Subtopics - Chemical Equilibrium (NEET)
Ten topic blocks: equilibrium concept with thermal-mechanical-chemical types, reversible reactions and characteristics, homogeneous versus heterogeneous equilibria, Law of Mass Action and active mass, equilibrium constants Kc and Kp with their derivation, Kp-Kc relation and units, factors affecting K (temperature, stoichiometry, mode of writing), reaction quotient Q for direction prediction, Le Chatelier's Principle with effects of concentration-pressure-temperature-inert gas on physical and chemical equilibria, degree of dissociation and vapour density, and free energy change with spontaneity.
1) Equilibrium
Introduces thermodynamic equilibrium as the state where temperature, pressure, and composition remain invariant with time. Decomposes into three sub-types: thermal (constant T), mechanical (constant P), and chemical (constant composition). Chemical equilibrium is the state where the net reaction rate is zero.
2) Reversible Reactions and Characteristics of Chemical Equilibrium
Defines reversible reactions as those proceeding in both forward and backward directions without going to completion. Lists five defining characteristics of chemical equilibrium: dynamic nature, equal forward and backward rates, invariant observable properties, catalyst effect, and closed vessel requirement.
3) Types of Equilibria
Classifies equilibria into homogeneous (all species in same phase) and heterogeneous (species in different phases). The classification determines which species appear in the equilibrium constant expression, since pure solids and pure liquids are excluded.
4) Law of Mass Action
States that at constant temperature, the rate of a reaction is directly proportional to the active masses of reactants raised to the power of their stoichiometric coefficients. Proposed by Guldberg and Waage. Active mass equals molarity for dilute solutions (activity coefficient gamma equals 1).
5) Equilibrium Constants Kc and Kp
Derives equilibrium constant Kc as the ratio of forward to backward rate constants (kf/kb). For gaseous equilibria, derives Kp using partial pressures. Establishes the Kp-Kc relation: Kp = Kc(RT)^delta-n. Covers units, conditions for Kp existence (at least one gas, no solution component), and the mixed constant Kpc.
6) Factors Affecting Equilibrium Constant
Identifies what changes the numerical value of K and what does not. Temperature is the only factor that changes K for a given reaction. Nature of reactants and products, stoichiometry, and the mode of writing the equation also affect the K value. Concentration, pressure, volume, catalyst, and inert gas do NOT change K.
7) Reaction Quotient
Defines the reaction quotient Qc as having the same form as Kc but using concentrations at any instant rather than at equilibrium. Comparing Q with K predicts the direction the reaction will proceed to reach equilibrium.
8) Le Chatelier's Principle
States that when a system at equilibrium is disturbed by an external factor, it shifts to nullify the disturbance and establishes a new equilibrium. Covers the effects of concentration, pressure (volume change), inert gas addition (constant volume vs constant pressure), and temperature. Extends to physical equilibria: melting point and boiling point under pressure changes.
9) Degree of Dissociation and Vapour Density
Defines degree of dissociation alpha as the fraction of moles dissociated out of initial moles. Derives the general formula relating alpha to initial and equilibrium vapour densities: alpha = (D - d) / ((n - 1) times d), valid for equilibria where Kp exists.
10) Free Energy Change and Spontaneity
Connects thermodynamics to equilibrium through Gibbs free energy. delta-G < 0 means spontaneous forward reaction; delta-G > 0 means spontaneous backward reaction; delta-G = 0 at equilibrium. The relation delta-G-zero = -RT ln K links standard free energy change to the equilibrium constant.
Chemical Equilibrium Download Notes & Weightage Plan
For each topic in the Chemical Equilibrium chapter below, you get (2) the exact resources to download and how to use them, and (3) a simple importance & time plan so NEET students know what to do first and what to revise last.
Foundation topic introducing three types of thermodynamic equilibrium: thermal, mechanical, and chemical. Establishes the concept that equilibrium means invariant macroscopic properties, not cessation of molecular processes.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Know that equilibrium is dynamic (forward and backward reactions continue) and can only be achieved in a closed vessel. A catalyst hastens approach to equilibrium but does not change the equilibrium position.
- High-risk Area: Students sometimes believe that at equilibrium the concentrations of reactants and products are equal. This is wrong. The concentrations become constant (invariant with time), but they are not necessarily equal to each other. The ratio is determined by the equilibrium constant K.
- Best Practice Style: When answering assertion-reason questions about equilibrium, test each statement independently: does equilibrium mean equal concentrations (no), does it mean reaction stops (no), does it require a closed vessel (yes).
Reversible Reactions and Characteristics of Chemical Equilibrium
Defines reversible reactions and lists five characteristics that define chemical equilibrium: dynamic nature, equal rates, invariant properties, catalyst effect, and closed-vessel requirement.
1) Download Packs For This Topic (And How To Use Them)
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2) Importance, Weightage & Time Allocation (Practical)
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- Scoring Focus: Catalyst effect is the highest-yield fact: a catalyst hastens the approach to equilibrium but does not change the equilibrium state, does not change K, and does not change product yield.
- High-risk Area: Incorrectly believing that a catalyst can increase the yield of products. The catalyst provides an alternative reaction pathway with lower activation energy for both forward and backward reactions equally. It does not shift the equilibrium position.
- Best Practice Style: For any question mentioning catalyst and equilibrium, immediately apply: catalyst changes rate, not position. Then eliminate options claiming yield increase or K change.
Classifies equilibria as homogeneous (single phase) or heterogeneous (multiple phases). The critical NEET application is knowing which species to include or exclude from K expressions.
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2) Importance, Weightage & Time Allocation (Practical)
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- Scoring Focus: For any equilibrium involving solids: the solid does not appear in K. For CaCO3(s) to CaO(s) + CO2(g), Kp = P(CO2). Students who include CaCO3 or CaO get the wrong expression.
- High-risk Area: Including pure solids or pure liquids in the equilibrium expression. NEET places the incorrect expression (with solids) as a distractor. The rule is absolute: if a species is a pure solid or pure liquid, its concentration is constant and absorbed into K.
- Best Practice Style: Before writing any K expression, draw a line under each species and label its phase (s, l, g, aq). Cross out every (s) and (l). Write K with only the remaining species.
The Guldberg-Waage law stating that the rate of reaction is proportional to the active masses of reactants raised to their stoichiometric coefficients. Active mass equals molarity for dilute solutions.
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2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Active mass is molarity for gases and solutions. For a given mass of substance in a given volume: active mass = (mass / molecular weight) / volume in litres.
- High-risk Area: Confusing active mass with mass concentration (g/L). Active mass is always in mol/L (molarity). Students who forget to divide mass by molecular weight before dividing by volume get the wrong answer by a factor of the molecular weight.
- Best Practice Style: For any active mass calculation: (1) find moles = mass / MW, (2) divide by volume in litres. Never skip the MW step.
Equilibrium Constants Kc and Kp
The central quantitative topic: derives Kc from the Law of Mass Action, derives Kp from ideal gas equation, establishes Kp = Kc(RT)^delta-n, and covers units and conditions for Kp existence.
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Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
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- Scoring Focus: The Kp-Kc relation Kp = Kc(RT)^delta-n is the highest-yield formula. Know how to count delta-n (gaseous species only) and when to apply R = 0.0821 L atm / (mol K).
- High-risk Area: Counting delta-n incorrectly by including non-gaseous species. For CaCO3(s) to CaO(s) + CO2(g), delta-n = 1 (only CO2 is gaseous). Students who count all species get delta-n = 1 + 1 - 1 = 1, which happens to be correct here, but for reactions with solution-phase species this approach fails.
- Best Practice Style: Circle all (g) species in the balanced equation. Count product gas moles minus reactant gas moles. That is delta-n. Ignore every species that is not (g).
Factors Affecting Equilibrium Constant
Identifies temperature as the only factor changing K. Covers stoichiometric effects: multiplying equation by n gives K^n, reversing gives 1/K. Covers the Van't Hoff equation for temperature dependence.
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2) Importance, Weightage & Time Allocation (Practical)
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- Scoring Focus: Combined-equation problems: if reaction 3 = reaction 1 + reaction 2, then K3 = K1 times K2. If reaction 3 = reverse of reaction 1, then K3 = 1/K1. These combination rules are tested every year.
- High-risk Area: Mixing up when to multiply K values versus when to take their reciprocal. Rule: adding equations multiplies K values; reversing an equation inverts its K; halving exponents takes the square root of K.
- Best Practice Style: Write down the target equation and express it as a combination of given equations (sum, reverse, multiply). Apply corresponding K operations step by step.
The reaction quotient Qc has the same form as Kc but uses instantaneous concentrations. Comparison of Q with K predicts the direction the reaction must proceed to reach equilibrium.
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2) Importance, Weightage & Time Allocation (Practical)
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- Scoring Focus: The three-way comparison: Q < K forward, Q > K reverse, Q = K equilibrium. No marks for the calculation alone; the comparison and conclusion carry the marks.
- High-risk Area: Confusing the direction: Q > K does NOT mean forward (it means reverse). The mnemonic: if Q is too big (too many products), the system must go backward to reduce product concentrations and bring Q down to K.
- Best Practice Style: Calculate Q using the same expression as K but with given concentrations. Compare numerically. State direction. Three steps, no shortcuts.
The qualitative prediction tool for equilibrium shifts. Covers effects of concentration, pressure, inert gas (constant volume versus constant pressure), and temperature. Extends to physical equilibria affecting melting and boiling points.
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Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Three scoring facts: (1) inert gas at constant volume has no effect, (2) temperature is the only factor that changes K, (3) pressure effect depends on delta-n. These three facts alone answer 80% of Le Chatelier questions.
- High-risk Area: Inert gas at constant pressure with delta-n > 0 shifts equilibrium forward, which is counterintuitive. Students often assume inert gas never affects equilibrium, which is true only at constant volume. At constant pressure, inert gas addition increases total volume, diluting all species, and the side with more moles of gas is favoured.
- Best Practice Style: For every Le Chatelier question: (1) identify the disturbance type (concentration, pressure, temperature, inert gas), (2) for inert gas, check whether it is constant V or constant P, (3) for pressure, calculate delta-n, (4) apply the appropriate rule.
Degree of Dissociation and Vapour Density
Derives the formula connecting degree of dissociation alpha with initial and equilibrium vapour densities. Key formula: alpha = (D - d) / ((n - 1) times d). Applicable only when Kp exists.
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2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: The formula alpha = (D-d)/((n-1)d) and the relation molar mass = 2 times vapour density. If the question gives molecular masses instead of vapour densities, divide by 2 to get VD, then apply the formula.
- High-risk Area: Incorrectly identifying n for reactions with fractional coefficients. For 2SO3 to 2SO2 + O2, if written per mole of SO3: SO3 to SO2 + (1/2)O2, n = 1.5. Students who use n = 3 (total product moles for 2 moles of SO3) get the wrong alpha.
- Best Practice Style: Always rewrite the equation for ONE mole of the dissociating species before identifying n. Then n = total moles of products from one mole of reactant.
Free Energy Change and Spontaneity
Links thermodynamics to equilibrium. delta-G determines spontaneity; delta-G-zero = -RT ln K quantifies the relationship between standard free energy change and the equilibrium constant.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Two facts for quick marks: (1) at equilibrium, delta-G = 0 (not delta-G-zero), (2) delta-G-zero = -RT ln K connects free energy to the equilibrium constant. These two facts answer every NEET question from this section.
- High-risk Area: Confusing delta-G with delta-G-zero. delta-G-zero is the free energy change at standard conditions (1 mol/L concentrations). delta-G is the free energy change at any given composition. At equilibrium, delta-G = 0, but delta-G-zero can be any value (it depends on K). Students who assume delta-G-zero = 0 at equilibrium get the wrong answer.
- Best Practice Style: When a question says 'at equilibrium,' set delta-G = 0 and derive delta-G-zero = -RT ln K from delta-G = delta-G-zero + RT ln Q (at equilibrium Q = K, so delta-G-zero = -RT ln K).
Chemical Equilibrium Chapter NEET Traps & Common Mistakes (Topic-Wise)
Each subtopic below is of the Chemical Equilibrium chapter and shows what NEET students usually do wrong in NEET examination, a short example of the mistake, and how NEET frames the question to trick you with close options are given below.
Mistake Snapshot (What Students Do Wrong)
- Including pure solids or pure liquids in K expressions: For CaCO3(s) to CaO(s) + CO2(g), the correct Kp = P(CO2). Students who include CaCO3 and CaO get Kp = P(CO2) times [CaO] / [CaCO3], which is the Keq expression, not Kc or Kp. NEET places this incorrect expression as a distractor.
- Counting delta-n with non-gaseous species: In the Kp-Kc relation, delta-n must count only gaseous species. For C(s) + CO2(g) to 2CO(g), delta-n = 2 - 1 = 1 (not 2 - 2 = 0). Including C(s) in the count gives an incorrect delta-n of zero.
For NH4HS(s) to NH3(g) + H2S(g): correct Kp = P(NH3) times P(H2S). delta-n = 2 - 0 = 2 (no gaseous reactants). Students who include NH4HS(s) get delta-n = 2 - 1 = 1, then calculate Kc incorrectly from Kp.
How NEET Frames The Trap
NEET poses heterogeneous equilibria and asks for Kp or Kc. The distractor includes solid species in the expression or miscounts delta-n by including solids.
Q. For the equilibrium CaCO3(s) to CaO(s) + CO2(g), which expression is correct?
A. Kp = P(CO2) B. Kp = P(CO2) P(CaO) / P(CaCO3) C. Kp = P(CaO) / P(CaCO3) D. Kp = 1 / P(CO2)
Trick: Pure solids CaCO3 and CaO are excluded from Kp. Only CO2 is gaseous. Kp = P(CO2) (Option A). Option B includes both solids as if they had partial pressures. Option C includes solids and omits CO2. Option D is K for the reverse reaction.
Mistake Snapshot (What Students Do Wrong)
- Using wrong units or value of R in Kp = Kc(RT)^delta-n: The relation requires R in litre-atm per mole per kelvin (0.0821). Using R = 8.314 J/(mol K) gives numerically wrong results because Kp is in atm units while 8.314 gives energy units.
- Forgetting that delta-n can be negative: For N2 + 3H2 to 2NH3, delta-n = 2 - 4 = -2. The (RT)^delta-n term becomes (RT)^(-2) = 1/(RT)^2, making Kc > Kp. Students who use delta-n = +2 get Kp/Kc inverted.
For PCl5(g) to PCl3(g) + Cl2(g) at 500 K: delta-n = 2 - 1 = +1. If Kc = 0.04 mol/L, then Kp = 0.04 times (0.0821 times 500)^1 = 0.04 times 41.05 = 1.642 atm. Using R = 8.314 instead gives 0.04 times 4157 = 166.3, which is off by a factor of approximately 100.
How NEET Frames The Trap
NEET gives Kc and asks for Kp (or vice versa). The incorrect R value produces a distractor that differs by a factor related to the ratio of R values.
Q. For N2(g) + 3H2(g) to 2NH3(g) at 500 K, if Kc = 0.5 (mol/L)^-2, what is the value of Kp?
A. 2.97 x 10^-4 atm^-2 B. 841 atm^-2 C. 0.5 atm^-2 D. 6.9 x 10^-5 atm^-2
Trick: delta-n = 2 - 4 = -2. Kp = 0.5 times (0.0821 times 500)^(-2) = 0.5 / (41.05)^2 = 0.5 / 1685 = 2.97 x 10^-4 atm^-2 (Option A). Option B uses delta-n = +2 instead of -2. Option C assumes Kp = Kc (delta-n = 0 error). Option D uses R = 8.314 instead of 0.0821.
Mistake Snapshot (What Students Do Wrong)
- Assuming inert gas never affects equilibrium: At constant volume, inert gas addition indeed has no effect. But at constant pressure, inert gas addition increases total volume, diluting all species. If delta-n > 0, equilibrium shifts forward (more gas moles favoured). Students who always apply the constant-volume rule miss this.
- Getting the direction wrong for inert gas at constant pressure: At constant pressure with delta-n < 0: the dilution effects push equilibrium backward (fewer moles side is disfavoured). Students who think dilution always favours forward direction get the wrong answer.
N2O4(g) to 2NO2(g), delta-n = +1. Add helium at constant pressure. Volume increases, diluting both species. Since products side has more moles, equilibrium shifts forward to produce more gas moles. At constant volume with the same helium addition: no effect because concentrations of N2O4 and NO2 are unchanged.
How NEET Frames The Trap
NEET asks what happens when an inert gas is added, sometimes specifying constant pressure, sometimes constant volume, sometimes not specifying (in which case assume constant total pressure if the container is flexible, or constant volume if rigid).
Q. For PCl5(g) to PCl3(g) + Cl2(g), an inert gas is added at constant pressure. What happens to the degree of dissociation?
A. Increases B. Decreases C. Remains unchanged D. First increases then decreases
Trick: delta-n = 2 - 1 = +1 (more gas moles on product side). At constant pressure, inert gas addition increases volume and dilutes all species. The equilibrium shifts forward to compensate, so degree of dissociation increases (Option A). Option C is the answer for constant volume. Option B reverses the direction.
Mistake Snapshot (What Students Do Wrong)
- Using K directly for a halved or reversed equation: If K for N2 + O2 to 2NO is K1, the K for NO to (1/2)N2 + (1/2)O2 is 1/sqrt(K1), not 1/K1 or sqrt(K1). Students who apply only one transformation (reverse OR halve) instead of both get the wrong answer.
- Adding K values instead of multiplying when combining equations: When two equations are added: K(total) = K1 times K2, not K1 + K2. The equilibrium expressions are multiplied because adding equations corresponds to multiplying the respective concentration ratios.
Given K1 for N2 + O2 to 2NO and K2 for 2NO + O2 to 2NO2. Find K for NO2 to (1/2)N2 + O2. Target = reverse of [(1/2)(reaction 1) + (1/2)(reaction 2)]. K for half of reaction 1 = sqrt(K1). K for half of reaction 2 = sqrt(K2). K for combined half-reactions = sqrt(K1) times sqrt(K2). K for reverse = 1/(sqrt(K1) times sqrt(K2)) = 1/sqrt(K1 K2). Answer: K = [1/(K1 K2)]^(1/2).
How NEET Frames The Trap
NEET gives K for two or three standard reactions and asks for K of a target reaction that is a combination of reversed, halved, or added versions of the given reactions.
Q. If K for A + B to C is 10 and K for C + D to E is 20, what is K for the reaction A + B + D to E?
A. 200 B. 30 C. 2 D. 0.5
Trick: The target equation = (A + B to C) + (C + D to E) = A + B + D to E, with C cancelling. K = K1 times K2 = 10 times 20 = 200 (Option A). Option B adds the K values (10 + 20 = 30). Option C divides (20/10). Option D takes the reciprocal.
Mistake Snapshot (What Students Do Wrong)
- Using n from the full equation instead of per-mole form: For 2SO3(g) to 2SO2(g) + O2(g), if alpha is calculated per mole of SO3, the equation becomes SO3 to SO2 + (1/2)O2 and n = 1.5. Students who use the full equation get moles = 2 + 1 = 3 and set n = 3/2 = 1.5 per mole of SO3. Either approach gives the same n, but confusion arises when students use n = 3 (total product moles from 2 moles of SO3).
- Confusing molar mass with vapour density: Molar mass = 2 times vapour density. The formula uses vapour density (D and d), not molar mass. If the question gives molecular masses, divide by 2 before substituting.
N2O4 (MW = 92, D = 46) dissociates at 400 K to give equilibrium MW = 72 (d = 36). alpha = (D - d)/((n - 1) times d) = (46 - 36)/((2 - 1) times 36) = 10/36 = 0.278 or 27.8%. If student uses MW directly (92 and 72) without dividing by 2: alpha = (92 - 72)/((2-1) times 72) = 20/72 = 0.278. The answer is the same because the factor of 2 cancels, but this only works if the student consistently uses MW or consistently uses VD.
How NEET Frames The Trap
NEET gives molecular mass data and asks for percentage dissociation. The conversion to vapour density is straightforward but forgetting it leads to confusion when formulas are memorised with D and d specifically.
Q. The vapour density of PCl5 at 250 degrees C is found to be 57.9. What is the degree of dissociation? (Molecular mass of PCl5 = 208.5)
A. 80% B. 40% C. 57.9% D. 20%
Trick: D (initial VD) = 208.5/2 = 104.25. d (equilibrium VD) = 57.9. n = 2 (PCl5 to PCl3 + Cl2). alpha = (104.25 - 57.9)/((2-1) times 57.9) = 46.35/57.9 = 0.80 = 80% (Option A). Option B halves the answer. Option C confuses VD with percent. Option D quarters the answer.