If a system at equilibrium is disturbed (concentration, pressure, temperature, catalyst), the system shifts to reduce the disturbance. (1) Adding reactant: forward shift. (2) Increase pressure: shift to fewer moles of gas. (3) Increase T: shift in endothermic direction.
-- NCERT Class 11 Chemistry, Ch. 6, p. 185Le Chatelier
8 MCQs9-step worked example
Source: NCERT EquilibriumPYQ coverage: NEET 2025Official key: NTA-verifiedLast updated: 24 Sep 2026
Le Chatelier, explained for NEET
Here is the trap that costs marks on Le Chatelier questions: students select "catalyst shifts equilibrium toward products" as a correct response. It is wrong. A catalyst lowers the activation energy of both the forward and reverse reactions equally. The system reaches equilibrium faster, but the equilibrium position does not change.
Le Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the equilibrium shifts in the direction that tends to counteract the imposed change (NCERT Class 11 Chemistry Chapter 6, page 181).
Concentration changes. Adding more reactant shifts equilibrium toward products. Removing a product also shifts toward products. The system consumes the excess species or compensates for the removed one.
Pressure changes (gaseous equilibria only). Increasing pressure shifts equilibrium toward the side with fewer moles of gas. Decreasing pressure favours the side with more moles. If both sides have the same total moles of gas, pressure change has no effect on equilibrium position. A common confusion: predicting the wrong direction because the student counts total moles rather than gas-phase moles only.
Temperature changes. For an exothermic forward reaction (ΔH < 0), increasing temperature shifts equilibrium toward reactants (the endothermic direction). For an endothermic forward reaction (ΔH > 0), increasing temperature shifts toward products. Temperature is the only factor that changes the value of the equilibrium constant K.
Inert gas addition. Adding an inert gas at constant volume does not change the partial pressures of reactants or products — no shift. At constant pressure, adding inert gas increases volume, effectively decreasing partial pressures, so the equilibrium shifts toward more moles of gas.
Catalyst. No shift. Faster attainment of the same equilibrium. This is a high-frequency NEET distractor.
Can you answer these Le Chatelier MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
According to Le Chatelier's principle, which change shifts a gaseous equilibrium toward the side with fewer moles of gas?
Show answer and why every option is right or wrong
Answer: A. Increasing pressure favours the side with fewer moles of gas, directly from the statement of Le Chatelier's principle (NCERT Class 11 Chemistry Chapter 6, page 181).
Why B is wrong: B: Decreasing pressure shifts equilibrium toward MORE moles of gas, not fewer — the system tries to restore pressure by producing more gas molecules.
Why C is wrong: C: A catalyst does not shift equilibrium in any direction; it only speeds up attainment of equilibrium (trap: catalyst-no-shift confusion).
Why D is wrong: D: Adding an inert gas at constant volume does not change partial pressures of reactants or products, so no shift occurs.
MCQ 2Easy RecallPractice
Which of the following factors changes the numerical value of the equilibrium constant K?
Show answer and why every option is right or wrong
Answer: C. Temperature is the only factor that changes the value of K. Concentration, pressure, and catalyst changes shift equilibrium position (or not, in the case of catalyst) but do not alter K (NCERT Class 11 Chemistry Chapter 6, page 181).
Why A is wrong: A: A catalyst speeds attainment of equilibrium but does not change K (trap: catalyst-no-shift).
Why B is wrong: B: Adding reactant shifts the equilibrium position but Q adjusts back to the same K; the constant itself is unchanged.
Why D is wrong: D: Pressure changes shift the equilibrium position for gaseous reactions with unequal moles, but K remains the same at a given temperature.
MCQ 3Easy RecallPractice
For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ/mol, increasing the temperature will:
Show answer and why every option is right or wrong
Answer: B. The forward reaction is exothermic. Increasing temperature favours the endothermic (reverse) direction, shifting equilibrium toward reactants N₂ and H₂. K decreases (NCERT Class 11 Chemistry Chapter 6, page 181).
Why A is wrong: A: This would be correct only if the forward reaction were endothermic. Since ΔH is negative (exothermic forward), heat increase drives the reverse reaction.
Why C is wrong: C: Temperature always affects equilibrium position — it is the one factor that changes K. Confusing temperature with catalyst addition leads to this error.
Why D is wrong: D: For an exothermic reaction, increasing temperature decreases K, not increases it. K increases with temperature only for endothermic forward reactions.
MCQ 4Direct ApplicationPractice
For the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), what happens to the equilibrium position when the total pressure is increased at constant temperature?
Show answer and why every option is right or wrong
Answer: D. Reactant side has 3 moles of gas (2 + 1); product side has 2 moles. Increasing pressure shifts equilibrium toward the side with fewer gas moles — toward SO₃ (NCERT Class 11 Chemistry Chapter 6, page 181).
Why A is wrong: A: This would occur if pressure were decreased. Increasing pressure favours fewer gas moles (product side here). Mistaking the direction is a common error with pressure-shift problems (mistake: predicting wrong direction on pressure change).
Why B is wrong: B: K is changed only by temperature, not by pressure. Confusing equilibrium shift with K change is a conceptual error.
Why C is wrong: C: The two sides have different total gas moles (3 vs 2), so pressure change does shift equilibrium. No-change applies only when moles are equal on both sides.
MCQ 5Direct ApplicationPractice
An inert gas is added to the equilibrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) at constant pressure. The equilibrium will:
Show answer and why every option is right or wrong
Answer: B. At constant pressure, adding inert gas increases total volume, reducing partial pressures of all species. The system shifts toward the side with more moles of gas to restore pressure. Products have 2 moles of gas vs 1 mole for reactant, so equilibrium shifts toward PCl₃ + Cl₂ (NCERT Class 11 Chemistry Chapter 6, page 181).
Why A is wrong: A: Shift toward fewer moles of gas occurs when pressure increases, not decreases. Here, partial pressures drop (volume increases at constant total P), so the system shifts toward more moles.
Why C is wrong: C: While the inert gas does not react chemically, at constant pressure it increases volume and lowers partial pressures, causing a shift. This no-shift reasoning applies only at constant volume.
Why D is wrong: D: The identity of the inert gas is irrelevant — what matters is the volume/pressure change. All ideal inert gases produce the same effect.
MCQ 6Direct ApplicationPractice
A catalyst is added to the equilibrium N₂O₄(g) ⇌ 2NO₂(g). Which statement is correct?
Show answer and why every option is right or wrong
Answer: A. A catalyst lowers the activation energy of both forward and reverse reactions equally. Equilibrium is attained faster, but the position (and K) remain unchanged (NCERT Class 11 Chemistry Chapter 6, page 181).
Why B is wrong: B: Same reasoning — a catalyst does not favour the reverse reaction either. Both directions speed up equally (mistake: catalyst shifts equilibrium).
Why C is wrong: C: A catalyst does not favour the forward reaction over the reverse. Believing a catalyst shifts equilibrium toward products is a high-frequency NEET trap (trap: catalyst-no-shift).
Why D is wrong: D: K depends only on temperature. A catalyst does not change temperature and therefore does not change K.
MCQ 7Concept TrapPractice
For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), if the total pressure is doubled at constant temperature, the equilibrium position will:
Show answer and why every option is right or wrong
Answer: C. Both sides have the same total moles of gas (2 moles on each side: 1 + 1 = 2 reactant, 2 product). When moles of gas are equal, pressure change does not shift the equilibrium position (NCERT Class 11 Chemistry Chapter 6, page 181).
Why A is wrong: A: This would apply only if the product side had fewer moles of gas. Here, both sides have 2 moles of gas, so no directional shift occurs (mistake: predicting wrong direction on pressure change).
Why B is wrong: B: Same error as option A but in the opposite direction. Equal gas moles on both sides means pressure has no effect on equilibrium position.
Why D is wrong: D: Temperature affects K and equilibrium position, but the question asks about pressure change. Whether the reaction is exothermic or endothermic is irrelevant to the pressure-shift analysis.
MCQ 8Concept TrapPractice
Consider the endothermic reaction: CaCO₃(s) ⇌ CaO(s) + CO₂(g). Which combination of changes will shift equilibrium most toward products?
Show answer and why every option is right or wrong
Answer: C. The forward reaction is endothermic — increasing temperature favours products. CO₂ is the only gas (1 mole product side, 0 on reactant side since CaCO₃ and CaO are solids). Decreasing pressure shifts toward more gas moles — toward products. Both changes reinforce the forward shift (NCERT Class 11 Chemistry Chapter 6, page 181).
Why A is wrong: A: Increasing temperature favours products (correct part), but increasing pressure shifts equilibrium toward fewer gas moles — toward reactant side (CaCO₃). The two effects oppose each other.
Why B is wrong: B: Decreasing temperature shifts equilibrium toward the exothermic (reverse) direction — toward CaCO₃. While decreasing pressure favours products, the temperature effect works against it. This combination does not maximally favour products.
Why D is wrong: D: Both changes work against product formation. Decreasing temperature favours the exothermic reverse reaction, and increasing pressure favours fewer gas moles (reactant side).
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How do you solve a Le Chatelier question? A worked example
Pattern: P.CHE.U06.LE_CHATELIER_SHIFT — Predict equilibrium shift on changing concentration, pressure, temperature, or adding catalyst (observed in NEET 2021, 2023, 2024).
- 1
Given
• Reaction: 2NO₂(g) ⇌ N₂O₄(g)• ΔH = −57 kJ/mol (exothermic forward)• Initial equilibrium at 300 K• Change (i): T increased to 350 K• Change (ii): total pressure doubled
- 2
Required
Net direction of equilibrium shift (toward products, toward reactants, or no net shift).
- 3
Concept
Le Chatelier's principle applied independently to each perturbation, then assess whether effects reinforce or oppose.
- 4
Formula
No quantitative formula needed — this is a qualitative application of Le Chatelier's principle. The relevant analysis is:• Temperature effect: endothermic direction favoured when T increases.• Pressure effect: side with fewer moles of gas favoured when P increases.
- 5
Substitution / Analysis
Temperature increase (300 K → 350 K):
Forward reaction is exothermic (ΔH < 0). Increasing temperature favours the reverse (endothermic) direction → shifts toward NO₂ (reactants).
Pressure doubled:
Reactant side: 2 moles of gas (2NO₂). Product side: 1 mole of gas (N₂O₄). Increasing pressure favours fewer gas moles → shifts toward N₂O₄ (products). - 6
Evaluation
The two effects oppose each other:• Temperature increase → shifts LEFT (toward 2NO₂)• Pressure increase → shifts RIGHT (toward N₂O₄)
Without numerical values of K at both temperatures and the exact pressure change, we cannot determine which effect dominates quantitatively. However, for a NEET-level qualitative answer: the two effects partially cancel. The question asks for the "net direction" — the answer is that the effects oppose, and the net shift depends on the magnitudes. - 7
Final answer
Temperature increase favours the reverse reaction (toward NO₂). Pressure increase favours the forward reaction (toward N₂O₄). The two effects oppose each other. Without quantitative data, the net shift cannot be definitively determined — a NEET question structured this way would typically ask about each factor separately or specify which dominates.
- 8
Common trap
Selecting "catalyst shifts equilibrium" when catalyst is listed among the options. Also: forgetting that pressure change has no effect when gas moles are equal on both sides (not this reaction, but a common paired question).
- 9
Similar NEET-style question
For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ/mol, predict the effect on equilibrium when (a) temperature is decreased, and (b) an inert gas is added at constant volume.
Answer: (a) Decrease in T favours exothermic forward reaction → shifts toward NH₃. (b) Inert gas at constant volume does not change partial pressures → no shift.
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What to remember before solving Le Chatelier questions
Where do students lose marks on Le Chatelier?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Overthinking
Student claims catalyst shifts equilibrium toward products. Catalyst speeds up forward AND reverse equally; equilibrium position unchanged.
When it triggers
Question lists catalyst addition among options for shifting equilibrium.
How to avoid
Catalyst lowers activation energy of BOTH forward and reverse reactions equally. Time to reach equilibrium decreases; equilibrium position is unchanged.
Root cause: concept gap
Correction
Catalyst speeds up forward AND reverse equally. Equilibrium position is unchanged. Time to reach equilibrium decreases.
Root cause: concept gap
Correction
Increase P → shift toward FEWER moles of gas. Decrease P → shift toward MORE moles. Reactions with same number of moles of gas: P doesn't shift equilibrium.
More in Equilibrium: 5 exam traps and mistakes · 5 formulas · 3 question patterns from its other lessons.
Le Chatelier questions from past NEET papers
1 question from NEET 2025. Answers verified against NTA official keys.
NEET 2025
1A, C, D only
2A, D only
3B, C only
4B, C, D only
NTA Answer: Option 1(final)
How does NEET ask about Le Chatelier?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Predict equilibrium shift on changing concentration, P, T, or adding catalyst.
InterpretationMedium
Common distractors
treats catalyst as shifting
Believes catalyst shifts equilibrium
Sources
NCERT refs: Class 11 Chemistry Chapter 6, p.181
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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