For reaction aA + bB ⇌ cC + dD: K_c = [C]^c[D]^d / ([A]^a[B]^b). K_p uses partial pressures: K_p = K_c (RT)^Δn where Δn = (c+d) - (a+b).
-- NCERT Class 11 Chemistry, Ch. 6, p. 8Law of Equilibrium Kp Kc
8 MCQs1 revision card9-step worked example
Source: NCERT EquilibriumPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
The trap that costs marks: you write Δn correctly as (products − reactants), but you count ALL species — solids, liquids, gases. NEET distractors are built on exactly this mistake. Δn counts gaseous moles only.
The law of chemical equilibrium states that for a reversible reaction at a given temperature, the ratio of product concentrations to reactant concentrations — each raised to their stoichiometric coefficients — is a constant. This constant is K_c when expressed in molar concentrations (NCERT Class 11 Chemistry Chapter 7, page 8).
For gas-phase reactions, we can also express the equilibrium constant in terms of partial pressures as K_p. The two are related by:
K_p = K_c (RT)^Δn
where Δn = (total moles of gaseous products) − (total moles of gaseous reactants), R = 0.0821 L atm mol⁻¹ K⁻¹ (when K_p is in atm units), and T is in kelvin.
Three cases to lock in:
- Δn > 0 (more gas moles in products): K_p > K_c at any temperature above 0 K.
- Δn < 0 (fewer gas moles in products): K_p < K_c.
- Δn = 0 (equal gas moles): K_p = K_c regardless of temperature.
Key properties of K_c and K_p:
- They depend only on temperature — not on initial concentrations or pressure.
- Pure solids and pure liquids do not appear in the equilibrium expression.
- If you reverse the reaction, the new K is the reciprocal (1/K).
- If you multiply the equation by a factor n, K becomes K^n.
Watch out: when a reaction involves a mix of gases and solids (e.g., thermal decomposition of CaCO₃), students often include the solid in Δn. Solids and liquids are excluded from both the equilibrium expression and the Δn count.
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 1Direct ApplicationPractice
For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the value of Δn used in the relation K_p = K_c(RT)^Δn is:
MCQ 2CalculationPractice
For the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), if K_c = 4.0 × 10² at 1000 K, what is the approximate value of K_p? (R = 0.0821 L atm mol⁻¹ K⁻¹)
MCQ 3Easy RecallPractice
The equilibrium constant K_c for a reaction depends on:
MCQ 4CalculationPractice
For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), if K_p = 1.8 at a certain temperature, what is the value of K_p for the reaction 2PCl₃(g) + 2Cl₂(g) ⇌ 2PCl₅(g) at the same temperature?
MCQ 5Direct ApplicationPractice
For the decomposition CaCO₃(s) ⇌ CaO(s) + CO₂(g), the equilibrium expression for K_p is:
MCQ 6Easy RecallPractice
For a gaseous equilibrium where Δn = 0, which statement is correct?
MCQ 7Direct ApplicationPractice
Consider the reaction: C(s) + CO₂(g) ⇌ 2CO(g). When computing Δn for the K_p = K_c(RT)^Δn relation, the value of Δn is:
MCQ 8Concept TrapPractice
The equilibrium constant for the reaction H₂(g) + I₂(g) ⇌ 2HI(g) is K_c. What is the equilibrium constant for the reaction HI(g) ⇌ ½H₂(g) + ½I₂(g)?
Quick recall before you leave
Worked Example
- 1
Given
For the reaction N₂O₄(g) ⇌ 2NO₂(g), K_c = 4.63 × 10⁻³ at 25°C (298 K). R = 0.0821 L atm mol⁻¹ K⁻¹.
- 2
Required
Find K_p at 25°C.
- 3
Concept
K_p and K_c are related through K_p = K_c(RT)^Δn. We need to determine Δn by counting gaseous moles on each side.
- 4
Formula
K_p = K_c × (RT)^Δn
- 5
Substitution
Δn = (moles gaseous products) − (moles gaseous reactants) = 2 − 1 = +1 K_p = 4.63 × 10⁻³ × (0.0821 × 298)^(+1)
- 6
Calculation
RT = 0.0821 × 298 = 24.47 K_p = 4.63 × 10⁻³ × 24.47 K_p = 0.1133 Note: 298 K is an exact temperature specification in this problem, and the stoichiometric coefficients (2, 1) are exact counting numbers. These do not limit significant figures. The answer is reported to three significant figures, matching K_c.
- 7
Final answer
K_p ≈ 0.113 (or 1.13 × 10⁻¹) Since Δn = +1 > 0, K_p > K_c as expected — a quick sanity check.
- 8
Common trap
The high-frequency trap here is computing Δn = 1 − 2 = −1 (reversing the subtraction order). That gives K_p = 4.63 × 10⁻³ / 24.47 ≈ 1.89 × 10⁻⁴ — a distractor that appears plausible. Another common error: including solid or liquid species in the Δn count when the reaction contains a mix of phases.
- 9
Similar NEET-style question
For the reaction 2SO₃(g) ⇌ 2SO₂(g) + O₂(g), K_c = 6.9 × 10⁻² at 727°C. Calculate K_p. (Hint: Δn = +1, T = 1000 K.) ---
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 7, p.8
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