.
🧪 CHAPTER 3 – CHEMICAL KINETICS (3 Marks Questions with Answers)
Q1. Explain the concept of reaction rate and how it can be determined experimentally.
Answer:
The rate of a chemical reaction is the change in concentration of a reactant or product per unit time.
It is expressed as Rate = −Δ[R]/Δt = Δ[P]/Δt.
It can be determined experimentally by:
-
Measuring volume of gas evolved or absorbed using a gas syringe.
-
Measuring change in colour or absorbance using a colorimeter.
-
Measuring change in concentration by titration (quenching aliquots at intervals).
The slope of concentration–time graph gives the reaction rate.
Q2. Differentiate between zero-order, first-order, and second-order reactions with suitable examples and mathematical expressions.
Answer:
| Order | Rate law | Integrated form | Half-life relation | Example |
|---|---|---|---|---|
| Zero order | Rate = k | [A] = [A]₀ − kt | t₁/₂ = [A]₀ / 2k | Decomposition of NH₃ on Pt |
| First order | Rate = k[A] | ln([A]₀/[A]) = kt | t₁/₂ = 0.693 / k | Radioactive decay |
| Second order | Rate = k[A]² | 1/[A] − 1/[A]₀ = kt | t₁/₂ = 1 / (k[A]₀) | 2NO₂ → 2NO + O₂ |
Q3. The rate constant is 2.5 × 10⁻² s⁻¹ at 300 K and 4.5 × 10⁻² s⁻¹ at 310 K. Calculate the activation energy. (R = 8.314 J mol⁻¹ K⁻¹)
Answer:
From Arrhenius equation:
ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂)
⇒ Eₐ = R ln(k₂/k₁) / (1/T₁ − 1/T₂)
= 8.314 × ln(4.5×10⁻² / 2.5×10⁻²) / (1/300 − 1/310)
= 8.314 × 0.5878 / (0.0001075) ≈ 45 kJ mol⁻¹.
Q4. A first-order reaction takes 23.1 min for 50 % completion. Calculate the time required for 75 % completion of this reaction.
Answer:
For first-order: t = (2.303/k) log([A]₀/[A]).
Half-life t₁/₂ = 0.693/k = 23.1 ⇒ k = 0.693/23.1 = 0.030 min⁻¹.
For 75 % completion, [A]₀/[A] = 4 ⇒ t = 2.303(1/0.030) log 4 = 2.303×33.3×0.602 ≈ 46 min.
Q5. The rate constant for a first-order reaction is 60 s⁻¹. How much time will it take to reduce the concentration of the reactant to 1/10 of its initial value?
Answer:
For first-order: t = (2.303/k) log([A]₀/[A]).
[A]₀/[A] = 10; k = 60 s⁻¹.
t = (2.303/60) × log 10 = 2.303/60 = 0.038 s.
Q6. The rate constant of a first-order reaction increases from 2.5 × 10⁻² s⁻¹ to 5.0 × 10⁻² s⁻¹ when the temperature is raised from 300 K to 310 K. Calculate Eₐ.
Answer:
ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂)
Eₐ = 8.314 ln(2.0)/(1/300 − 1/310) = 8.314 × 0.693 / 0.0001075 ≈ 53.6 kJ mol⁻¹.
Q7. The decomposition of N₂O₅ follows first-order kinetics. At 318 K, the rate constant is 6.2 × 10⁻³ s⁻¹. Calculate: (a) Half-life of the reaction (b) Time for 75 % decomposition of N₂O₅.
Answer:
(a) t₁/₂ = 0.693 / k = 0.693 / 6.2×10⁻³ = 112 s.
(b) t₇₅% = (2.303 / k) log ([A]₀/[A]) = (2.303 / 6.2×10⁻³) log 4 = 371 × 0.602 ≈ 224 s.
⚗️ CHAPTER 4 – d AND f BLOCK ELEMENTS (3 Marks Questions with Answers)
Q1. Explain the following with reasons: (a) Transition metals have high melting and boiling points. (b) Transition metals and their compounds show magnetic properties. (c) Transition metals form alloys easily.
Answer:
(a) Strong metallic bonding due to delocalised s and d electrons → high enthalpy of atomisation.
(b) Presence of unpaired d electrons produces paramagnetism; magnetic moment µ = √n(n+2) B.M.
(c) Similar atomic sizes and crystal structures allow atoms of different metals to mix freely, forming solid solutions (alloys).
Q2. Discuss the general trends in the following properties of first transition-series elements: (a) Atomic radius (b) Ionization enthalpy (c) Oxidation states.
Answer:
(a) Atomic radii decrease slightly from Sc to Cr then remain almost constant (Fe–Ni) and increase slightly at Cu–Zn due to added electron–electron repulsion.
(b) Ionization enthalpy increases irregularly; extra stability of d⁵ and d¹⁰ configurations causes anomalies at Cr and Cu.
(c) Variable oxidation states common; maximum near Mn because participation of both 4s and 3d electrons.
Q3. Compare the catalytic behaviour of transition metals with reference to their electronic configuration and surface activity.
Answer:
Transition metals act as good catalysts because:
-
Variable oxidation states enable formation of intermediates in redox cycles.
-
Partially filled d orbitals allow adsorption of reactant molecules, weakening bonds and lowering Eₐ.
-
Large surface area provides numerous active sites. Examples: Fe in Haber process, V₂O₅ in Contact process, Pt in hydrogenation.
Q4. Explain why actinoids show greater range of oxidation states than lanthanoids.
Answer:
Actinoids have 5f, 6d and 7s orbitals of similar energy, so variable numbers of electrons can participate in bonding.
In lanthanoids only 4f and 5d orbitals are involved and 4f are well shielded, giving mainly +3 state.
Hence actinoids exhibit oxidation states from +2 to +7.
Q5. Give reasons for the following: (a) Lanthanoids are called inner transition elements. (b) Lanthanoids form trivalent ions commonly. (c) Actinoids are more reactive than lanthanoids.
Answer:
(a) Because their differentiating electrons enter the inner 4f orbitals.
(b) Loss of two 6s and one 5d electron leads to the stable Ln³⁺ state.
(c) Actinoids have larger atomic size, smaller ionisation energy, and less shielding of 5f electrons, making them more electropositive and reactive.
Q6. Describe three important uses of transition metals and their compounds in industry.
Answer:
-
Catalysts: Fe (Haber process), V₂O₅ (Contact process), Ni (hydrogenation).
-
Alloys: Cr, Ni in stainless steel, Cu in bronze/brass.
-
Pigments and compounds: TiO₂ as white pigment, KMnO₄ and K₂Cr₂O₇ as oxidising agents.
Q7. Explain the factors responsible for the formation of complexes by transition metals.
Answer:
-
Small ionic size and high charge density favour strong field interactions.
-
Availability of vacant d orbitals to accept lone pairs from ligands.
-
Variable oxidation states enable stable coordination compounds with ligands like NH₃, CN⁻, H₂O.
Q8. Why do transition metals have high densities? Relate your answer to atomic structure.
Answer:
Transition metals have closely packed atoms and strong metallic bonds from delocalised s and d electrons.
High atomic mass combined with small atomic volume leads to high density values.
Q9. Explain why the stability of +2 oxidation state decreases from Mn to Zn in the first transition series.
Answer:
As we move from Mn (d⁵) to Zn (d¹⁰), the tendency to lose further d electrons decreases.
Effective nuclear charge increases and pairing energy rises, making higher oxidation states less stable; hence +2 state becomes progressively less stable.
Q10. What is lanthanide contraction? Discuss its causes and consequences on the properties of elements following the lanthanide series.
Answer:
Lanthanide contraction is the gradual decrease in atomic and ionic radii from La³⁺ to Lu³⁺ due to poor shielding of 4f electrons.
Consequences:
-
Similar sizes of Zr and Hf → similar chemistry.
-
Increased hardness and density across the series.
-
Difficulty in separation of lanthanoids due to close ionic radii.
