40 M, [O₂] = 0.30 M, [SO₃] = 1.2 M. Calculate Q and predict the direction of reaction.

💡 Show Solution

Given:

Reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
K_c = 4.0
Current: [SO₂] = 0.40 M, [O₂] = 0.30 M, [SO₃] = 1.2 M

Calculate Q:

K/Q expression:

$Q = \frac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]}$

Substitute values:

$Q = \frac{(1.2)^2}{(0.40)^2(0.30)}$

$Q = \frac{1.44}{(0.16)(0.30)}$

$Q = \frac{1.44}{0.048}$

$Q = 30$

Answer: Q = 30

Compare Q and K:

Q = 30, K = 4.0

Q > K

Meaning:

Too many products (SO₃) relative to equilibrium
System will shift LEFT (reverse)
Consume SO₃, make more SO₂ and O₂

Direction: ← Reverse (toward reactants)

Summary:

| Value | Amount | |-------|--------| | Q | 30 | | K | 4.0 | | Comparison | Q > K | | Direction | Reverse (←) |

Until Q = K, reaction proceeds in reverse direction.

2Problem 2medium

❓ Question:

For the equilibrium: N₂O₄(g) ⇌ 2NO₂(g), ΔH° = +57.2 kJ. Predict how each change affects the equilibrium: (a) Add N₂O₄, (b) Decrease volume, (c) Increase temperature, (d) Add catalyst.

💡 Show Solution

Reaction: N₂O₄(g) ⇌ 2NO₂(g) ΔH° = +57.2 kJ (endothermic)

(a) Add N₂O₄

Type: Concentration change

Effect:

Adding reactant
Shifts RIGHT (toward products)
System consumes excess N₂O₄
Makes more NO₂

Answer: Shifts right, increases [NO₂]

(b) Decrease volume

Type: Pressure/volume change

Count moles gas:

Left: 1 mole N₂O₄
Right: 2 moles NO₂
Δn = 2 - 1 = +1

Decrease volume = increase pressure

Effect:

Shifts toward fewer moles
Shifts LEFT (toward N₂O₄)
Relieves pressure by reducing total moles

Answer: Shifts left, decreases [NO₂]

(c) Increase temperature

Type: Temperature change (changes K!)

Reaction is endothermic: Heat is reactant

N₂O₄(g) + heat ⇌ 2NO₂(g)

Increase T:

More heat available
Shifts RIGHT (toward products)
Uses added heat
K increases

Answer: Shifts right, increases [NO₂], K increases

(d) Add catalyst

Type: Catalyst

Effect:

Speeds up both forward and reverse equally
NO shift in position
NO change in K
Reaches equilibrium faster only

Answer: No shift, no change in concentrations or K

Summary:

| Change | Direction | [NO₂] | K Changes? | |--------|-----------|-------|------------| | (a) Add N₂O₄ | Right → | Increases | No | | (b) Decrease volume | Left ← | Decreases | No | | (c) Increase T | Right → | Increases | Yes (↑) | | (d) Add catalyst | No shift | No change | No |

3Problem 3hard

❓ Question:

For the exothermic reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH° = -198 kJ. To maximize SO₃ production, should you use: (a) high or low temperature? (b) high or low pressure? (c) Explain any trade-offs.

💡 Show Solution

Reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH° = -198 kJ (exothermic)

Goal: Maximize SO₃ production

(a) Temperature choice

Reaction is exothermic: Heat is product

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) + heat

Low temperature:

Shifts RIGHT (toward products)
Produces heat (relieves cold stress)
Favors SO₃ formation
K is larger at low T

High temperature:

Shifts LEFT (toward reactants)
Consumes heat
Favors SO₂ and O₂
K is smaller at high T

Answer (a): Low temperature favors SO₃

(b) Pressure choice

Count moles gas:

Left: 2 + 1 = 3 moles
Right: 2 moles
Δn = 2 - 3 = -1

High pressure:

Shifts toward fewer moles
Shifts RIGHT (toward SO₃)
Relieves pressure
Favors products

Low pressure:

Shifts toward more moles
Shifts LEFT
Favors reactants

Answer (b): High pressure favors SO₃

(c) Trade-offs

Temperature trade-off:

Low T (thermodynamics):

✓ Better yield (more SO₃)
✓ Higher K
✗ Reaction VERY slow
✗ Takes too long industrially

High T (kinetics):

✗ Lower yield (less SO₃)
✗ Lower K
✓ Much faster reaction
✓ Reaches equilibrium quickly

Industrial compromise:

Use moderate temperature (~450°C)
Acceptable yield
Reasonable rate
Use catalyst (V₂O₅) to speed up
Multiple passes through reactor

Pressure:

High pressure favored (1-2 atm)
No major trade-off
Equipment cost for very high pressure

Summary - Contact Process for SO₃:

| Factor | Ideal (equilibrium) | Industrial choice | Reason | |--------|---------------------|-------------------|---------| | Temperature | Low | Moderate (450°C) | Balance yield vs rate | | Pressure | High | Moderate (1-2 atm) | Good yield, low cost | | Catalyst | Yes | V₂O₅ | Speeds reaction at moderate T |

Key lesson: Industrial processes balance thermodynamics (K) and kinetics (rate)!

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