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. 175Law of Equilibrium Kp Kc
8 MCQs3 revision cards9-step worked example
Source: NCERT EquilibriumPYQ coverage: NEET 2022, 2024, 2025Official key: NTA-verifiedLast updated: 25 Sep 2026
Law of Equilibrium Kp Kc, explained for NEET
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 6, page 175).
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.
Can you answer these Law of Equilibrium Kp Kc 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:
Show answer and why every option is right or wrong
Answer: C. Δn = moles of gaseous products − moles of gaseous reactants = 2 − (1 + 3) = 2 − 4 = −2 (NCERT Class 11 Chemistry Chapter 6, page 175).
Why A is wrong: A is wrong because +2 reverses the subtraction order (reactants − products instead of products − reactants). This is the sign-convention trap where Δn is computed backwards.
Why B is wrong: B is wrong because +4 is the total moles of gaseous reactants (1 + 3) taken as Δn, with the product moles never subtracted. Δn is a difference: 2 − 4 = −2.
Why D is wrong: D is wrong because −4 likely subtracts product moles from some incorrect total. Δn = products − reactants = 2 − 4 = −2, not −4.
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)
Show answer and why every option is right or wrong
Answer: B. Δn = 2 − (2 + 1) = −1. K_p = K_c × (RT)^Δn = 4.0 × 10² × (0.0821 × 1000)^(−1) = 400 / 82.1 ≈ 4.87 (NCERT Class 11 Chemistry Chapter 6, page 175).
Why A is wrong: A is wrong because it uses Δn = +1 instead of −1, giving K_p = 400 × 82.1 = 3.28 × 10⁴. This is the Δn sign-convention trap — products minus reactants, not the reverse.
Why C is wrong: C is wrong because it assumes K_p = K_c, which is only true when Δn = 0. Here Δn = −1, so the (RT)^Δn factor matters.
Why D is wrong: D is wrong because it uses Δn = −2 instead of −1: K_p = 400/(82.1)² ≈ 5.9 × 10⁻². Δn = 2 − (2 + 1) = −1.
MCQ 3Easy RecallPractice
The equilibrium constant K_c for a reaction depends on:
Show answer and why every option is right or wrong
Answer: D. K_c is a function of temperature alone. Changing concentrations, pressure, or adding a catalyst does not change the value of K_c (NCERT Class 11 Chemistry Chapter 6, page 175).
Why A is wrong: A is wrong because changing initial concentrations changes the reaction quotient Q, not K_c. The system adjusts to restore Q = K_c, but K_c itself remains constant at a given temperature.
Why B is wrong: B is wrong because a catalyst speeds up both forward and reverse reactions equally, reducing the time to reach equilibrium without changing K_c.
Why C is wrong: C is wrong because pressure changes shift the equilibrium position (per Le Chatelier's principle) but do not alter the numerical value of K_c.
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?
Show answer and why every option is right or wrong
Answer: A. Reversing the reaction gives K' = 1/1.8. Multiplying the reversed reaction by 2 gives K'' = (1/1.8)² = 1/3.24 ≈ 0.31 (NCERT Class 11 Chemistry Chapter 6, page 175).
Why B is wrong: B is wrong because it computes only 1/1.8 ≈ 0.56 (the reciprocal for reversal) but forgets to raise it to the power of 2 for the coefficient doubling.
Why C is wrong: C is wrong because it doubles K_p (1.8 × 2 = 3.6), treating stoichiometric multiplication as arithmetic multiplication. When a balanced equation is multiplied by n, K becomes K^n, not n × K.
Why D is wrong: D is wrong because 1/1.8 accounts for reversing the reaction but ignores the coefficient multiplication by 2. The correct result is (1/1.8)² ≈ 0.31.
MCQ 5Direct ApplicationPractice
For the decomposition CaCO₃(s) ⇌ CaO(s) + CO₂(g), the equilibrium expression for K_p is:
Show answer and why every option is right or wrong
Answer: A. Pure solids (CaCO₃ and CaO) are excluded from the equilibrium expression. The only gaseous species is CO₂, so K_p = p(CO₂) (NCERT Class 11 Chemistry Chapter 6, page 175).
Why B is wrong: B is wrong because it includes partial pressures of CaO and CaCO₃, which are pure solids. Solids do not appear in equilibrium expressions — their activities are taken as 1.
Why C is wrong: C is wrong because it places CO₂ in the denominator, as if CO₂ were a reactant. CO₂ is a product of the forward decomposition.
Why D is wrong: D is wrong because it excludes CO₂ (the only gas) and includes only the two solids. Pure solids have activity = 1 and do not appear in K expressions.
MCQ 6Easy RecallPractice
For a gaseous equilibrium where Δn = 0, which statement is correct?
Show answer and why every option is right or wrong
Answer: D. When Δn = 0, (RT)^Δn = (RT)⁰ = 1, so K_p = K_c × 1 = K_c regardless of temperature (NCERT Class 11 Chemistry Chapter 6, page 175).
Why A is wrong: A is wrong because K_p > K_c only when Δn > 0 (since RT > 1 at any practical temperature). When Δn = 0, the (RT) factor cancels to 1.
Why B is wrong: B is wrong because K_p < K_c only when Δn < 0. When Δn = 0, K_p equals K_c exactly.
Why C is wrong: C is wrong because (RT)⁰ = 1 at every temperature, not just 273 K. Temperature is irrelevant when Δn = 0.
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:
Show answer and why every option is right or wrong
Answer: B. Count gaseous moles only. Products: 2 mol CO(g). Reactants: 1 mol CO₂(g) — C is solid, excluded. Δn = 2 − 1 = +1 (NCERT Class 11 Chemistry Chapter 6, page 175).
Why A is wrong: A is wrong because +2 counts only the gaseous product (2 mol CO) as Δn, without subtracting the 1 mol of CO₂ on the reactant side. Solids are ignored: Δn = 2 − 1 = +1.
Why C is wrong: C is wrong because Δn = 0 would result from counting C(s) as 1 gaseous mole on the reactant side (2 − 2 = 0). C is a solid and must be excluded from the Δn count.
Why D is wrong: D is wrong because −1 reverses the subtraction. Δn = gaseous products − gaseous reactants = 2 − 1 = +1, not 1 − 2 = −1.
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)?
Show answer and why every option is right or wrong
Answer: C. Reversing the original reaction gives K' = 1/K_c. Halving the coefficients means the new K = (1/K_c)^(1/2) = 1/√K_c (NCERT Class 11 Chemistry Chapter 6, page 175).
Why A is wrong: A is wrong because 1/K_c only accounts for reversing the reaction. The halved coefficients require an additional square-root operation: (1/K_c)^(1/2) = 1/√K_c.
Why B is wrong: B is wrong because √K_c would apply if you halved the original forward reaction without reversing it. Here the reaction is both reversed and halved.
Why D is wrong: D is wrong because K_c² would apply if you doubled the forward reaction. This is the opposite direction and the opposite coefficient manipulation.
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Law of Equilibrium Kp Kc: quick recall before you leave
How do you solve a Law of Equilibrium Kp Kc question? A 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.)
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What to remember before solving Law of Equilibrium Kp Kc questions
Which Law of Equilibrium Kp Kc formulas do you need for NEET?
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
Where do students lose marks on Law of Equilibrium Kp Kc?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
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.
Root cause: concept gap
Correction
Δn = (moles GAS product) - (moles GAS reactant). Solids and liquids excluded.
More in Equilibrium: 6 exam traps and mistakes · 4 formulas · 3 question patterns from its other lessons.
Law of Equilibrium Kp Kc questions from past NEET papers
5 questions from NEET 2022, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
10.021
283.1
32.077 × 10⁵
40.033
NTA Answer: Option 4(final)
NEET 2024Revised key
In which of the following equilibria, K_p and K_c are NOT equal?
1PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
2H₂(g) + I₂(g) ⇌ 2HI(g)
3CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)
42BrCl(g) ⇌ Br₂(g) + Cl₂(g)
NTA Answer: Option 1(revised_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 2024Revised key
10.00889
20.0889
30.8889
40.717
NTA Answer: Option 4(revised_final)
NEET 2022
11.2 × 10²¹
24.38 × 10⁻³²
31.9 × 10⁻⁶³
42.4 × 10³¹
NTA Answer: Option 2(final)
How does NEET ask about Law of Equilibrium Kp Kc?
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
Sources
NCERT refs: Class 11 Chemistry Chapter 6, p.175
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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