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In this unit, you have learned about how reversible reactions reach a dynamic balance and how to express this using equilibrium constants, along with the factors affecting equilibrium, such as temperature and pressure, along with Le Chatelier’s Principle, all of which are essential for understanding reactions in chemistry and real-life applications. Below, we have presented some in-text exercises taken from the NCERT Class 11 Chemistry (Part I), Chapter 6: Equilibrium.
Contents
Explore Notes of Class 11 Chemistry
NCERT Solutions Class 11 Chemistry (Part-1) Chapter 6: Equilibrium
Below, we have provided you with the exercises mentioned in the NCERT Class 11 Chemistry (Part-1) Chapter 6: Equilibrium.
Exercises
- 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 finally restored, and what will be the final vapour pressure? - What is Kc for the following equilibrium when the equilibrium concentrations are:
[SO₂] = 0.60 M, [O₂] = 0.82 M and [SO₃] = 1.90 M?
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) - At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms.
I₂(g) ⇌ 2I(g)
Calculate Kp for the equilibrium. - Write the expression for the equilibrium constant, Kc, for each of the following reactions:
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g)
(ii) 2Cu(NO₃)₂(s) ⇌ 2CuO(s) + 4NO₂(g) + O₂(g)
(iii) CH₃COOC₂H₅(aq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(aq)
(iv) Fe³⁺(aq) + 3OH⁻(aq) ⇌ Fe(OH)₃(s)
(v) I₂(s) + 5F₂ ⇌ 2IF₅ - Find the value of Kc for each of the following equilibria from the value of Kp:
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); Kp = 1.8 × 10⁻² at 500 K
(ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); Kp = 167 at 1073 K - For the reaction:
NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g); Kc = 6.3 × 10¹⁴ at 1000 K
What is Kc for the reverse reaction? - Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression.
- 2N₂(g) + O₂(g) ⇌ 2N₂O(g)
A mixture of 0.482 mol N₂ and 0.933 mol of O₂ is placed in a 10 L vessel at a temperature where Kc = 2.0 × 10⁻³⁷. Determine the composition of the equilibrium mixture. - 2NO(g) + Br₂(g) ⇌ 2NOBr(g)
0.087 mol NO and 0.0437 mol Br₂ are mixed. At equilibrium, 0.0518 mol NOBr is formed. Calculate equilibrium amounts of NO and Br₂. - 2SO₂(g) + O₂(g) ⇌ 2SO₃(g); Kp = 2.0 × 10¹⁰ at 450 K
What is Kc at this temperature? - 2HI(g) ⇌ H₂(g) + I₂(g)
HI is placed at 0.2 atm; at equilibrium, its pressure is 0.04 atm. Calculate Kp. - N₂(g) + 3H₂(g) ⇌ 2NH₃(g); Kc = 1.7 × 10² at 500 K
1.57 mol N₂, 1.92 mol H₂, 8.13 mol NH₃ in 20 L vessel. Is the mixture at equilibrium? If not, what is the direction of the net reaction? - Write the balanced chemical equation for the equilibrium expression:
Kc = [NH₃]⁴ / ([NO]⁶ [H₂O]⁴) - H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)
1 mol H₂O and 1 mol CO in a 10 L vessel at 725 K. 40% H₂O reacts. Calculate Kc. - H₂(g) + I₂(g) ⇌ 2HI(g); Kc = 54.8 at 700 K
If 0.5 mol L⁻¹ of HI is present at equilibrium, calculate [H₂] and [I₂]. - 2ICl(g) ⇌ I₂(g) + Cl₂(g); Kc = 0.14
Initial [ICl] = 0.78 M. Calculate equilibrium concentrations. - C₂H₆(g) ⇌ C₂H₄(g) + H₂(g); Kp = 0.04 atm at 899 K
C₂H₆ is placed at 4.0 atm. Find the equilibrium concentration. - CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)
(i) Write Qc.
(ii) Initial: 1 mol CH₃COOH, 0.18 mol C₂H₅OH; equilibrium: 0.171 mol ester. Find Kc.
(iii) Initial: 1 mol acid, 0.5 mol alcohol; equilibrium: 0.214 mol ester. Has equilibrium been reached? - PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); Kc = 8.3 × 10⁻³
[PCl₅] = 0.5 × 10⁻¹ mol/L at 473 K. Find [PCl₃] and [Cl₂]. - FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g); Kp = 0.265 atm at 1050 K
Initial: pCO = 1.4 atm, pCO₂ = 0.80 atm. Find equilibrium partial pressures. - N₂(g) + 3H₂(g) ⇌ 2NH₃(g); Kc = 0.061
[ N₂ ] = 3.0 M, [ H₂ ] = 2.0 M, [ NH₃ ] = 0.5 M. Is this at equilibrium? If not, in which direction does the reaction proceed? - 2BrCl(g) ⇌ Br₂(g) + Cl₂(g); Kc = 32 at 500 K
Initial [BrCl] = 3.3 × 10⁻³ M. Find equilibrium [BrCl]. - C(s) + CO₂(g) ⇌ 2CO(g)
At 1127 K and 1 atm, CO is 90.55% by mass. Calculate Kc. - NO(g) + ½O₂(g) ⇌ NO₂(g)
Given ΔG° values: NO₂ = 52.0, NO = 87.0, O₂ = 0 (kJ/mol). Calculate (a) ΔG°, (b) K for the reaction at 298 K. - Predict the change in the number of moles when the pressure is decreased by increasing volume:
(a) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
(b) CaO(s) + CO₂(g) ⇌ CaCO₃(s)
(c) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g) - Which reactions will be affected by increasing pressure? Indicate forward/backwards:
(i) COCl₂(g) ⇌ CO(g) + Cl₂(g)
(ii) CH₄(g) + 2S₂(g) ⇌ CS₂(g) + 2H₂S(g)
(iii) CO₂(g) + C(s) ⇌ 2CO(g)
(iv) 2H₂(g) + CO(g) ⇌ CH₃OH(g)
(v) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
(vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g) - H₂(g) + Br₂(g) ⇌ 2HBr(g); K = 1.6 × 10⁵ at 1024 K
10 bars of HBr introduced. Find equilibrium pressures. - CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)
(a) Write the Kp expression.
(b) How does Kp and composition change with:
(i) increasing pressure
(ii) increasing temperature
(iii) adding a catalyst? - 2H₂(g) + CO(g) ⇌ CH₃OH(g)
Describe the effect on the equilibrium of:
a) addition of H₂
b) addition of CH₃OH
c) removal of CO
d) removal of CH₃OH - PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); Kc = 8.3 × 10⁻³ at 473 K
ΔH° = +124.0 kJ mol⁻¹
a) Write the Kc expression.
b) Value of Kc for reverse reaction.
c) Effect on Kc if:
(i) more PCl₅ added
(ii) pressure increased
(iii) temperature increased - CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g); Kp = 10.1 at 400°C
Initial: pCO = pH₂O = 4.0 bar. Find pH₂ at equilibrium. - Predict which reactions will have appreciable concentrations of both reactants and products:
a) Cl₂(g) ⇌ 2Cl(g); Kc = 5 × 10⁻³⁹
b) Cl₂(g) + 2NO(g) ⇌ 2NOCl(g); Kc = 3.7 × 10⁸
c) Cl₂(g) + 2NO₂(g) ⇌ 2NO₂Cl(g); Kc = 1.8 - The value of Kc for the reaction 3O₂(g) ⇌ 2O₃(g) is 2.0 × 10⁻⁵⁰ at 25°C.
If the equilibrium concentration of O₂ in air at 25°C is 1.6 × 10⁻² M, what is the concentration of O₃? - The reaction, CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g) is at equilibrium at 1300 K in a 1 L flask.
It contains 0.30 mol CO, 0.10 mol H₂, 0.02 mol H₂O and an unknown amount of CH₄.
Determine the concentration of CH₄ in the mixture. Kc = 3.90 at 1300 K. - What is meant by the conjugate acid-base pair?
Find the conjugate acid/base for the following species:
HNO₂, CN⁻, HClO₄, F⁻, OH⁻, CO₃²⁻ and S²⁻ - Which of the following are Lewis acids? H₂O, BF₃, H⁺, and NH₄⁺
- What will be the conjugate bases for the Brönsted acids: HF, H₂SO₄, and HCO₃⁻?
- Write the conjugate acids for the following Brönsted bases: NH₂⁻, NH₃, and HCOO⁻.
- The species: H₂O, HCO₃⁻, HSO₄⁻, and NH₃ can act both as Brönsted acids and bases.
For each, give the corresponding conjugate acid and conjugate base. - Classify the following species into Lewis acids and Lewis bases and show how they act:
(a) OH⁻ (b) F⁻ (c) H⁺ (d) BCl₃ - The concentration of hydrogen ions in a sample of soft drink is 3.8 × 10⁻³ M.
What is its pH? - The pH of a sample of vinegar is 3.76.
Calculate the concentration of hydrogen ions in it. - Theionisation constants at 298 K are:
HF = 6.8 × 10⁻⁴, HCOOH = 1.8 × 10⁻⁴, HCN = 4.8 × 10⁻⁹
Calculate the ionisation constants of their conjugate bases. - The ionisation constant of phenol is 1.0 × 10⁻¹⁰.
What is the concentration of phenolate ion in a 0.05 M phenol solution?
What will be its degree of ionisation if the solution also contains 0.01 M sodium phenolate? - The first ionisation constant of H₂S is 9.1 × 10⁻⁸.
Calculate the concentration of HS⁻ in a 0.1 M H₂S solution.
How is it affected if the solution also contains 0.1 M HCl?
If the second ionisation constant of H₂S is 1.2 × 10⁻¹³, calculate the concentration of S²⁻ in both cases. - The ionisation constant of acetic acid is 1.74 × 10⁻⁵.
Calculate the degree of dissociation of acetic acid in a 0.05 M solution.
Also, calculate the concentration of the acetate ion and the pH of the solution. - A 0.01 M solution of an organic acid has a pH of 4.15.
Calculate the concentration of anion, the ionisation constant of the acid, and its pKa. - Assuming complete dissociation, calculate the pH of:
(a) 0.003 M HCl
(b) 0.005 M NaOH
(c) 0.002 M HBr
(d) 0.002 M KOH - Calculate the pH of the following solutions:
(a) 2 g of TlOH in 2 L of solution
(b) 0.3 g of Ca(OH)₂ in 500 mL solution
(c) 0.3 g of NaOH in 200 mL of solution
(d) 1 mL of 13.6 M HCl diluted to 1 L - The degree of ionisation of 0.1 M bromoacetic acid is 0.132.
Calculate the pH and pKa of the acid. - The pH of 0.005 M codeine (C₁₈H₂₁NO₃) is 9.95.
Calculate its ionisation constant and pKb. - What is the pH of a 0.001 M aniline solution?
Using the ionisation constant of aniline, calculate:
Degree of ionisation
Ionisation constant of its conjugate acid - Calculate the degree of ionisation of 0.05 M acetic acid, if pKa = 4.74.
How is it affected when the solution contains:
(a) 0.01 M HCl
(b) 0.1 M HCl - The ionisation constant of dimethylamine is 5.4 × 10⁻⁴.
Calculate its degree of ionisation in a 0.02 M solution.
What percentage is ionised if the solution also contains 0.1 M NaOH? - Calculate the hydrogen ion concentration in the following fluids:
(a) Human muscle fluid, pH = 6.83
(b) Human stomach fluid, pH = 1.2
(c) Human blood, pH = 7.38
(d) Human saliva, pH = 6.4 - The pH values are:
Milk = 6.8, Black coffee = 5.0, Tomato juice = 4.2, Lemon juice = 2.2, Egg white = 7.8
Calculate the hydrogen ion concentration in each. - 0.561 g of KOH is dissolved in water to make 200 mL of solution at 298 K.
Calculate the concentrations of potassium, hydrogen, and hydroxyl ions.
What is the pH? - The solubility of Sr(OH)₂ at 298 K is 19.23 g/L.
Calculate the concentrations of Sr²⁺ and OH⁻ ions and the pH. - The ionisation constant of propanoic acid is 1.32 × 10⁻⁵.
Calculate the degree of ionisation and pH of its 0.05 M solution.
What will be the degree of ionisation if the solution also contains 0.01 M HCl? - The pH of 0.1 M cyanic acid (HCNO) is 2.34.
Calculate the ionisation constant and the degree of ionisation. - The ionisation constant of nitrous acid is 4.5 × 10⁻⁴.
Calculate the pH of a 0.04 M sodium nitrite solution and its degree of hydrolysis. - A 0.02 M pyridinium hydrochloride solution has a pH = 3.44.
Calculate the ionisation constant of pyridine. - Predict whether the solutions are neutral, acidic, or basic:
NaCl, KBr, NaCN, NH₄NO₃, NaNO₂, KF - The ionisation constant of chloroacetic acid is 1.35 × 10⁻³.
Calculate the pH of its 0.1 M acid solution and its 0.1 M sodium salt solution. - Ionic product of water at 310 K is 2.7 × 10⁻¹⁴.
What is the pH of neutral water at this temperature? - Calculate the pH of the resulting mixtures:
(a) 10 mL of 0.2 M Ca(OH)₂ + 25 mL of 0.1 M HCl
(b) 10 mL of 0.01 M H₂SO₄ + 10 mL of 0.01 M Ca(OH)₂
(c) 10 mL of 0.1 M H₂SO₄ + 10 mL of 0.1 M KOH - Determine the solubilities of:
Ag₂CrO₄, BaCrO₄, Fe(OH)₃, PbCl₂, Hg₂I₂ at 298 K from their Ksp values.
Find the molarities of individual ions. - Ksp of Ag₂CrO₄ = 1.1 × 10⁻¹²; Ksp of AgBr = 5.0 × 10⁻¹³
Calculate the ratio of molarities of their saturated solutions. - Equal volumes of 0.002 M sodium iodate and cupric chlorate are mixed.
Will Cu(IO₃)₂ precipitate? (Ksp = 7.4 × 10⁻⁸) - Ka of benzoic acid = 6.46 × 10⁻⁵; Ksp of silver benzoate = 2.5 × 10⁻¹³
How much more soluble is silver benzoate in a buffer of pH 3.19 compared to pure water? - What is the maximum concentration of equimolar FeSO₄ and Na₂S so that no FeS precipitates? (Ksp = 6.3 × 10⁻¹⁸)
- What is the minimum volume of water required to dissolve 1 g of CaSO₄ at 298 K? (Ksp = 9.1 × 10⁻⁶)
- The concentration of S²⁻ in 0.1 M HCl saturated with H₂S is 1.0 × 10⁻¹⁹ M.
If 10 mL of this is added to 5 mL of 0.04 M FeSO₄, MnCl₂, ZnCl₂, and CdCl₂, in which solution will precipitation occur?
Solutions
- 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) The initial effect of an increase in volume at constant temperature is a decrease in pressure. Since the vapour pressure depends only on temperature and not on volume, the vapour pressure remains initially unchanged.
b) The rate of condensation decreases immediately because fewer vapour molecules now strike the liquid surface per unit time. The rate of evaporation remains unchanged at first.
c) As the rate of evaporation is now greater than the rate of condensation, more molecules pass into the vapour phase until a new equilibrium is established. The final vapour pressure will be equal to the original vapour pressure at the same temperature. - What is Kc for the following equilibrium when the equilibrium concentrations are:
[SO₂] = 0.60 M, [O₂] = 0.82 M and [SO₃] = 1.90 M
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Kc = [SO₃]² / ([SO₂]² × [O₂])
= (1.90)² / [(0.60)² × (0.82)] = 3.61 / (0.36 × 0.82) = 3.61 / 0.2952 ≈ 12.23 - At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms.
I₂(g) ⇌ 2I(g)
Let the initial pressure of I₂ = 100 Pa.
At equilibrium, 40% dissociates to atoms → 40 Pa of I atoms and 60 Pa of I₂ remain.
I₂ ⇌ 2I
Initial: 100 Pa → 0
Change: –x → +2x
Equilibrium: 100 – x → 2x
Given: 2x = 40 → x = 20 → [I₂] = 80 Pa
Kp = (PI)² / PI₂ = (40)² / 65 = 24.6 Pa - Write the expression for the equilibrium constant, K,c, for the following reactions:
(i) Kc = [NO]²[Cl₂] / [NOCl]²
(ii) Kc = [NO₂]^4[O₂] / 1 (pure solids are omitted)
(iii) Kc = [CH₃COOH][C₂H₅OH] / [CH₃COOC₂H₅]
(iv) Kc = 1 / ([Fe³⁺][OH⁻]³)
(v) Kc = [IF₅]² / [F₂]⁵ (I₂ is a solid and not included) - Find the value of Kc for each of the following equilibria from the value of Kp:
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); Kp = 1.8 × 10⁻² at 500 K
Δn = 3 – 2 = 1
Kc = Kp × (RT)^–Δn = 1.8 × 10⁻² / (0.0821 × 500) = 1.8 × 10⁻² / 41.05 ≈ 4.38 × 10⁻⁴ - (ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); Kp = 167 at 1073 K
Δn = 1 – 0 = 1
Kc = Kp / (RT) = 167 / (0.0821 × 1073) = 167 / 88.09 ≈ 1.896 mol L⁻¹ - For the reaction: NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g); Kc = 6.3 × 10¹⁴
Kc (reverse) = 1 / Kc = 1 / (6.3 × 10¹⁴) ≈ 1.59 × 10⁻¹⁵ - Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression.
In the expression for the equilibrium constant, the concentrations (or activities) of pure solids and pure liquids are taken as constant because their molar concentration remains unchanged. Therefore, they are not included in the equilibrium constant expression. - 2N₂(g) + O₂(g) ⇌ 2N₂O(g)
Initial moles: [N₂] = 0.482 / 10 = 0.0482 M, [O₂] = 0.933 / 10 = 0.0933 M
Let x be the amount of N₂O formed at equilibrium:
Change: –x for N₂ and –½x for O₂
Kc = [N₂O]² / ([N₂]²[O₂]) = (x/10)² / ((0.0482 – x/10)² × (0.0933 – x/20)) = 2.0 × 10⁻³⁷
Due to extremely small Kc, x ≈ 0, i.e., negligible formation of N₂O → composition remains essentially unchanged. - 2NO(g) + Br₂(g) ⇌ 2NOBr(g)
Initial: [NO] = 0.087 mol, [Br₂] = 0.0437 mol
At equilibrium: [NOBr] = 0.0518 mol
From the balanced equation:
2NO + Br₂ ⇌ 2NOBr
0.087 – x + 0.0437 – x + x = 0.0518 × 2 → x = 0.0518
[NO] = 0.087 – 0.0518 = 0.0352 mol, [Br₂] = 0.0437 – 0.0259 = 0.0178 mol - 2SO₂(g) + O₂(g) ⇌ 2SO₃(g); Kp = 2.0 × 10¹⁰ at 450 K
Δn = –1 → Kc = Kp / (RT)^Δn = Kp / (RT)⁻¹ = Kp × RT
= 2.0 × 10¹⁰ × 0.0821 × 450 = 7.389 × 10¹¹ mol L⁻¹ - 2HI(g) ⇌ H₂(g) + I₂(g)
Initial pressure of HI = 0.2 atm; Equilibrium pressure = 0.04 atm
Let dissociation = x
2HI ⇌ H₂ + I₂
Initial: 0.2, 0, 0
Change: –2x, +x, +x
At eq: 0.2 – 2x = 0.04 → x = 0.08
Kp = (x)² / (0.2 – 2x)² = (0.08)² / (0.04)² = 0.0064 / 0.0016 = 4.0 - N₂(g) + 3H₂(g) ⇌ 2NH₃(g); Kc = 1.7 × 10²
[ N₂ ] = 1.57 / 20 = 0.0785 M
[ H₂ ] = 1.92 / 20 = 0.096 M
[ NH₃ ] = 8.13 / 20 = 0.4065 M
Qc = [NH₃]² / ([N₂][H₂]³) = (0.4065)² / (0.0785 × (0.096)³)
Qc = 0.1652 / (0.0785 × 0.0008847) = 0.1652 / 6.94 × 10⁻⁵ ≈ 2380
Since Qc > Kc, the reaction will proceed in the reverse direction to attain equilibrium. - Kc = [NH₃]⁴ / ([NO]⁶[H₂O]⁴)
Balanced equation: 4NH₃ + 6NO ⇌ 5N₂ + 6H₂O - 1 mol H₂O and 1 mol CO in 10 L → 0.1 M each
40% of H₂O reacts → 0.04 mol → [H₂] = [CO₂] = 0.04 mol / 10 = 0.004 M
[H₂O] = [CO] = 0.06 M
Kc = [H₂][CO₂] / [H₂O][CO] = (0.004 × 0.004) / (0.06 × 0.06) = 0.000016 / 0.0036 = 4.44 × 10⁻³ - H₂(g) + I₂(g) ⇌ 2HI(g); Kc = 54.8
[HI] = 0.5 M
Let x be [H₂] = [I₂] at equilibrium
Then Kc = [HI]² / ([H₂][I₂]) = 0.25 / x² → x² = 0.25 / 54.8 = 0.00456
x = √0.00456 ≈ 0.068 M
[H₂] = [I₂] ≈ 0.068 M - 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) Vapour pressure initially decreases due to an increase in volume.
b) Rate of condensation decreases; rate of evaporation exceeds rate of condensation.
c) Eventually, a new equilibrium is established at the same temperature, but the final vapour pressure remains the same as before, since vapour pressure is temperature dependent. - What is Kc for the following equilibrium when the equilibrium concentrations are: [SO₂] = 0.60 M, [O₂] = 0.82 M and [SO₃] = 1.90 M?
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Kc = [SO₃]² / ([SO₂]² × [O₂])
= (1.90)² / ((0.60)² × 0.82)
= 3.61 / (0.36 × 0.82)
= 3.61 / 0.2952
≈ 12.23 - At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms.
I₂(g) ⇌ 2I(g)
Let the initial amount of I₂ = 1 mol (by volume)
At equilibrium: I₂ = 0.60 mol, I = 0.40 mol
Total moles = 0.60 + 0.40 = 1.00 mol
Mole fraction of I = 0.40
Partial pressure of I = 0.40 × 105 Pa = 42 Pa
Partial pressure of I₂ = 0.60 × 105 Pa = 63 Pa
Kp = (PI)² / PI₂ = (42)² / 63 = 1764 / 63 ≈ 28 Pa - Write the expression for the equilibrium constant, Kc, for each of the following reactions:
(i) Kc = [NO]²[Cl₂] / [NOCl]²
(ii) Kc = [NO₂]⁴[O₂] (since solids are not included)
(iii) Kc = [CH₃COOH][C₂H₅OH] / [CH₃COOC₂H₅]
(iv) Kc = 1 / [Fe³⁺][OH⁻]³ (or omit solids)
(v) Kc = [IF₅]² / [F₂]⁵ (omit I₂ solid) - Find the value of Kc for each of the following equilibria from the value of Kp:
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); Kp = 1.8 × 10⁻² at 500 K
Δng = (2 + 1) – 2 = 1
Kc = Kp × (RT)^(-Δng)
Kc = 1.8 × 10⁻² / (0.0821 × 500)
= 1.8 × 10⁻² / 41.05 ≈ 4.4 × 10⁻⁴ - (ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); Kp = 167 at 1073 K
Δng = 1 – 0 = 1
Kc = 167 / (0.0821 × 1073)
= 167 / 88.1 ≈ 1.90 mol L⁻¹ - For the reaction: NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g); Kc = 6.3 × 10¹⁴ at 1000 K
Kc for reverse = 1 / Kc = 1 / 6.3 × 10¹⁴ ≈ 1.59 × 10⁻¹⁵ - Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression.
The concentration of pure solids and pure liquids remains constant and does not change with the extent of reaction. Therefore, they are incorporated in the equilibrium constant and are not explicitly written in the expression. - 2HI(g) ⇌ H₂(g) + I₂(g)
Initial pressure of HI = 0.2 atm; equilibrium pressure = 0.04 atm
Decrease in HI = 0.2 – 0.04 = 0.16 atm
Since 2 mol HI ⇌ 1 mol H₂ + 1 mol I₂,
H₂ = I₂ = 0.16 / 2 = 0.08 atm
Kp = (PH₂ × PI₂) / (PHI)² = (0.08 × 0.08) / (0.04)² = 0.0064 / 0.0016 = 4.0 - N₂(g) + 3H₂(g) ⇌ 2NH₃(g); Kc = 1.7 × 10² at 500 K
[N₂] = 1.57 mol / 20 L = 0.0785 M
[H₂] = 1.92 mol / 20 L = 0.096 M
[NH₃] = 8.13 mol / 20 L = 0.4065 M
Q = [NH₃]² / ([N₂][H₂]³) = (0.4065)² / (0.0785 × (0.096)³)
≈ 0.165 / (0.0785 × 8.84×10⁻⁴) ≈ 0.165 / 6.94×10⁻⁵ ≈ 2378.38
Since Q > Kc, the reaction will proceed in the reverse direction (toward reactants). - Write the balanced chemical equation for the equilibrium expression:
Kc = [NH₃]⁴ / ([NO]⁶ [H₂O]⁴)
Balanced equation: 6NO(g) + 4H₂O(g) ⇌ 4NH₃(g) + 3O₂(g) - H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)
1 mol of H₂O and CO in a 10 L vessel → 0.1 M each
40% of H₂O reacts → 0.04 mol reacts
H₂O = 0.06 mol = 0.006 M
CO = 0.06 mol = 0.006 M
H₂ = CO₂ = 0.04 mol = 0.004 M
Kc = [H₂][CO₂] / [H₂O][CO] = (0.004 × 0.004) / (0.006 × 0.006) = 1.78 × 10⁻¹ - H₂(g) + I₂(g) ⇌ 2HI(g); Kc = 54.8 at 700 K
[HI] = 0.5 M
Let [H₂] = [I₂] = x
Kc = [HI]² / ([H₂][I₂]) = (0.5)² / x² = 0.25 / x²
54.8 = 0.25 / x² → x² = 0.25 / 54.8 = 0.00456 → x = √0.00456 ≈ 0.068 M
So, [H₂] = [I₂] = 0.068 M - 2ICl(g) ⇌ I₂(g) + Cl₂(g); Kc = 0.14
Initial [ICl] = 0.78 M, let x be the amount dissociated
At equilibrium: [ICl] = 0.78 – 2x, [I₂] = x, [Cl₂] = x
Kc = (x × x) / (0.78 – 2x)²
Let’s solve using approximation: assume x is small.
Kc ≈ x² / (0.78)² = 0.14
x² ≈ 0.14 × 0.6084 = 0.08518
x ≈ √0.08518 ≈ 0.292 M
[ICl] ≈ 0.78 – 2(0.292) ≈ 0.196 M
[I₂] = [Cl₂] ≈ 0.292 M - C₂H₆(g) ⇌ C₂H₄(g) + H₂(g); Kp = 0.04 atm at 899 K
Initial pressure of C₂H₆ = 4.0 atm
Let x atm decompose
At equilibrium:
PC₂H₆ = 4.0 – x
PC₂H₄ = PH₂ = x
Kp = x² / (4.0 – x) = 0.04
Solve: x² = 0.04(4.0 – x) = 0.16 – 0.04x
x² + 0.04x – 0.16 = 0
Using the quadratic formula:
x = [-0.04 ± √(0.0016 + 0.64)] / 2 ≈ 0.38 atm
PC₂H₆ = 4.0 – 0.38 = 3.62 atm - CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)
(i) Qc = [CH₃COOC₂H₅][H₂O] / [CH₃COOH][C₂H₅OH]
(ii) Initial: CH₃COOH = 1 mol, C₂H₅OH = 0.18 mol
At equilibrium, ester = 0.171 mol → [ester] = 0.171
So, [CH₃COOH] = 1 – 0.171 = 0.829
[C₂H₅OH] = 0.18 – 0.171 = 0.009
Kc = (0.171 × 0.171) / (0.829 × 0.009) ≈ 0.0292 / 0.007461 ≈ 3.91
(iii) Initial acid = 1 mol, alcohol = 0.5 mol; ester = 0.214 mol
[acid] = 1 – 0.214 = 0.786
[alcohol] = 0.5 – 0.214 = 0.286
Qc = (0.214 × 0.214) / (0.786 × 0.286) = 0.0458 / 0.2248 ≈ 0.204
Since Qc < Kc (3.91), the reaction is not at equilibrium and will proceed forward - PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); Kc = 8.3 × 10⁻³
[PCl₅] = 0.05 mol/L
Let x mol/L dissociate.
At equilibrium: [PCl₅] = 0.05 – x, [PCl₃] = x, [Cl₂] = x
Kc = x² / (0.05 – x)
Assume x is small:
Kc ≈ x² / 0.05 = 8.3 × 10⁻³
x² = 8.3 × 10⁻³ × 0.05 = 4.15 × 10⁻⁴
x ≈ √4.15 × 10⁻⁴ ≈ 0.0204 mol/L
[PCl₃] = [Cl₂] ≈ 0.0204 mol/L
[PCl₅] ≈ 0.05 – 0.0204 = 0.0296 mol/L - FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g); Kp = 0.265 atm
Initial: pCO = 1.4 atm, pCO₂ = 0.80 atm
Qp = pCO₂ / pCO = 0.80 / 1.4 ≈ 0.571
Since Qp > Kp, the reaction will shift in the reverse direction.
Let x be the change
At equilibrium: pCO = 1.4 + x, pCO₂ = 0.80 – x
Kp = (0.80 – x) / (1.4 + x) = 0.265
0.80 – x = 0.265 × (1.4 + x)
0.80 – x = 0.371 + 0.265x
0.429 = 1.265x → x ≈ 0.339 atm
pCO = 1.4 + 0.339 = 1.739 atm
pCO₂ = 0.80 – 0.339 = 0.461 atm - N₂(g) + 3H₂(g) ⇌ 2NH₃(g); Kc = 0.061
[ N₂ ] = 3.0 M, [ H₂ ] = 2.0 M, [ NH₃ ] = 0.5 M
Qc = [NH₃]² / ([N₂][H₂]³) = (0.5)² / (3.0 × 8) = 0.25 / 24 = 0.0104
Since Qc < Kc, the reaction proceeds in the forward direction - 2BrCl(g) ⇌ Br₂(g) + Cl₂(g); Kc = 32 at 500 K
Initial [BrCl] = 3.3 × 10⁻³ M
Let x be the amount dissociated:
[Br₂] = x, [Cl₂] = x, [BrCl] = 3.3 × 10⁻³ – 2x
Kc = x² / (3.3 × 10⁻³ – 2x)² = 32
Use approximation: if x is small, (3.3 × 10⁻³)² ≈ 1.089 × 10⁻⁵
So x² ≈ 32 × 1.089 × 10⁻⁵ ≈ 3.485 × 10⁻⁴ → x ≈ √(3.485 × 10⁻⁴) ≈ 0.0187
But this value is greater than the initial concentration, so no valid solution using approximation. Solve using full quadratic or assume 100% dissociation. In this case, [BrCl] at equilibrium is negligible, and [Br₂] ≈ [Cl₂] ≈ 0.00165 M each (approx). - C(s) + CO₂(g) ⇌ 2CO(g)
Given 90.55% CO by mass at 1127 K and 1 atm
Let total gas = 100 g
CO = 90.55 g, CO₂ = 9.45 g
Moles CO = 90.55 / 28 = 3.23 mol
Moles CO₂ = 9.45 / 44 = 0.215 mol
Total moles = 3.445 mol
Mole fraction CO = 3.23 / 3.445 = 0.9375 → pCO = 0.9375 atm
pCO₂ = 0.0625 atm
Kp = (pCO)² / pCO₂ = (0.9375)² / 0.0625 = 0.879 / 0.0625 = 14.06 - NO(g) + ½O₂(g) ⇌ NO₂(g)
ΔG° = ΣΔG°(products) – ΣΔG°(reactants)
= [52.0] – [87.0 + 0] = –35.0 kJ/mol
K = e^(–ΔG°/RT) = e^[35,000 / (8.314 × 298)] ≈ e^14.1 ≈ 1.34 × 10⁶ - Change in number of moles with pressure decrease (volume increase):
(a) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g): forward reaction increases moles (1 → 2), so forward reaction favoured
(b) CaO(s) + CO₂(g) ⇌ CaCO₃(s): gas → solid, decrease in moles, backward favoured
(c) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g): 4 mol gas → 4 mol gas → no change - (i) COCl₂(g) ⇌ CO(g) + Cl₂(g) — 1 mole ⇌ 2 moles → increasing pressure favours backward
(ii) CH₄(g) + 2S₂(g) ⇌ CS₂(g) + 2H₂S(g) — 3 ⇌ 3 → no effect
(iii) CO₂(g) + C(s) ⇌ 2CO(g) — 1 ⇌ 2 → increasing pressure favours backward
(iv) 2H₂(g) + CO(g) ⇌ CH₃OH(g) — 3 ⇌ 1 → increasing pressure favours forward
(v) CaCO₃(s) ⇌ CaO(s) + CO₂(g) — solids ignored, gas on RHS → increasing pressure favours backward
(vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g) — 9 ⇌ 10 → increasing pressure favours backward - H₂(g) + Br₂(g) ⇌ 2HBr(g); K = 1.6 × 10⁵ at 1024 K
Total pressure of HBr introduced = 10 bar.
Let dissociation be x → [HBr] = 10 – 2x, [H₂] = x, [Br₂] = x
K = (10 – 2x)² / (x²) = 1.6 × 10⁵
Solving: (10 – 2x)² = 1.6 × 10⁵ x²
(100 – 40x + 4x²) = 1.6 × 10⁵ x²
1.6 × 10⁵ x² – 4x² + 40x – 100 = 0
Use quadratic formula or approximate: x ≈ 0.0226 ba.r
Then, [H₂] = [Br₂] = 0.0226 bar, [HBr] = 10 – 2(0.0226) ≈ 9.9548 bar - CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)
(a) Kp = (PCO)(PH₂)³ / (PCH₄)(PH₂O)
(b) (i) Increasing pressure → decrease volume → reaction shifts toward fewer gas molecules → backward
(ii) Increasing temperature → endothermic reaction → forward reaction favoured → Kp increases
(iii) Adding catalyst → no effect on Kp, only speeds up equilibrium attainment - 2H₂(g) + CO(g) ⇌ CH₃OH(g)
a) Addition of H₂ → reaction shifts forward
b) Addition of CH₃OH → reaction shifts backward
c) Removal of CO → reaction shifts backwards
d) Removal of CH₃OH → reaction shifts forward - PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); Kc = 8.3 × 10⁻³ at 473 K; ΔH° = +124.0 kJ mol⁻¹
a) Kc = [PCl₃][Cl₂] / [PCl₅]
b) Kc for reverse = 1 / 8.3 × 10⁻³ ≈ 120.5
c) (i) Adding PCl₅ → no change in Kc
(ii) Increasing pressure → reaction shifts to fewer moles → backward → no change in Kc
(iii) Increasing temperature → endothermic → favours forward → Kc increases - CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g); Kp = 10.1 at 400°C
Initial pCO = pH₂O = 4.0 bar
Let x be the change in pressure:
pCO = 4 – x, pH₂O = 4 – x, pCO₂ = x, pH₂ = x
Kp = (x × x) / ((4 – x)²) = 10.1
x² / (4 – x)² = 10.1 → solving gives x ≈ 3.2
Thus, pH₂ = 3.2 bar - Predict which reactions will have appreciable concentrations of both reactants and products:
a) Cl₂(g) ⇌ 2Cl(g); Kc = 5 × 10⁻³⁹ → lies far to the left (reactants dominate)
b) Cl₂(g) + 2NO(g) ⇌ 2NOCl(g); Kc = 3.7 × 10⁸ → lies far to the right (products dominate)
c) Cl₂(g) + 2NO₂(g) ⇌ 2NO₂Cl(g); Kc = 1.8 → value near 1, appreciable concentrations of both - 3O₂(g) ⇌ 2O₃(g); Kc = 2.0 × 10⁻⁵⁰ at 25°C
[O₂] = 1.6 × 10⁻² M
Kc = [O₃]² / [O₂]³ → 2.0 × 10⁻⁵⁰ = x² / (1.6 × 10⁻²)³
x² = 2.0 × 10⁻⁵⁰ × 4.096 × 10⁻⁶ = 8.192 × 10⁻⁵⁶
x = √(8.192 × 10⁻⁵⁶) = 9.05 × 10⁻²⁸ M
[O₃] = 9.05 × 10⁻²⁸ M - CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g); Kc = 3.90 at 1300 K
Initial: CO = 0.30 mol, H₂ = 0.10 mol, H₂O = 0.02 mol, CH₄ = x mol
At equilibrium: CO = 0.30 – x, H₂ = 0.10 – 3x, H₂O = 0.02 + x
Kc = x(0.02 + x) / [(0.30 – x)(0.10 – 3x)³] = 3.90
Solving this equation gives x ≈ 0.016 M (approximate from the textbook) - Conjugate acid-base pairs:
A conjugate acid-base pair consists of two species that differ by a proton (H⁺)
Examples:
HNO₂ → conjugate base: NO₂⁻
CN⁻ → conjugate acid: HCN
HClO₄ → conjugate base: ClO₄⁻
F⁻ → conjugate acid: HF
OH⁻ → conjugate acid: H₂O
CO₃²⁻ → conjugate acid: HCO₃⁻
S²⁻ → conjugate acid: HS⁻ - Lewis Acids: BF₃ and H⁺ are Lewis acids because they can accept a pair of electrons. H₂O and NH₄⁺ are not Lewis acids.
- Conjugate Bases for Brønsted Acids:
HF → F⁻
H₂SO₄ → HSO₄⁻
HCO₃⁻ → CO₃²⁻ - Conjugate Acids for Brønsted Bases:
NH₂⁻ → NH₃
NH₃ → NH₄⁺
HCOO⁻ → HCOOH - Amphoteric Species and their Conjugates:
H₂O → conjugate acid: H₃O⁺; conjugate base: OH⁻
HCO₃⁻ → conjugate acid: H₂CO₃; conjugate base: CO₃²⁻
HSO₄⁻ → conjugate acid: H₂SO₄; conjugate base: SO₄²⁻
NH₃ → conjugate acid: NH₄⁺; conjugate base: NH₂⁻ - Classification of Species:
OH⁻: Lewis base (donates an electron⁻ pair)
F⁻: Lewis base (donates an electron⁻ pair)
H⁺: Lewis acid (accepts an electron⁻ pair)
BCl₃: Lewis acid (electron-deficient) - pH from [H⁺] = 3.8 × 10⁻³ M (soft drink):
pH = –log(3.8 × 10⁻³) ≈ 2.42 - [H⁺] from pH = 3.76 (vinegar):
[H⁺] = 10⁻³.⁷⁶ ≈ 1.74 × 10⁻⁴ M - Ionisation Constants of Conjugate Bases:
K_b = K_w / K_a
HF: 1.0 × 10⁻¹⁴ / 6.8 × 10⁻⁴ = 1.47 × 10⁻¹¹
HCOOH: 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁴ = 5.56 × 10⁻¹¹
HCN: 1.0 × 10⁻¹⁴ / 4.8 × 10⁻⁹ = 2.08 × 10⁻⁶ - Phenol Solution:
Ka = 1 × 10⁻¹⁰, [Phenol] = 0.05 M
x² / (0.05 – x) = 1 × 10⁻¹⁰ → x ≈ 7.07 × 10⁻⁶ M
With 0.01 M sodium phenolate: [C₆H₅O⁻] = 0.01 M
α = x / (0.05) = 1.41 × 10⁻⁴ without salt; suppressed with salt - Ionisation of H₂S:
Ka₁ = 9.1 × 10⁻⁸
x² / (0.1 – x) = 9.1 × 10⁻⁸ → x ≈ 9.5 × 10⁻⁵ M (HS⁻)
With HCl, the common ion effect suppresses ionisation.
Ka₂ = 1.2 × 10⁻¹³ → [S²⁻] = Ka₂ × [HS⁻] / [H⁺]
Without HCl: [S²⁻] ≈ 1.14 × 10⁻¹² M
With HCl: [S²⁻] is even lower - 0.05 M Acetic Acid (Ka = 1.74 × 10⁻⁵):
x² / (0.05 – x) = 1.74 × 10⁻⁵ → x ≈ 9.3 × 10⁻³
α = 0.186; [CH₃COO⁻] = 9.3 × 10⁻³ M
pH = –log(9.3 × 10⁻³) ≈ 2.03 - 0.01 M Organic Acid, pH = 4.15:
[H⁺] = 7.1 × 10⁻⁵; Ka = (7.1 × 10⁻⁵)² / 0.01 ≈ 5.0 × 10⁻⁷
pKa = –log(5.0 × 10⁻⁷) ≈ 6.30 - Strong Acids/Bases (Complete dissociation):
HCl (0.003 M): pH = 2.52
NaOH (0.005 M): pOH = 2.3 → pH = 11.7
HBr (0.002 M): pH = 2.70
KOH (0.002 M): pOH = 2.7 → pH = 11.3 - pH from Mass + Volume:
(a) TlOH: 2 g / 204.38 g/mol = 0.00979 mol in 2 L → 0.0049 M → pOH ≈ 2.31 → pH = 11.69
(b) Ca(OH)₂: 0.3 g / 74.09 g/mol = 0.00405 mol = 0.0081 mol OH⁻ in 0.5 L → 0.0162 M → pOH ≈ 1.79 → pH = 12.21
(c) NaOH: 0.3 g / 40 = 0.0075 mol in 200 mL → 0.0375 M → pOH ≈ 1.42 → pH = 12.58
(d) 1 mL of 13.6 M HCl in 1 L = 0.0136 M → pH = 1.87 - 0.1 M Bromoacetic Acid, α = 0.132:
[H⁺] = 0.0132 M → pH = 1.88
Ka = α² × C = (0.132)² × 0.1 = 1.74 × 10⁻³
pKa = 2.76 - Codeine, 0.005 M, pH = 9.95:
[OH⁻] = 1.12 × 10⁻⁵ M
Kb = x² / (0.005 – x) = 2.5 × 10⁻⁸
pKb = 7.6 - Aniline, 0.001 M:
Kb = 4.27 × 10⁻¹⁰
x² / 0.001 = Kb → x ≈ 2.07 × 10⁻⁶ → pOH = 5.68 → pH = 8.32
α = x / C = 0.00207
Ka = Kw / Kb = 2.34 × 10⁻⁵ - 0.05 M Acetic Acid, pKa = 4.74:
Ka = 1.82 × 10⁻⁵ → x = 9.5 × 10⁻³
With 0.01 M HCl, [H⁺] from HCl suppresses dissociation
With 0.1 M HCl, even more suppression; α decreases further - Dimethylamine, 0.02 M, Kb = 5.4 × 10⁻⁴:
x² / 0.02 = 5.4 × 10⁻⁴ → x ≈ 3.3 × 10⁻³
α = x / 0.02 = 0.165
With 0.1 M NaOH: OH⁻ common ion → α suppressed - [H⁺] from pH of fluids:
pH = –log[H⁺] → [H⁺] = 10^–pH
Muscle: 1.48 × 10⁻⁷
Stomach: 6.3 × 10⁻²
Blood: 4.17 × 10⁻⁸
Saliva: 3.98 × 10⁻⁷ - pH from Food pH Values:
Milk: 1.58 × 10⁻⁷
Coffee: 1.0 × 10⁻⁵
Tomato: 6.3 × 10⁻⁵
Lemon: 6.3 × 10⁻³
Egg: 1.58 × 10⁻⁸ - KOH: 0.561 g in 200 mL:
Moles = 0.561 / 56.1 = 0.01 mol → 0.05 M
[OH⁻] = 0.05 → pOH = 1.3 → pH = 12.7
[K⁺] = 0.05 M, [H⁺] = 2 × 10⁻¹³ - Sr(OH)₂ Solubility = 19.23 g/L:
Moles = 0.227 mol
[Sr²⁺] = 0.227 M, [OH⁻] = 2 × 0.227 = 0.454 M
pOH ≈ 0.34 → pH ≈ 13.66 - 0.05 M Propanoic Acid, Ka = 1.32 × 10⁻⁵:
x ≈ 8.1 × 10⁻³ → pH = 2.09
With 0.01 M HCl, α decreases
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