This MCQ module is based on: Le Chatelier Shifts
TOPIC 26 OF 28
Le Chatelier Shifts
🎓 Class 11
Chemistry
CBSE
Theory
Ch 6 – Equilibrium
⏱ ~14 min
🌐 Language: [gtranslate]
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Le Chatelier's Principle, Haber Process and Contact Process
6.10 Le Chatelier's Principle
What happens to a system at equilibrium when we disturb it? The French chemist Henry Louis Le Chatelier answered this in 1884:
Le Chatelier's Principle: When a system at equilibrium is subjected to a change in concentration, pressure, volume, or temperature, the equilibrium shifts in the direction that partially counteracts the imposed change.
6.11 Factors Affecting Equilibrium
6.11.1 Effect of Concentration
If you ADD a reactant or remove a product → equilibrium shifts FORWARD (→) to make more product.
If you ADD a product or remove a reactant → equilibrium shifts REVERSE (←) to make more reactant.
For Fe³⁺(aq) + SCN⁻(aq) ⇌ [FeSCN]²⁺(aq) (red):
| Action | Shift | Observation |
|---|---|---|
| Add FeCl₃ | → | Red colour deepens |
| Add KSCN | → | Red colour deepens |
| Add NaOH (removes Fe³⁺ as Fe(OH)₃) | ← | Red colour fades |
| Add Na₂HPO₄ (removes Fe³⁺) | ← | Red colour fades |
6.11.2 Effect of Pressure / Volume
For gas-phase reactions, an INCREASE in pressure (decrease in volume) shifts equilibrium toward the side with FEWER moles of gas.
| Reaction | Δn_g | ↑P shifts |
|---|---|---|
| N₂(g) + 3H₂(g) ⇌ 2NH₃(g) | −2 | → (toward NH₃) |
| 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) | −1 | → (toward SO₃) |
| PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) | +1 | ← (toward PCl₅) |
| H₂(g) + I₂(g) ⇌ 2HI(g) | 0 | No shift |
6.11.3 Effect of Temperature
This is the only factor that changes the value of K itself.
| Reaction type | ↑T shifts | K vs T |
|---|---|---|
| Exothermic (ΔH < 0) | ← (reverse) | K decreases with T |
| Endothermic (ΔH > 0) | → (forward) | K increases with T |
Quantitative relation (Van't Hoff equation, Class 12):
\[\ln\frac{K_2}{K_1} = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\]
For an exothermic reaction (ΔH < 0), as T rises, K falls. Heat is treated as a "product".
6.11.4 Effect of a Catalyst
A catalyst does NOT shift the position of equilibrium. It speeds up both forward and reverse reactions equally, so equilibrium is reached faster but at the same composition. K remains unchanged.
6.11.5 Effect of Inert Gas
Adding an inert gas (like Ar) at constant volume: no effect (partial pressures of reacting gases unchanged).
Adding inert gas at constant pressure: total volume increases, partial pressures of reactants/products decrease — shift toward more moles of gas (just like decreasing P).
6.12 Industrial Applications
6.12.1 Haber Process — Synthesis of Ammonia
\[\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \qquad \Delta H = -92.4\,\text{kJ/mol}\]Le Chatelier predicts: maximum yield needs high P, low T. But low T means the rate is too slow to be commercially useful! Compromise:
| Condition | Industrial value | Why? |
|---|---|---|
| Pressure | 200–250 atm | Higher P shifts → (Δn = −2) but ultra-high P expensive |
| Temperature | ~700 K (425–450 °C) | Lower T favours product, but rate would be too slow; compromise temperature |
| Catalyst | Iron with K₂O & Al₂O₃ promoters | Speeds both forward and reverse equally — equilibrium reached fast |
| Removal | NH₃ liquefied & removed | Le Chatelier: continuously removing product drives → forward |
Historical impact: Fritz Haber developed the process (1909) and won the 1918 Nobel Prize. Carl Bosch scaled it up industrially. Estimates: today, the Haber–Bosch process feeds nearly half of humanity by producing nitrogen fertilisers.
6.12.2 Contact Process — Synthesis of Sulfuric Acid
Key step: oxidation of SO₂ to SO₃.
\[2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \qquad \Delta H = -198\,\text{kJ/mol}\]| Condition | Value | Reason |
|---|---|---|
| Pressure | 1–2 atm | Δn = −1 favours SO₃ at high P, but yield is already ~98% at 1 atm — no need for high P |
| Temperature | ~720 K (450 °C) | Compromise; below 720 K rate is too slow with V₂O₅ |
| Catalyst | V₂O₅ (vanadium pentoxide) | Used since 1930s; cheaper and less easily poisoned than Pt |
| Excess O₂ | 1:1 SO₂:O₂ → 1:1.5 | Le Chatelier: extra reactant pushes → |
Subsequent steps: SO₃ is absorbed in 98% H₂SO₄ to give oleum (H₂S₂O₇), which is then diluted with water to give H₂SO₄.
🎯 Interactive: Le Chatelier Shift Predictor
Pick a stress applied to N₂ + 3H₂ ⇌ 2NH₃ + heat. Predict the shift.
Shift: FORWARD (→)
Adding N₂ raises [N₂]; system consumes some N₂ to produce more NH₃.
K remains unchanged (only T affects K).
🧪 Activity 6.3 — NO₂ ⇌ N₂O₄ Tube Demonstration
Setup: A sealed glass tube contains brown NO₂ in equilibrium with colourless N₂O₄: 2NO₂(g) ⇌ N₂O₄(g); ΔH = −58 kJ/mol. Two identical tubes are placed in (a) ice-water bath, (b) hot water bath.
Predict: What colour change will you observe in each tube? Explain using Le Chatelier's principle.
Cold tube: Equilibrium shifts to the LEFT? No — wait! The forward reaction (2NO₂ → N₂O₄) is exothermic. Lowering T removes "heat product" → equilibrium shifts FORWARD (toward N₂O₄). N₂O₄ is colourless. Tube becomes lighter / pale.
Hot tube: Adding heat (raising T) shifts equilibrium REVERSE (toward NO₂). NO₂ is dark brown. Tube becomes darker / deep brown.
K_eq: falls as T rises (exothermic). At 273 K K ≈ 13; at 350 K K ≈ 0.6.
Worked Example 6.6: Predicting Shifts
For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) (ΔH = +93 kJ), predict the effect on equilibrium of (a) increasing T (b) doubling V (c) adding Cl₂ (d) adding catalyst.
(a) ↑T: endothermic forward → shift forward (→); more dissociation; K increases.
(b) ↑V (↓P): Δn_g = +1 → shift to side with more moles → forward (→); more dissociation.
(c) Add Cl₂ (product): shift reverse (←); less dissociation.
(d) Catalyst: no shift; equilibrium reached faster.
(b) ↑V (↓P): Δn_g = +1 → shift to side with more moles → forward (→); more dissociation.
(c) Add Cl₂ (product): shift reverse (←); less dissociation.
(d) Catalyst: no shift; equilibrium reached faster.
Worked Example 6.7: Optimum Conditions
For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g); ΔH = −198 kJ. What conditions of T and P maximize the yield of SO₃?
For maximum yield (Le Chatelier alone): low T, high P, excess O₂.
Reality: low T = slow rate. Industrially, ~720 K with V₂O₅ catalyst is the compromise. Δn = −1, but yield is already ~98% at modest pressure (~1 atm). High pressure is unnecessary.
Reality: low T = slow rate. Industrially, ~720 K with V₂O₅ catalyst is the compromise. Δn = −1, but yield is already ~98% at modest pressure (~1 atm). High pressure is unnecessary.
🎯 Competency-Based Questions
Q1. A catalyst alters: L1 Remember
Answer: (c). A catalyst speeds up both directions equally, so equilibrium is reached faster but the equilibrium composition (and K) are unchanged.
Q2. For an exothermic reaction, K decreases on raising T. Why? L2 Understand
Answer: Heat is a "product" of an exothermic reaction. Adding heat (raising T) drives the equilibrium backward (Le Chatelier) so [products]/[reactants] decreases; hence K decreases. Quantitatively, the Van't Hoff equation gives ln(K₂/K₁) = (−ΔH/R)(1/T₂ − 1/T₁); ΔH < 0 → K falls as T rises.
Q3. For 2NO₂(g) ⇌ N₂O₄(g), if pressure is doubled at constant T, in which direction will the equilibrium shift? L3 Apply
Answer: Δn_g = 1 − 2 = −1. Higher P favours fewer moles → shift forward (→), more N₂O₄ formed. Total mol decrease, so observed pressure rises by less than the factor of 2.
Q4. Why does adding inert gas (Ar) at constant volume not shift the equilibrium of N₂ + 3H₂ ⇌ 2NH₃? L4 Analyse
Answer: At constant V, the partial pressures (and concentrations) of N₂, H₂, NH₃ are unchanged. Total P rises, but Q = (p_NH₃)² / (p_N₂ × p_H₂³) is built only from active species. Q remains equal to K → no shift. (At constant P, however, adding Ar increases total V; partial pressures of reacting gases drop → shift toward more moles, i.e., reverse for Haber.)
Q5. HOT (Evaluate): Why does the Haber process not use 1000 atm pressure even though Le Chatelier suggests it would maximize NH₃ yield? L5 Evaluate
Answer: Several practical constraints: (i) very-high-pressure reactor walls become extremely thick and expensive; (ii) energy cost of compression rises sharply; (iii) NH₃ yield improves only marginally beyond ~250 atm (diminishing returns); (iv) safety risk of leakage. The compromise of ~200–250 atm balances yield, capital cost, energy use, and safety. Engineering optimisation always involves trade-offs that a thermodynamic prediction alone misses.
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: A catalyst doesn't shift the position of equilibrium.
R: A catalyst speeds up the forward and reverse reactions to the same extent.
Answer: (A). Both true; R explains A. Catalyst lowers activation energy of both directions equally.
A: Increasing pressure shifts the equilibrium of N₂ + 3H₂ ⇌ 2NH₃ to the right.
R: The product side has fewer moles of gas.
Answer: (A). Both true; R explains A. Le Chatelier shifts toward the side with fewer moles when P is increased.
A: Increasing temperature increases K for an exothermic reaction.
R: Higher T provides more energy for forward bond-making.
Answer: (D). A is FALSE — for an exothermic reaction, K DECREASES with rising T. R is also flawed (heat is treated as a product). Strictly, both are false; mark (D) — A is the wrong claim.
Frequently Asked Questions — Le Chatelier's Principle, Haber Process and Contact Process
What is Le Chatelier's principle?
Le Chatelier's principle (1884) states that if a system at equilibrium is subjected to a change in concentration, pressure, volume or temperature, the equilibrium shifts in the direction that partially counteracts the change, re-establishing equilibrium at a new position. NCERT Class 11 Chemistry Chapter 6 applies this principle to predict the direction of equilibrium shifts in industrially important reactions like the Haber process (NH₃ synthesis) and Contact process (H₂SO₄ manufacture). The principle does not change K but changes the equilibrium concentrations within the constraint of K being constant at a given temperature.
How does concentration affect equilibrium?
Adding more reactant shifts equilibrium to the right (forward), forming more products to consume the added reactant. Adding more product shifts equilibrium to the left (reverse), forming more reactants. Removing reactant shifts equilibrium to the left; removing product shifts it to the right. The equilibrium constant K remains unchanged because temperature is constant. In NCERT Class 11 Chemistry Chapter 6, this principle is applied to drive reactions forward by continuously removing one product (e.g., distilling off water in esterification) or by adding excess of a cheap reactant in industrial processes.
How does pressure affect equilibrium of gaseous reactions?
For a gaseous equilibrium, increasing pressure (or decreasing volume) shifts the equilibrium toward the side with fewer moles of gas to reduce pressure. Decreasing pressure shifts it toward the side with more moles. For reactions with equal moles of gas on both sides (e.g., H₂ + I₂ ⇌ 2HI), pressure has no effect. NCERT Class 11 Chemistry Chapter 6 applies this in the Haber process: N₂ + 3H₂ ⇌ 2NH₃ has 4 moles → 2 moles gas, so high pressure (200 atm) favours NH₃ formation. Temperature is kept moderate (450°C) to balance K and rate.
How does temperature affect equilibrium?
Increasing temperature shifts an exothermic equilibrium (ΔH < 0) backward (reactants favoured) and shifts an endothermic equilibrium (ΔH > 0) forward (products favoured). Decreasing temperature does the opposite. Unlike concentration and pressure changes, temperature changes also change K itself — K increases for endothermic and decreases for exothermic reactions when temperature rises. The van't Hoff equation quantifies this: d(ln K)/dT = ΔH°/RT². NCERT Class 11 Chemistry Chapter 6 uses temperature effects to optimise industrial reactions and to predict spontaneity at different temperatures.
What is the Haber process and its industrial conditions?
The Haber process synthesises ammonia from atmospheric nitrogen and hydrogen: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = −92.4 kJ/mol. Optimal conditions (NCERT Class 11 Chemistry Chapter 6): pressure 200 atm (high pressure favours fewer moles → NH₃), temperature 450°C (compromise: low temperature shifts equilibrium toward NH₃ but reduces rate; 450°C is a kinetics-thermodynamics balance), and an iron catalyst with K₂O and Al₂O₃ promoters to speed up equilibrium attainment. NH₃ produced is liquefied and removed, driving the equilibrium further forward. This process produces fertilisers and is essential for global food security.
What is the Contact process and its conditions?
The Contact process manufactures sulfuric acid in three stages: (1) S + O₂ → SO₂; (2) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −197 kJ/mol; (3) SO₃ + H₂O → H₂SO₄ (absorbed in conc. H₂SO₄ as oleum, then diluted). The critical equilibrium step is the second one. Optimal conditions in NCERT Class 11 Chemistry Chapter 6: pressure 1–2 atm (moderate; high pressure not very effective since K is large), temperature 450°C (compromise as in Haber), V₂O₅ catalyst, and excess air to favour SO₃ formation. Conversion exceeds 95% under these conditions.
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