Definition
Dynamic equilibrium
State in which forward and reverse reactions occur at equal rates; concentrations remain constant. Reaction continues at molecular level despite no net macroscopic change.
-- NCERT Class 11 Chemistry, Ch. 6, p. 4Dynamic Equilibrium
8 MCQs9-step worked example
Source: NCERT EquilibriumPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
Dynamic equilibrium — what NEET actually tests
The concept sounds simple, but NEET uses it to filter students who memorised a definition from those who understand what "dynamic" means at the molecular level.
The core idea (NCERT Class 11 Chemistry Chapter 6, page 4): In a closed system, when a reversible reaction reaches equilibrium, the forward and reverse reactions continue at equal rates. Macroscopic properties — concentrations, pressure, colour — stop changing, but molecular-level transformations never stop.
Three non-negotiable features of dynamic equilibrium:
- Both reactions run simultaneously. Equilibrium is not "reaction stopped." Reactant molecules still convert to products, and products still revert to reactants — at the same rate.
- Closed system required. If products escape (open vessel, gas released), the system cannot reach equilibrium. NEET stems sometimes describe open containers to test whether you notice.
- Observable properties are constant, not zero. Concentrations are constant because the rate of formation equals the rate of consumption — not because nothing is happening.
The distinction that costs marks: Static vs. dynamic. A saturated salt solution with undissolved solid at the bottom is at dynamic equilibrium — dissolution and crystallisation occur at equal rates. A sealed bottle of pure water with no dissolved gas is simply unchanging (no competing process). NEET stems test whether you can identify which scenario is genuinely dynamic.
Watch-out: If a question says "equilibrium is reached," that tells you rates are equal and net change is zero. It does NOT tell you the reaction has stopped, that concentrations are equal, or that the amounts of reactant and product are the same.
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
At dynamic equilibrium in a closed system, which statement is correct?
MCQ 2Easy RecallPractice
Which of the following is an essential condition for a system to attain dynamic equilibrium?
MCQ 3Direct ApplicationPractice
A sealed container holds N₂O₄(g) in equilibrium with NO₂(g). The brown colour of NO₂ remains constant over time. This observation indicates that:
MCQ 4Direct ApplicationPractice
In a saturated solution of NaCl with undissolved solid NaCl at the bottom, the process of dissolution:
MCQ 5Easy RecallPractice
Which of the following correctly distinguishes a static situation from dynamic equilibrium?
MCQ 6Direct ApplicationPractice
At equilibrium, the concentration of a reactant in a closed vessel is 0.5 mol/L. A student claims this means 0.5 mol/L of products must also be present. This claim is:
MCQ 7Concept TrapPractice
A reversible reaction A(g) ⇌ B(g) is carried out in an open flask. After some time, only product B is detected. This observation suggests:
MCQ 8Direct ApplicationPractice
Consider the reversible reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) in a sealed vessel at constant temperature. At equilibrium, which of the following must be true?
Worked Example
- 1
Given
A closed container holds liquid water in equilibrium with water vapour at 100°C and 1 atm: H₂O(l) ⇌ H₂O(g) The rate of evaporation is measured as 2.5 × 10⁻³ mol/s.
- 2
Required
Find the rate of condensation at equilibrium.
- 3
Concept
At dynamic equilibrium, the rate of the forward process (evaporation) equals the rate of the reverse process (condensation). The system is closed, so water vapour cannot escape.
- 4
Formula
Rate_forward = Rate_reverse (at dynamic equilibrium)
- 5
Substitution
Rate_condensation = Rate_evaporation = 2.5 × 10⁻³ mol/s
- 6
Calculation
No arithmetic needed. The equality follows directly from the definition of dynamic equilibrium.
- 7
Final answer
The rate of condensation is **2.5 × 10⁻³ mol/s**.
- 8
Common trap
The most common error is claiming the rate of condensation is zero ("the water has finished evaporating") or that it differs from the evaporation rate. At equilibrium, both processes continue at the same rate — that is why the water level and vapour pressure remain constant.
- 9
Similar NEET-style question
In a sealed flask, liquid bromine is in equilibrium with bromine vapour. If the rate of vaporisation is r mol/s, what is the rate of condensation? (Answer: r mol/s, by the definition of dynamic equilibrium in a closed system.) ---
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.4
Test yourself on this topic with real past-paper questions:
Practice this topic →Free NEET study resources
Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.