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. 14Le Chatelier
8 MCQs9-step worked example
Source: NCERT EquilibriumPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
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 14).
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.
Practice 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?
MCQ 2Easy RecallPractice
Which of the following factors changes the numerical value of the equilibrium constant K?
MCQ 3Easy RecallPractice
For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ/mol, increasing the temperature will:
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?
MCQ 5Direct ApplicationPractice
An inert gas is added to the equilibrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) at constant pressure. The equilibrium will:
MCQ 6Direct ApplicationPractice
A catalyst is added to the equilibrium N₂O₄(g) ⇌ 2NO₂(g). Which statement is correct?
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:
MCQ 8Concept TrapPractice
Consider the endothermic reaction: CaCO₃(s) ⇌ CaO(s) + CO₂(g). Which combination of changes will shift equilibrium most toward products?
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. ---
Before solving, remember these
Formulas
Henderson-Hasselbalch (buffer)
pH of acidic buffer in terms of conjugate base/acid concentrations. For basic buffer: pOH = pKb + log10([salt]/[base]).
| Symbol | Quantity | SI Unit |
|---|---|---|
| pKa | -log Ka | - |
| [salt] | conjugate base conc | mol/L |
| [acid] | weak acid conc | mol/L |
Valid when
- Buffer (weak acid + conjugate base)
- Concentrations not too dilute
- Approximate (assumes negligible dissociation)
Ka, Kb, Kw relationship
Stronger acid → weaker conjugate base, and vice versa. pKa + pKb = 14.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Ka | acid dissociation | - |
| Kb | base dissociation | - |
| Kw | water 10^-14 | - |
Valid when
- Conjugate acid-base pair
- 25°C
Equilibrium constant K_p and K_c
Convert between pressure-based and concentration-based equilibrium constants. T in K; R = 0.0821 L atm/mol/K (when K_p in atm).
| Symbol | Quantity | SI Unit |
|---|---|---|
| K_p | pressure constant | - |
| K_c | concentration constant | - |
| Δn | mole change | - |
Valid when
- Gas-phase equilibrium
- Same temperature
Solubility product
Equilibrium constant for sparingly soluble salt. Q < K_sp: dissolves; Q > K_sp: precipitates.
| Symbol | Quantity | SI Unit |
|---|---|---|
| K_sp | solubility product | - |
Valid when
- Sparingly soluble salt
- Saturated solution
pH and pOH
Logarithmic acidity scale. Pure water at 25°C: pH = 7 = pOH.
| Symbol | Quantity | SI Unit |
|---|---|---|
| [H+] | hydrogen ion conc | mol/L |
| [OH-] | hydroxide conc | mol/L |
Valid when
- Aqueous solution
- Use Kw = 10^-14 at 25°C
Exam Traps & Common Mistakes
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.
Category: Sign Convention
Δn = (mol gas product) - (mol gas reactant). Sign matters; K_p = K_c (RT)^Δn.
When it triggers
Convert K_p ↔ K_c for gas-phase reaction.
How to avoid
Count moles of gas only (ignore solids/liquids). Δn = product - reactant. If Δn = 0, K_p = K_c. If Δn = +1, K_p = K_c × RT.
Category: Similar Terms
For salt M_aX_b: K_sp = [M⁺]^a [X⁻]^b. Student uses [M⁺][X⁻] regardless of stoichiometry.
When it triggers
K_sp problem with non-1:1 salt (e.g. CaF₂, Mg(OH)₂, Ag₂CrO₄).
How to avoid
Write dissolution: M_aX_b → aM + bX. Then K_sp = [M]^a · [X]^b. For CaF₂ ↔ Ca + 2F: K_sp = s · (2s)² = 4s³.
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
Δn = (moles GAS product) - (moles GAS reactant). Solids and liquids excluded.
Root cause: formula misuse
Correction
For M_aX_b: K_sp = [M]^a · [X]^b. CaF₂: K_sp = s · (2s)² = 4s³.
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.
Root cause: concept gap
Correction
Weak acid: only fraction ionises. [H+] ≈ √(Ka · C). For 0.1 M acetic acid (Ka ~10⁻⁵): [H+] ~10⁻³ (pH ~3), not 1 (pH=1).
Past Year Questions
15 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
1B, D, and E only
2A, B, and C only
3A, B, and D only
4B, C, and D only
NTA Answer: Option 3(final)
NEET 2025
1Sucrose
2D-Glucose
3D-Fructose
4Maltose
NTA Answer: Option 3(final)
NEET 2025
10.021
283.1
32.077 × 105
40.033
NTA Answer: Option 4(final)
NEET 2024Revised key
C [A] = [B] = [C] = 2 × 10–3 M. Then, which of the following is correct?
1Reaction is at equilibrium.
2Reaction has a tendency to go in forward direction.
3Reaction has a tendency to go in backward direction.
4Reaction has gone to completion in forward direction.
NTA Answer: Option 3(revised_final)
NEET 2023
1l = 2n + 1
2n = 2l2 + 1 m m n –1
3n = l + 2
4l m m 2
NTA Answer: Option 4(final)
NEET 2023
Amongst the given options which of the following molecules/ ion acts as a Lewis acid?
1H O
2BF 2 3
3OH–
4NH 3
NTA Answer: Option 2(final)
NEET 2023
Which complex compound is most stable?
1CoNH NO 3 3 3 3
2CoCl en NO 2 2 3
3CoNH SO 3 6 2 4 3
4CoNH H OBrNO 3 4 2 3 2
NTA Answer: Option 2(final)
NEET 2022
1Both diamond and graphite are used as dry lubricants.
2Diamond and graphite have two dimensional network.
3Diamond is covalent and graphite is ionic.
4Diamond is sp3 hybridised and graphite is sp2 hybridized.
NTA Answer: Option 4(final)
NEET 2022
1p =p, where = mole fraction of ith gas in gaseous mixture i i i i p = pressure of ith gas in pure state i
2p = p + p + p 1 2 3 RT RT RT
3p=n +n +n 1 V 2 V 3 V
4p = p, where p = partial pressure of ith gas i i i = mole fraction of ith gas in gaseous mixture i
NTA Answer: Option 1(final)
NEET 2022
14.9
22.5
3498.6
449.8
NTA Answer: Option 1(final)
NEET 2022
11.2 × 1021
24.38 × 10–32
31.9 × 10–63
42.4 × 1031
NTA Answer: Option 2(final)
NEET 2022
11.05 V
21.0385 V
31.385 V
40.9615 V Answer (NA)
NTA Answer: Option 5(final)
NEET 2021
1and 8
2and 4 sp3 sp2
3and 6
4and 6
NTA Answer: Option 4(final)
NEET 2021
1Beryllium chloride
2Calcium chloride
3Strontium chloride
4Magnesium chloride
NTA Answer: Option 1(final)
NEET 2021
126.02
22.518
32.602
425.18
NTA Answer: Option 4(final)
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Convert between K_p and K_c using K_p = K_c (RT)^Δn. Identify Δn carefully.
Multi StepMedium
Common distractors
wrong delta n sign
Counts moles incorrectly
Predict equilibrium shift on changing concentration, P, T, or adding catalyst.
InterpretationMedium
Common distractors
treats catalyst as shifting
Believes catalyst shifts equilibrium
Compute pH of weak acid/base solution using Ka or Kb. Approximation: [H+] ≈ √(Ka·C).
Multi StepMedium
Common distractors
ignores weak vs strong
Treats weak acid as fully dissociated
Compute solubility from K_sp or vice versa. Apply common-ion effect when relevant.
Multi StepMedium
Common distractors
forgets stoichiometric coefficient power
Uses [M+]^1 [X-]^1 for M(X)_2 salt
Sources
NCERT refs: Class 11 Chemistry Chapter 6, p.14
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