Welcome to our complete guide on Class 11 Chemistry Chapter 6 Equilibrium NCERT Solutions. This chapter explains important concepts such as chemical equilibrium, equilibrium constant, Le Chatelier’s principle, and ionic equilibrium including pH, buffer solutions, and solubility product.
All the in-text and numerical problems are solved clearly to help CBSE students strengthen their understanding and perform well in exams. These Class 11 Chemistry Chapter 6 Equilibrium NCERT Solutions are prepared as per the latest syllabus and are ideal for revision and concept clarity.
Class 11 Chemistry Chapter 6 Equilibrium NCERT Textbook Answers
6.1
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
(a) What is the initial effect of the change on vapour pressure?
(b) How do rates of evaporation and condensation change initially?
(c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Answer 6.1 (step-by-step)
Context: At the given temperature the liquid and its vapour were in dynamic equilibrium. That means vapour pressure = equilibrium (saturation) pressure at that temperature, and the rates of evaporation and condensation are equal.
(a) Initial effect on vapour pressure:
When the volume of the container is suddenly increased, the same number of vapour molecules now occupy a larger volume. The partial pressure of the vapour therefore falls immediately (instantaneous drop in vapour pressure).
(b) How the rates change initially:
Because the vapour pressure has dropped below the saturation pressure at that temperature, there are now fewer vapour molecules per unit volume to strike the liquid surface. So the rate of condensation decreases. At the same time, the liquid “sees” that the vapour pressure is lower than equilibrium and so net evaporation increases (more molecules leave the liquid than return). Therefore: evaporation rate increases, condensation rate decreases (initially).
(c) What happens finally and final vapour pressure:
Evaporation continues at a higher rate than condensation until enough liquid has evaporated that the vapour partial pressure rises back to the saturation (equilibrium) value at that temperature. At that point the rates of evaporation and condensation become equal again and dynamic equilibrium is restored. The final vapour pressure equals the original equilibrium (saturation) vapour pressure at that temperature — it does not change with the temporary volume change (so long as temperature is fixed and some liquid remains).
Summary in one line: volume increase → immediate drop in vapour pressure → net evaporation (evaporation > condensation) → vapour pressure rises back → finally vapour pressure = original saturation pressure.
6.2
What is Kc for the following equilibrium when the equilibrium concentrations of each substance are: [SO2] = 0.60 M, [O2] = 0.82 M and [SO3] = 1.90 M?
Reaction: 2 SO2(g) + O2(g) ⇌ 2 SO3(g)
Answer 6.2 (step-by-step)
Definition: For a reaction aA + bB ⇌ cC, the equilibrium constant Kc (in terms of concentration) is
Kc = [C]^c / ([A]^a [B]^b).
Apply to the given reaction:
Kc = [SO3]^2 / ( [SO2]^2 × [O2] ).
Now substitute numbers carefully (digit-by-digit arithmetic):
[SO3] = 1.90 ; so [SO3]^2 = 1.90 × 1.90 = 3.61.
[SO2] = 0.60 ; so [SO2]^2 = 0.60 × 0.60 = 0.36.
[O2] = 0.82.
Denominator = 0.36 × 0.82 = 0.2952.
So Kc = 3.61 / 0.2952.
Do the division: 3.61 ÷ 0.2952 ≈ 12.23 (rounded to three significant figures).
Final answer: Kc ≈ 12.23.
(You can report Kc = 12.2 if you want 3 s.f.)
6.3
At a certain temperature and at a total pressure of 10^5 Pa, iodine vapour contains 40% by volume of I atoms.
Reaction: I2(g) ⇌ 2 I(g)
Calculate Kp for the equilibrium.
Answer 6.3 (step-by-step)
Interpretation: “40% by volume of I atoms” means that at equilibrium 40% of the gas (by mole fraction) consists of atomic iodine I, and the remaining 60% (by mole fraction) is molecular iodine I2. For gases, volume percent = mole fraction percent.
Given: total pressure P_total = 10^5 Pa.
Therefore partial pressures:
p(I) = 0.40 × 10^5 Pa = 4.0 × 10^4 Pa.
p(I2) = 0.60 × 10^5 Pa = 6.0 × 10^4 Pa.
Reaction equilibrium expression in terms of pressures (Kp):
For I2 ⇌ 2 I, Kp = (p_I)^2 / p_I2.
Substitute values:
(p_I)^2 = (4.0 × 10^4)^2 = 1.6 × 10^9 (Pa^2).
Divide by p_I2 = 6.0 × 10^4 Pa.
Kp = 1.6 × 10^9 / 6.0 × 10^4 = (1.6 / 6.0) × 10^(9−4) = 0.266666… × 10^5 = 2.6667 × 10^4 Pa.
Final answer: Kp ≈ 2.67 × 10^4 Pa.
(If you prefer Kp in kPa: Kp ≈ 26.7 kPa.)
6.4
Write the expression for the equilibrium constant, Kc, for each of the following reactions.
(i) 2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)
(ii) 2 Cu(NO3)2 (s) ⇌ 2 CuO (s) + 4 NO2 (g) + O2 (g)
(iii) CH3COOC2H5(aq) + H2O(l) ⇌ CH3COOH(aq) + C2H5OH(aq)
(iv) Fe3+(aq) + 3 OH−(aq) ⇌ Fe(OH)3(s)
(v) I2(s) + 5 F2(g) ⇌ 2 IF5(g)
Answer 6.4 (expressions and notes)
General rule: In Kc expression include only concentrations (activities) of species in the gaseous or aqueous phase. Pure solids and pure liquids are omitted (their activity ≈ 1), so they do not appear explicitly in Kc expressions.
(i) 2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)
Kc = [NO]^2 × [Cl2] / [NOCl]^2
(ii) 2 Cu(NO3)2 (s) ⇌ 2 CuO (s) + 4 NO2 (g) + O2 (g)
Both Cu(NO3)2 and CuO are solids → omit them.
Kc = [NO2]^4 × [O2]
(iii) CH3COOC2H5(aq) + H2O(l) ⇌ CH3COOH(aq) + C2H5OH(aq)
H2O is liquid → omit. Include aqueous species:
Kc = [CH3COOH] × [C2H5OH] / [CH3COOC2H5]
(iv) Fe3+(aq) + 3 OH−(aq) ⇌ Fe(OH)3(s)
Product is a solid → omit product. Therefore Kc is inverse of the ion product:
Kc = 1 / ( [Fe3+] × [OH−]^3 )
(Alternative wording: If the reaction were written as Fe(OH)3(s) ⇌ Fe3+ + 3 OH−, then the equilibrium constant (solubility product Ksp) would be Ksp = [Fe3+] × [OH−]^3. But as written here (ions → solid), Kc = 1/Ksp.)
(v) I2(s) + 5 F2(g) ⇌ 2 IF5(g)
I2 is a solid → omit it. Then Kc = [IF5]^2 / [F2]^5
6.5
Find out the value of Kc for each of the following equilibria from the value of Kp given:
(i) 2 NOCl(g) ⇌ 2 NO(g) + Cl2(g); Kp = 1.8 × 10^−2 at 500 K.
(ii) CaCO3(s) ⇌ CaO(s) + CO2(g); Kp = 167 at 1073 K.
Answer 6.5 (complete)
Concept and formula used:
The relation between Kp and Kc is
Kp = Kc × (R T)^{Δn}
where Δn = (sum of stoichiometric coefficients of gaseous products) − (sum of stoichiometric coefficients of gaseous reactants),
R is the gas constant and T is temperature in kelvin.
Rearranged:
Kc = Kp ÷ (R T)^{Δn}.
Important note about units: the R used must match the pressure units implied by Kp. A common and convenient choice is R = 0.08206 L·atm·mol^−1·K^−1 and to treat Kp in atm. If Kp is reported in different pressure units, convert it to atm before using this R. Below we assume the given Kp values are expressed with the usual convention consistent with using R = 0.08206 L·atm·mol^−1·K^−1.
(i) Reaction: 2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)
Step 1 — Compute Δn: gaseous products = 2 (NO) + 1 (Cl2) = 3; gaseous reactants = 2 (NOCl) = 2.
So Δn = 3 − 2 = 1.
Step 2 — Compute R T: R = 0.08206 L·atm·mol^−1·K^−1, T = 500 K.
R T = 0.08206 × 500 = 41.03 (units: L·atm·mol^−1).
Step 3 — Use formula Kc = Kp ÷ (R T)^{Δn} = Kp ÷ (R T) because Δn = 1.
Kc = 1.8 × 10^−2 ÷ 41.03
= 0.018 ÷ 41.03
≈ 4.39 × 10^−4
Answer (i): Kc ≈ 4.39 × 10^−4 (in units of mol L^−1, consistent with Δn = 1).
(ii) Reaction: CaCO3(s) ⇌ CaO(s) + CO2(g)
Step 1 — Identify gases and Δn: only CO2 appears as gas. Gaseous products = 1 (CO2). Gaseous reactants = 0. So Δn = 1 − 0 = 1.
Step 2 — Compute R T at T = 1073 K:
R T = 0.08206 × 1073 = 88.04 (approx).
Step 3 — Kc = Kp ÷ (R T) because Δn = 1.
Kc = 167 ÷ 88.04 ≈ 1.897 ≈ 1.90
Answer (ii): Kc ≈ 1.90 (units consistent with one mole of gas produced; often called mol L^−1).
Short remark about units: Because Δn = 1 in both parts, Kc has units of concentration (mol L^−1) in the way it is obtained here. If your source reports Kp in units other than atm, convert Kp to atm before using R = 0.08206. The numerical answers above assume Kp values are compatible with R = 0.08206.
6.6
For the following equilibrium, Kc = 6.3 × 10^14 at 1000 K:
NO(g) + O3(g) ⇌ NO2(g) + O2(g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc for the reverse reaction?
Answer 6.6 (complete)
If for the forward reaction A ⇌ B we have equilibrium constant K(forward) = Kf, then for the reverse reaction B ⇌ A the equilibrium constant is the reciprocal:
K(reverse) = 1 / K(forward).
Given Kc(forward) = 6.3 × 10^14, therefore
Kc(reverse) = 1 ÷ (6.3 × 10^14) = 1.5873 × 10^−15 ≈ 1.59 × 10^−15.
Answer: Kc for the reverse reaction ≈ 1.59 × 10^−15.
(Units: same units as Kc for forward reaction — the reciprocal is unit-consistent.)
6.7
Explain why pure liquids and pure solids can be ignored while writing the equilibrium constant expression.
Answer 6.7 (complete)
Idea / physical reason: The equilibrium constant expression is written in terms of activities of species. The activity of a pure solid or a pure liquid (in its standard state) is essentially constant and equal to 1 at a given temperature and pressure. This is because the “effective concentration” (activity) of a pure phase does not change when its amount changes, as long as some of that pure phase remains present.
Practical consequence: Because their activity = 1 (constant), pure solids and pure liquids do not change the value of the equilibrium expression and therefore can be omitted from the algebraic form of Kc or Kp. Only species whose concentrations (or partial pressures) can change (gases and solutes in solution) are included explicitly.
Example: For the reaction CaCO3(s) ⇌ CaO(s) + CO2(g), the equilibrium constant involves only the gaseous CO2:
K = p(CO2) (or Kc = [CO2] if written in concentration form using appropriate conversion), because activities of CaCO3(s) and CaO(s) are taken as 1 and omitted.
One-sentence summary: Pure solids and liquids are omitted from equilibrium expressions because their activities remain essentially constant (≈1) and therefore do not affect the equilibrium ratio.
Reaction between N2 and O2 takes place as follows:
2 N2(g) + O2(g) ⇌ 2 N2O(g).
If a mixture of 0.482 mol N2 and 0.933 mol O2 is placed in a 10.0 L reaction vessel and allowed to form N2O at a temperature for which Kc = 2.0 × 10^−37, determine the composition of the equilibrium mixture.
Answer 6.8 (complete)
Step 1 — Initial concentrations (mol per L):
Volume = 10.0 L.
[N2]_initial = 0.482 mol / 10.0 L = 0.0482 M.
[O2]_initial = 0.933 mol / 10.0 L = 0.0933 M.
[N2O]_initial = 0 (none initially).
Step 2 — Let y = moles of N2O formed at equilibrium, so concentration [N2O] = y/10 = 0.1 y (M).
Stoichiometry: 2 N2 + O2 → 2 N2O
So change in concentration:
[N2]_eq = 0.0482 − 2 × (y/10) = 0.0482 − 0.2 y.
[O2]_eq = 0.0933 − (y/10) = 0.0933 − 0.1 y.
[N2O]_eq = 0.1 y.
Step 3 — Write Kc expression for the reaction:
Kc = [N2O]^2 / ( [N2]^2 × [O2] ).
Step 4 — Because Kc is extremely small (2.0 × 10^−37), the extent of reaction will be negligibly small. We therefore approximate by taking denominator at initial values (y ≈ 0 in denominator). This gives a very good approximation.
Approximate denominator = (0.0482)^2 × 0.0933.
Compute (digit by digit):
0.0482^2 = 0.00232324.
0.00232324 × 0.0933 ≈ 0.00021676 (approx).
So with numerator (0.1 y)^2 = 0.01 y^2,
Kc ≈ 0.01 y^2 / 0.00021676.
Solve for y^2:
y^2 ≈ Kc × 0.00021676 / 0.01
= (2.0 × 10^−37) × 0.00021676 / 0.01
≈ (2.0 × 10^−37) × 0.021676
≈ 4.335 × 10^−39.
Therefore y ≈ sqrt(4.335 × 10^−39) ≈ 6.58 × 10^−20 mol.
Step 5 — Equilibrium amounts (moles, to show practically no reaction):
N2O formed ≈ 6.6 × 10^−20 mol (effectively zero).
N2 remaining ≈ 0.482 − 2×(6.6×10^−20) ≈ 0.482 mol.
O2 remaining ≈ 0.933 − (6.6×10^−20) ≈ 0.933 mol.
Step 6 — Equilibrium concentrations (mol L^−1):
[N2]_eq ≈ 0.0482 M, [O2]_eq ≈ 0.0933 M, [N2O]_eq ≈ 6.6 × 10^−21 M.
Conclusion: Because Kc is vanishingly small, practically no N2O forms; the equilibrium mixture is essentially the original reactants.
6.9
Nitric oxide reacts with bromine and gives nitrosyl bromide as per the reaction:
2 NO(g) + Br2(g) ⇌ 2 NOBr(g)
When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate the equilibrium amounts (moles) of NO and Br2.
Answer 6.9 (complete)
Step 1 — Reaction stoichiometry: For every 2 mol NO consumed and 1 mol Br2 consumed, 2 mol NOBr are formed.
Step 2 — Let extent ξ be such that 2 ξ = moles of NOBr formed. Given NOBr formed = 0.0518 mol, so
ξ = 0.0518 / 2 = 0.0259 mol.
Step 3 — Changes in reactants:
NO consumed = 2 ξ = 0.0518 mol.
Br2 consumed = ξ = 0.0259 mol.
Step 4 — Equilibrium amounts:
NO_eq = NO_initial − NO_consumed = 0.087 − 0.0518 = 0.0352 mol.
Br2_eq = Br2_initial − Br2_consumed = 0.0437 − 0.0259 = 0.0178 mol.
NOBr_eq = 0.0518 mol (given).
Final answer: At equilibrium: NO = 0.0352 mol, Br2 = 0.0178 mol, NOBr = 0.0518 mol.
6.10
At 450 K, Kp = 2.0 × 10^10 per bar for the reaction at equilibrium:
2 SO2(g) + O2(g) ⇌ 2 SO3(g).
What is Kc at this temperature?
Answer 6.10 (complete)
Step 1 — Relation between Kp and Kc:
Kp = Kc × (R T)^{Δn} where Δn = (moles gas products) − (moles gas reactants).
For 2 SO2 + O2 ⇌ 2 SO3, Δn = 2 − (2 + 1) = −1.
Therefore Kp = Kc × (R T)^{−1} = Kc / (R T). Rearranged: Kc = Kp × (R T).
Step 2 — Use R appropriate to the pressure units. Kp is per bar, so use R = 0.08314 L·bar·mol^−1·K^−1 (often given as 0.08314).
T = 450 K, so R T = 0.08314 × 450 = 37.413 L·bar·mol^−1.
Step 3 — Compute Kc:
Kc = Kp × (R T) = (2.0 × 10^10 per bar) × 37.413 ≈ 7.48 × 10^11.
Final answer: Kc ≈ 7.48 × 10^11 (in the concentration units consistent with Δn = −1 — effectively L·mol^−1 units built into the value).
6.11
A sample of HI(g) is placed in a flask at an initial pressure of 0.20 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. For the equilibrium:
2 HI(g) ⇌ H2(g) + I2(g)
What is Kp for this equilibrium?
Answer 6.11 (complete)
Step 1 — Initial partial pressures: P_HI(initial) = 0.20 atm. At equilibrium P_HI(eq) = 0.04 atm. The decrease in HI pressure = 0.20 − 0.04 = 0.16 atm.
Step 2 — Stoichiometry: 2 HI → H2 + I2. If the extent of decomposition corresponds to x in partial pressure units, the change in HI pressure = 2 x = 0.16 atm → x = 0.08 atm. Then at equilibrium:
P_H2 = x = 0.08 atm.
P_I2 = x = 0.08 atm.
P_HI = 0.04 atm (given).
Step 3 — Expression for Kp:
Kp = (P_H2 × P_I2) / (P_HI)^2.
Substitute values:
Kp = (0.08 × 0.08) / (0.04)^2 = 0.0064 / 0.0016 = 4.
Final answer: Kp = 4.0.
6.12
A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20.0 L reaction vessel at 500 K. At this temperature the equilibrium constant Kc for the reaction
N2(g) + 3 H2(g) ⇌ 2 NH3(g)
is 1.7 × 10^2. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
Answer 6.12 (complete)
Step 1 — Compute initial concentrations (mol L^−1):
Volume = 20.0 L.
[N2]_0 = 1.57 / 20 = 0.0785 M.
[H2]_0 = 1.92 / 20 = 0.0960 M.
[NH3]_0 = 8.13 / 20 = 0.4065 M.
Step 2 — Reaction quotient Qc (same form as Kc):
For N2 + 3 H2 ⇌ 2 NH3,
Kc = [NH3]^2 / ( [N2] × [H2]^3 ).
Compute Qc with initial concentrations:
Numerator: [NH3]^2 = (0.4065)^2 ≈ 0.16524 (compute: 0.4065 × 0.4065 ≈ 0.16524).
Denominator: [N2] × [H2]^3 = 0.0785 × (0.0960)^3.
Compute (0.0960)^2 = 0.009216; (0.0960)^3 = 0.009216 × 0.0960 = 0.000884736.
Now denominator = 0.0785 × 0.000884736 ≈ 0.000069452 (compute precisely: ≈ 6.9452 × 10^−5).
Therefore Qc = 0.16524 / 0.000069452 ≈ 2379.6 ≈ 2.38 × 10^3.
Step 3 — Compare Qc with Kc:
Kc = 1.7 × 10^2 = 170. Qc ≈ 2380, which is much larger than Kc.
Step 4 — Conclusion on direction:
If Qc > Kc, the reaction will shift to the left (toward reactants) to reach equilibrium. That means NH3 will decompose into N2 and H2 until Qc decreases to Kc.
Final answer: The mixture is not at equilibrium. Since Qc (≈ 2.38 × 10^3) > Kc (1.7 × 10^2), the net reaction proceeds in the reverse direction (toward formation of N2 and H2).
6.13
The equilibrium constant expression for a gas reaction is:
Kc = [NH3]^4 [O2]^5 / ( [NO]^4 [H2O]^6 ).
Write the balanced chemical equation corresponding to this expression.
Answer 6.13 (complete)
Step 1 — Read exponents as stoichiometric coefficients. The numerator has NH3 to the power 4 and O2 to the power 5 (these are products). The denominator has NO^4 and H2O^6 (these are reactants).
Thus the balanced equation is:
4 NO(g) + 6 H2O(g) ⇌ 4 NH3(g) + 5 O2(g).
Step 2 — Quick atom check:
Nitrogen: left 4 (from 4 NO), right 4 (from 4 NH3) → OK.
Hydrogen: left 6×2 = 12, right 4×3 = 12 → OK.
Oxygen: left from NO: 4×1 = 4, from H2O: 6×1 = 6 → total left = 10; right: 5×2 = 10 → OK.
Final balanced equation: 4 NO(g) + 6 H2O(g) ⇌ 4 NH3(g) + 5 O2(g).
Problem 6.1
Q: Write the expression for the equilibrium constant for the following reactions:
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g)
(ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
A:
(i) Kc = [NO]²[Cl₂] / [NOCl]²
(ii) Kc = [CO₂] (solids are not included in the equilibrium expression)
Problem 6.2
Q: What will be the value of Kp for the reaction:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Given: Kc = 4.2 × 10¹⁰ at 727°C.
A:
Kp = Kc(RT)^Δn
Δn = 2 – (2 + 1) = –1
T = 727 + 273 = 1000 K
Kp = 4.2 × 10¹⁰ / (0.0821 × 1000)
Kp ≈ 5.12 × 10⁸
Problem 6.3
Q: At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms. Calculate Kp for the equilibrium:
I₂(g) ⇌ 2I(g)
A:
Let total moles = 1 mol I₂ initially
At equilibrium:
I₂ = 0.6 mol
I = 0.4 × 2 = 0.8 mol
Total = 1.4 mol
P(I₂) = 0.6 / 1.4 = 0.428 atm
P(I) = 0.8 / 1.4 = 0.571 atm
Kp = (0.571)² / 0.428 ≈ 0.76
Problem 6.4
Q: Write the expressions for the equilibrium constants for the dissociation of the following acids:
(i) HF
(ii) HCl
(iii) HCN
(iv) HNO₃
A:
(i) HF ⇌ H⁺ + F⁻; K = [H⁺][F⁻] / [HF]
(ii) HCl ⇌ H⁺ + Cl⁻; K = [H⁺][Cl⁻] / [HCl]
(iii) HCN ⇌ H⁺ + CN⁻; K = [H⁺][CN⁻] / [HCN]
(iv) HNO₃ ⇌ H⁺ + NO₃⁻; K = [H⁺][NO₃⁻] / [HNO₃]
Problem 6.5
Q: Mention two factors that affect the equilibrium constant.
A:
Only temperature affects the equilibrium constant.
Catalysts and pressure do not affect the equilibrium constant.
Problem 6.6
Q: For the equilibrium:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
What will be the effect on the equilibrium if:
(i) More N₂ is added
(ii) Pressure is increased
(iii) Temperature is decreased
A:
(i) Equilibrium shifts to the right (forward direction)
(ii) Equilibrium shifts to the right (fewer gas molecules)
(iii) Equilibrium shifts to the right (exothermic reaction)
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