This MCQ module is based on: Law Mass Action Kc Kp
TOPIC 25 OF 28
Law Mass Action Kc Kp
🎓 Class 11
Chemistry
CBSE
Theory
Ch 6 – Equilibrium
⏱ ~14 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📝 Worksheet / Assessment
▲
This assessment will be based on: Law Mass Action Kc Kp
Upload images, PDFs, or Word documents to include their content in assessment generation.
Law of Mass Action, Kc and Kp
6.5 Law of Chemical Equilibrium — Law of Mass Action
In 1864, the Norwegian chemists Cato Guldberg and Peter Waage proposed the Law of Mass Action:
Law of Mass Action: The rate of a chemical reaction is proportional to the product of the molar concentrations of the reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation.
For the elementary reaction aA + bB → cC + dD:
Forward rate: r_f = k_f [A]^a [B]^b
Reverse rate: r_r = k_r [C]^c [D]^d
At equilibrium, r_f = r_r, so:
\[k_f[A]^a[B]^b = k_r[C]^c[D]^d \quad\Rightarrow\quad \frac{k_f}{k_r} = \frac{[C]^c[D]^d}{[A]^a[B]^b} = K_c\]6.6 Equilibrium Constant K_c
Equilibrium Constant (in terms of concentration):
\[K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]
where the square brackets denote equilibrium molar concentrations (mol/L). At a given T, K_c is constant for a given reaction.
6.6.1 Significance of the Magnitude of K
| K range | Implication | Example |
|---|---|---|
| K > 10³ | Reaction nearly complete (products dominate) | 2H₂ + O₂ → 2H₂O at 298 K (K ≈ 10⁸⁰) |
| 10⁻³ < K < 10³ | Significant amounts of both reactants and products | H₂ + I₂ ⇌ 2HI at 700 K (K ≈ 50) |
| K < 10⁻³ | Reaction barely occurs (reactants dominate) | N₂ + O₂ ⇌ 2NO at 298 K (K ≈ 5 × 10⁻³¹) |
6.6.2 Properties of K
- K depends only on temperature, not on initial concentrations or pressure (for ideal systems).
- K for the reverse reaction = 1/K_forward.
- If a reaction is multiplied by n, K becomes K^n.
- For a sum of reactions, K_total = K₁ × K₂ × … (multiplication rule).
6.7 Equilibrium Constant K_p (Gas-Phase Reactions)
For reactions involving gases, partial pressures are often more convenient than concentrations:
K_p: For aA(g) + bB(g) ⇌ cC(g) + dD(g),
\[K_p = \frac{p_C^c\,p_D^d}{p_A^a\,p_B^b}\]
where p_X is the partial pressure of X at equilibrium (in atm or bar).
6.7.1 Relationship between K_p and K_c
From ideal gas law, p = (n/V) RT = c RT. Substituting:
\[\boxed{K_p = K_c (RT)^{\Delta n_g}}\]where Δn_g = (moles of gaseous products) − (moles of gaseous reactants), and R = 0.0831 L·bar K⁻¹ mol⁻¹ (or 0.0821 L·atm K⁻¹ mol⁻¹).
| Reaction | Δn_g | Relation |
|---|---|---|
| H₂(g) + I₂(g) ⇌ 2HI(g) | 0 | K_p = K_c |
| N₂(g) + 3H₂(g) ⇌ 2NH₃(g) | −2 | K_p = K_c (RT)⁻² |
| PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) | +1 | K_p = K_c (RT)¹ |
| 2NO₂(g) ⇌ N₂O₄(g) | −1 | K_p = K_c (RT)⁻¹ |
6.7.2 Units of K
K is sometimes treated as dimensionless (when concentrations/pressures are divided by their standard values 1 mol/L or 1 bar). Strictly, it can have units depending on Δn:
| Δn | Units of K_c | Units of K_p |
|---|---|---|
| 0 | dimensionless | dimensionless |
| +1 | mol L⁻¹ | bar (or atm) |
| −1 | L mol⁻¹ | bar⁻¹ |
| +2 | mol² L⁻² | bar² |
6.8 Reaction Quotient Q
The reaction quotient (Q) has the same formula as K but is evaluated at any (non-equilibrium) state:
\[Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]| If… | Then… | Reaction shifts |
|---|---|---|
| Q < K | Too few products | FORWARD (→) until Q = K |
| Q = K | At equilibrium | No net change |
| Q > K | Too many products | REVERSE (←) until Q = K |
6.9 Homogeneous & Heterogeneous Equilibria
6.9.1 Homogeneous
All species in the same phase. Examples: 2NO(g) + O₂(g) ⇌ 2NO₂(g) (all gas); CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq) (all aqueous, with water solvent).
6.9.2 Heterogeneous
Species in different phases. The activities of pure solids and pure liquids are taken as 1 (constant), so they don't appear in K:
Examples:
CaCO₃(s) ⇌ CaO(s) + CO₂(g) → K_c = [CO₂], K_p = p_{CO₂}
H₂O(l) ⇌ H₂O(g) → K_p = p_{H₂O} (at given T, the saturated VP)
3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g) → K_p = p_{H₂}⁴ / p_{H₂O}⁴
CaCO₃(s) ⇌ CaO(s) + CO₂(g) → K_c = [CO₂], K_p = p_{CO₂}
H₂O(l) ⇌ H₂O(g) → K_p = p_{H₂O} (at given T, the saturated VP)
3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g) → K_p = p_{H₂}⁴ / p_{H₂O}⁴
🎯 Interactive: K_c ↔ K_p Converter
Pick a reaction (Δn_g auto-set) and a temperature. The calculator gives you both K_c and K_p.
K_p = K_c × (RT)^Δn_g = —
Using R = 0.0831 L·bar K⁻¹ mol⁻¹ and pressures in bar.
🧪 Activity 6.2 — Compute K from ICE-Table Data
Setup: 1.00 mol N₂ and 3.00 mol H₂ are mixed in a 1.0-L vessel at 700 K. At equilibrium, 0.40 mol NH₃ is found. Calculate K_c.
Predict: Use an ICE (Initial–Change–Equilibrium) table to fill in the equilibrium concentrations.
| N₂(g) | 3H₂(g) | 2NH₃(g) | |
|---|---|---|---|
| I (mol/L) | 1.00 | 3.00 | 0 |
| C (mol/L) | −x | −3x | +2x |
| E (mol/L) | 1.00−x | 3.00−3x | 2x = 0.40 → x = 0.20 |
So [N₂] = 0.80 M, [H₂] = 2.40 M, [NH₃] = 0.40 M.
K_c = [NH₃]² / ([N₂][H₂]³) = (0.40)² / (0.80 × 2.40³) = 0.16 / (0.80 × 13.824) = 0.0145 mol⁻²·L²
Worked Example 6.3: Direction Using Q
For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), K_c = 280 at 1000 K. The mixture initially contains [SO₂] = 0.20 M, [O₂] = 0.50 M, [SO₃] = 0.10 M. In which direction will the reaction proceed?
Q_c = [SO₃]² / ([SO₂]² [O₂]) = (0.10)² / ((0.20)² × 0.50) = 0.01 / 0.02 = 0.5
Q (0.5) << K (280), so Q < K → reaction proceeds forward (→) producing more SO₃ until Q = K.
Q (0.5) << K (280), so Q < K → reaction proceeds forward (→) producing more SO₃ until Q = K.
Worked Example 6.4: K_p from K_c
For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), K_c = 0.50 mol⁻²·L² at 750 K. Calculate K_p. (Use R = 0.0831 L·bar K⁻¹ mol⁻¹)
Δn_g = 2 − (1 + 3) = −2
K_p = K_c × (RT)^Δn = 0.50 × (0.0831 × 750)^(−2) = 0.50 × (62.325)⁻² = 0.50 × 1/3884.4 = 1.29 × 10⁻⁴ bar⁻²
K_p < K_c because Δn is negative (gas moles decrease).
K_p = K_c × (RT)^Δn = 0.50 × (0.0831 × 750)^(−2) = 0.50 × (62.325)⁻² = 0.50 × 1/3884.4 = 1.29 × 10⁻⁴ bar⁻²
K_p < K_c because Δn is negative (gas moles decrease).
Worked Example 6.5: Heterogeneous Equilibrium
For CaCO₃(s) ⇌ CaO(s) + CO₂(g), the equilibrium pressure of CO₂ at 1100 K is 1.42 bar. Write K_p and find its value.
Pure solids do not appear in K. So:
K_p = p_{CO₂} = 1.42 bar.
This is why a closed lime kiln reaches a constant CO₂ pressure regardless of how much solid is loaded.
K_p = p_{CO₂} = 1.42 bar.
This is why a closed lime kiln reaches a constant CO₂ pressure regardless of how much solid is loaded.
🎯 Competency-Based Questions
Q1. The equilibrium constant K depends on: L1 Remember
Answer: (c) Temperature only. K is independent of initial concentrations, total pressure (in ideal systems), and catalyst.
Q2. For 2NO(g) + O₂(g) ⇌ 2NO₂(g), Δn_g = ____ and K_p / K_c = ____. L2 Understand
Answer: Δn_g = 2 − 3 = −1. K_p = K_c × (RT)⁻¹ → K_p / K_c = 1/(RT). At 298 K with R = 0.0821, K_p/K_c = 1/24.45 ≈ 0.041 atm⁻¹.
Q3. For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), K_c = 1.8 × 10⁻³ at 250 °C. Initially: [PCl₅] = 0.5 M; [PCl₃] = [Cl₂] = 0. Will reaction proceed? L3 Apply
Answer: Q = (0)(0)/(0.5) = 0 < K → reaction proceeds forward. PCl₅ dissociates until equilibrium is established.
Q4. The K of N₂ + 3H₂ ⇌ 2NH₃ is K₁. What is K for ½N₂ + (3/2)H₂ ⇌ NH₃? L4 Analyse
Answer: When the equation is divided by 2, K becomes √K₁ (raised to the power ½). So K_new = K₁^(1/2). And for the reverse 2NH₃ ⇌ N₂ + 3H₂, K_rev = 1/K₁.
Q5. HOT (Apply): 0.5 mol H₂ + 0.5 mol I₂ heated to 700 K in a 1-L flask. K_c = 49 for H₂ + I₂ ⇌ 2HI. Find equilibrium concentrations. L3 Apply
Solution (ICE): Let x = mol/L of H₂ reacted.
[H₂]_eq = 0.5 − x, [I₂]_eq = 0.5 − x, [HI]_eq = 2x.
K_c = (2x)² / ((0.5 − x)²) = 49
Take √: 2x / (0.5 − x) = 7 → 2x = 3.5 − 7x → 9x = 3.5 → x = 0.389 M
[H₂]_eq = [I₂]_eq = 0.111 M, [HI]_eq = 0.778 M. Verify: (0.778)² / (0.111)² = 0.605 / 0.0123 ≈ 49 ✓
[H₂]_eq = 0.5 − x, [I₂]_eq = 0.5 − x, [HI]_eq = 2x.
K_c = (2x)² / ((0.5 − x)²) = 49
Take √: 2x / (0.5 − x) = 7 → 2x = 3.5 − 7x → 9x = 3.5 → x = 0.389 M
[H₂]_eq = [I₂]_eq = 0.111 M, [HI]_eq = 0.778 M. Verify: (0.778)² / (0.111)² = 0.605 / 0.0123 ≈ 49 ✓
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: The equilibrium constant K_c does not include the concentrations of pure solids or pure liquids.
R: The activity (effective concentration) of any pure solid or liquid is 1.
Answer: (A). Both true; R explains A. This is why heterogeneous equilibria have simpler K expressions.
A: K_p = K_c when Δn_g = 0.
R: K_p = K_c (RT)^Δn_g.
Answer: (A). Both true; R explains A. Substituting Δn_g = 0 gives (RT)⁰ = 1.
A: If Q > K, the reaction proceeds in the forward direction.
R: When Q > K, [products] is too high relative to equilibrium.
Answer: (D). A is FALSE — when Q > K, products are in excess, so reaction proceeds in the REVERSE direction. R is TRUE — products are too high.
Frequently Asked Questions — Law of Mass Action, Kc and Kp
What is the law of mass action?
The law of mass action, proposed by Guldberg and Waage (1864), states that the rate of a chemical reaction is directly proportional to the product of the active masses (molar concentrations) of the reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation. For aA + bB → products, rate ∝ [A]^a [B]^b. At equilibrium, the forward and reverse rates are equal, leading to the equilibrium constant expression K_c = [C]^c [D]^d / [A]^a [B]^b. NCERT Class 11 Chemistry Chapter 6 uses this law as the foundation of all equilibrium calculations.
What is the difference between Kc and Kp?
K_c is the equilibrium constant expressed in terms of molar concentrations (mol/L) of reactants and products. K_p is the equilibrium constant expressed in terms of partial pressures (usually in bar or atm) of gaseous reactants and products. K_p is used only for reactions involving gases. The two are related by K_p = K_c (RT)^Δn_g where Δn_g = (moles of gaseous products) − (moles of gaseous reactants) and R = 0.0831 L·bar·K⁻¹·mol⁻¹. For reactions where Δn_g = 0 (e.g., H₂ + I₂ ⇌ 2HI), K_p = K_c. NCERT Class 11 Chemistry Chapter 6 covers both extensively.
What is the reaction quotient Q?
The reaction quotient (Q) has the same mathematical form as the equilibrium constant K but uses concentrations or pressures at any moment, not necessarily at equilibrium. Comparing Q with K predicts the direction in which the reaction will proceed to reach equilibrium: (1) if Q < K, the forward reaction predominates (more products form); (2) if Q > K, the reverse reaction predominates (more reactants form); (3) if Q = K, the system is already at equilibrium. NCERT Class 11 Chemistry Chapter 6 uses Q frequently to predict reaction direction from initial concentrations.
What is a heterogeneous equilibrium?
A heterogeneous equilibrium involves substances in two or more phases (gas, liquid, solid). Examples from NCERT Class 11 Chemistry Chapter 6: CaCO₃(s) ⇌ CaO(s) + CO₂(g); H₂O(l) ⇌ H₂O(g); NH₄Cl(s) ⇌ NH₃(g) + HCl(g); FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g). In equilibrium expressions, the concentrations (or activities) of pure solids and pure liquids are taken as constant and incorporated into K. So for CaCO₃ decomposition: K_p = P_CO₂. Only gases and aqueous solutes appear in K expressions for heterogeneous equilibria.
What is the significance of the magnitude of K?
The magnitude of the equilibrium constant K tells us how far the reaction proceeds before reaching equilibrium: (1) K > 10³ — reaction goes nearly to completion, mostly products; (2) K < 10⁻³ — reaction barely proceeds, mostly reactants; (3) K between 10⁻³ and 10³ — significant amounts of both reactants and products at equilibrium. A large K does not mean fast — it indicates product favourability, not rate. NCERT Class 11 Chemistry Chapter 6 emphasises this distinction: kinetics (rate) and thermodynamics (equilibrium position) are separate concepts. K depends only on temperature, not on initial concentrations.
What are the units of Kc and Kp?
The units of K_c and K_p depend on Δn (change in moles) of the balanced reaction. K_c has units of (mol/L)^Δn where Δn = (moles of products) − (moles of reactants); K_p has units of (bar)^Δn_g. For example, for N₂ + 3H₂ ⇌ 2NH₃ (Δn = 2 − 4 = −2), K_c has units (mol/L)⁻² = L²/mol². For H₂ + I₂ ⇌ 2HI (Δn = 0), K_c is dimensionless. NCERT Class 11 Chemistry strictly speaking treats K as dimensionless when concentrations are divided by standard concentration (1 mol/L). Always state units when asked.
AI Tutor
Chemistry Class 11 Part I – NCERT (2025-26)
Ready