Le Chatelier's Principle: Shifting Equilibrium

For $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) + 92 \text{ kJ}$:

Predict the effect of:

1. Adding N₂
2. Removing NH₃
3. Increasing pressure
4. Increasing temperature

Solution:

1. Add N₂ → Shifts right (forward) to consume added N₂
2. Remove NH₃ → Shifts right (forward) to replace removed NH₃
3. Increase pressure → Shifts right (4 moles → 2 moles, favors fewer moles)
4. Increase temperature → Shifts left (backward, exothermic so heat is product; adding heat shifts away from it)

Which reaction’s equilibrium is NOT affected by pressure change?

A) $2SO_2 + O_2 \rightleftharpoons 2SO_3$ B) $N_2 + O_2 \rightleftharpoons 2NO$ C) $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ D) $N_2 + 3H_2 \rightleftharpoons 2NH_3$

Solution: Calculate Δn for each:

A) Δn = 2 - 3 = -1 ✗ B) Δn = 2 - 2 = 0 ✓ C) Δn = 2 - 1 = +1 ✗ D) Δn = 2 - 4 = -2 ✗

Answer: B ($N_2 + O_2 \rightleftharpoons 2NO$) - Equal moles on both sides

For $2NO_2(g) \rightleftharpoons N_2O_4(g)$, ΔH = -58 kJ/mol (NO₂ is brown, N₂O₄ is colorless)

What happens when the mixture is cooled?

Solution: Reaction is exothermic (ΔH negative, heat is product):

Cooling = removing heat → Shifts right (forward) to replace removed heat

Result: More N₂O₄ forms → Color becomes lighter/less brown

For the equilibrium: $CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)$, ΔH = -92 kJ

To maximize methanol production, which conditions are best?

A) High temperature, high pressure B) Low temperature, high pressure C) High temperature, low pressure D) Low temperature, low pressure

Solution: For maximum CH₃OH:

1. Pressure consideration:

   • Δn = 1 - 3 = -2 (reactants have more moles)
   • High pressure favors products → High pressure ✓

2. Temperature consideration:

   • Reaction is exothermic (ΔH negative)
   • Low temperature favors products → Low temperature ✓

Answer: B) Low temperature, high pressure

(This is why industrial methanol synthesis uses ~300°C and 50-100 atm - a compromise between rate and yield)

For $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$:

Helium gas is added at:

1. Constant volume
2. Constant pressure

Predict the effect on equilibrium in each case.

Solution:

Case 1: Constant volume

• Partial pressures of PCl₅, PCl₃, Cl₂ remain unchanged
• Only total pressure increases
• No effect on equilibrium position

Case 2: Constant pressure

• Volume must increase to maintain constant total pressure
• All partial pressures decrease
• Equivalent to decreasing pressure
• Δn = 2 - 1 = +1 (more moles on right)
• Shifts right (toward more moles)
• Degree of dissociation increases

For the reaction: $A(g) + 3B(g) \rightleftharpoons 2C(g)$, ΔH < 0

Initially at equilibrium at pressure P and temperature T. The pressure is doubled and temperature is increased. Which statement is correct?

A) Equilibrium shifts right, K increases B) Equilibrium shifts left, K decreases C) Equilibrium shifts right, K decreases D) Equilibrium shifts left, K increases

Solution:

Effect of doubling pressure:

• Δn = 2 - 4 = -2 (reactants have more moles)
• High pressure favors products
• Shifts right ⟶

Effect of increasing temperature:

• Reaction is exothermic (ΔH < 0)
• High temperature favors reactants
• Shifts left ⟵
• K decreases (exothermic: K decreases with temperature increase)

Net effect: Temperature dominates for K value

Answer: C or B depending on which effect dominates for position, but K definitely decreases

If asked only about K: K decreases (temperature effect)

At 700 K, the equilibrium constant for $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ is 54.

Initially, 1 mol H₂, 1 mol I₂, and 1 mol HI are placed in a 1L flask. After equilibrium:

1. Is the system initially at equilibrium?
2. If not, in which direction will it shift?
3. Calculate final concentrations.

Solution:

Step 1: Check if at equilibrium

Initial concentrations: [H₂] = 1M, [I₂] = 1M, [HI] = 1M

$Q = 1 < K = 54$ → Not at equilibrium, will shift right (forward)

Step 2: ICE Table

Step 3: Equilibrium expression

Taking square root both sides:

Final concentrations:

• [H₂] = 1 - 0.68 = 0.32M
• [I₂] = 1 - 0.68 = 0.32M
• [HI] = 1 + 2(0.68) = 2.36M

The Contact Process for sulfuric acid:

ΔH = -197 kJ/mol

Industrial conditions: 450°C, 1-2 atm pressure, V₂O₅ catalyst

Questions:

1. Why not use higher pressure for better yield?
2. Why not use lower temperature for better yield?
3. What is the role of V₂O₅?

Solution:

1. Why low pressure (1-2 atm)?

• Δn = 2 - 3 = -1 (high pressure would favor products)
• BUT: Even at 1 atm, conversion is ~98%
• Higher pressure requires stronger (expensive) equipment
• Cost-benefit analysis: Not worth the extra expense for 2% more yield

2. Why not very low temperature?

• Low T would give better yield (exothermic reaction)
• BUT: Rate would be too slow at low T
• At 300°C: Great yield but takes days!
• At 450°C: Slightly lower yield (~98%) but much faster
• Compromise: 450°C balances rate and yield

3. Role of V₂O₅ catalyst:

• Does NOT change equilibrium position or yield
• Makes reaction reach equilibrium faster
• Allows use of lower temperature (better yield) while maintaining reasonable rate
• Economic benefit: Time is money in industry!

Key Insight: Industrial chemistry is about economic optimization, not just maximum yield. Le Chatelier helps predict, but economics decides!

Oxygen transport in blood involves equilibrium:

(Hb = hemoglobin)

At high altitude:

• Lower O₂ partial pressure
• Body produces more red blood cells

Explain using Le Chatelier’s Principle.

Solution:

At sea level:

• High O₂ partial pressure
• Equilibrium lies toward right (Hb(O₂)₄ forms easily)
• Oxygen binds efficiently to hemoglobin

At high altitude:

• Low O₂ partial pressure (less O₂ available)
• Equilibrium shifts left (less Hb(O₂)₄ formation)
• Less oxygen carried per hemoglobin molecule

Body’s response:

• Produce more RBCs (more hemoglobin total)
• Compensates for lower efficiency per Hb
• Restores oxygen-carrying capacity

Le Chatelier application: Can’t increase O₂ pressure, so increase Hb concentration instead!

JEE Connection: Carbon monoxide poisoning works similarly—CO binds Hb stronger than O₂, shifting equilibrium away from oxygen binding.