PSEB/NCERT CLASS 12 CHEMISTRY CHAPTER 3 AND 4 IMPORTANT QUESTIONS WITH ANSWERS (CEP 2)
Chapter 3 – Chemical Kinetics
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Rate of Chemical Reaction:
It is the change in molar concentration of the species taking part in the reaction per unit time. -
Moderate Speed Reaction:
These are chemical reactions which neither proceed very fast nor very slow. -
Average Rate of Reaction:
It is the change in molar concentration of the reactant divided by the time taken for the change.
In general, the average rate is expressed as Δ[C]/Δt. -
Instantaneous Rate of Reaction:
It is defined as the rate of reaction at any instant of time.
In general, the instantaneous rate is expressed as d[C]/dt. -
Law of Mass Action:
According to this law, at a given temperature, the rate of a chemical reaction is directly proportional to the product of molar concentrations of the reacting species, each raised to the power of their stoichiometric coefficients. -
Rate Law:
It is an expression that describes the reaction rate in terms of molar concentrations of the reactants as determined experimentally. -
Rate Constant (k):
It is the rate of reaction when the molar concentration of each reactant is taken as unity. -
Molecularity of a Reaction:
It is the number of reacting species that collide simultaneously in a chemical reaction. -
Order of Chemical Reaction:
It is the sum of the powers of the concentrations of the reactants that appear in the rate equation of the reaction.
Competency enhancement Plan Class 6 to 12 Solution
2 Marks questions
🧪 CHAPTER 3 – CHEMICAL KINETICS (2 Marks Questions with Answers)
Q1. Derive the formula for the half-life of a first-order reaction and calculate the half-life for a reaction with a rate constant of 0.693 min⁻¹.
Answer:
For a first-order reaction:
ln([A]₀/[A]) = kt.
At half-life, [A] = [A]₀/2 → ln2 = kt₁/₂ →
t₁/₂ = 0.693/k.
Substituting k = 0.693 min⁻¹:
t₁/₂ = 0.693 / 0.693 = 1 minute.
Q2. Calculate half-life of a first-order reaction having K = 200 s⁻¹.
Answer:
t₁/₂ = 0.693 / k = 0.693 / 200 = 3.47 × 10⁻³ s (≈ 3.47 milliseconds).
Q3. Calculate half-life of a first-order reaction having K = 4.93 × 10⁻⁴ s⁻¹.
Answer:
t₁/₂ = 0.693 / 4.93×10⁻⁴ = 1.41 × 10³ s ≈ 23.4 minutes.
Q4. A catalyzed reaction shows rate constant k = 5.2 × 10⁻² s⁻¹ and uncatalyzed reaction has k = 2.1 × 10⁻³ s⁻¹ at 300 K. Calculate the difference in activation energy.
Answer:
From Arrhenius relation:
ln(k_c / k_u) = (Eₐ(u) − Eₐ(c)) / (R T)
⇒ ΔEₐ = RT ln(k_c / k_u)
= (8.314 × 300) × ln(5.2×10⁻² / 2.1×10⁻³)
≈ 8.3 × 10³ J mol⁻¹ = ≈ 8 kJ mol⁻¹.
Q5. Define the Pseudo first order reaction. Give its example.
Answer:
A pseudo first-order reaction is one that is higher order overall but becomes effectively first order because one reactant is in large excess (its concentration remains constant).
Example: Hydrolysis of sucrose:
C₁₂H₂₂O₁₁ + H₂O → glucose + fructose.
Q6. Write the difference between the order and molecularity of a reaction.
Answer:
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Order: Sum of powers of concentration terms in the experimentally determined rate law. Can be zero, fractional, or integer.
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Molecularity: Number of reacting species colliding in one elementary step. Always a whole number.
Q7. Write difference between the rate of reaction and rate constant.
Answer:
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Rate of reaction: Change in concentration per unit time; depends on reactant concentration and temperature.
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Rate constant (k): Proportionality constant in rate law; independent of concentration at fixed temperature.
Q8. What is the effect of catalyst on: (a) Activation energy (b) Rate constant.
Answer:
(a) Activation energy: Decreases because the catalyst provides an alternative path.
(b) Rate constant: Increases (since k = A e⁻ᴱᵃ⁄ᴿᵀ).
Q9. How can you determine the order of a reaction experimentally?
Answer:
Methods:
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Initial rate method – compare rate changes when varying one reactant.
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Half-life method – observe how t₁/₂ varies with [A]₀.
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Integrated rate-law plots – check which plot gives a straight line (zero, first, or second order).
Q10. Why do most reactions slow down with time as they proceed?
Answer:
Because the concentration of reactants decreases, reducing collision frequency and hence reaction rate. In reversible reactions, the backward rate also rises, further slowing the net rate.
Q11. What is meant by energy profile diagram of a reaction? Sketch it roughly and label activation energy.
Answer:
An energy profile diagram shows potential energy versus reaction progress. The curve peaks at the activated complex; the energy gap between reactants and this peak is the activation energy (Eₐ).
Q12. Explain briefly the difference between rate constant and rate of reaction.
Answer:
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Rate constant: Fixed value for a given temperature; shows how fast a reaction can proceed.
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Rate: Actual reaction speed at a given moment; changes with reactant concentration.
⚗️ CHAPTER 4 – d AND f BLOCK ELEMENTS (2 Marks Questions with Answers)
Q1. Name any three coinage metals and state whether they are transition metals. Give reason.
Answer:
Cu, Ag, Au are coinage metals. They are considered transition metals because they form cations with incomplete d-subshells (e.g., Cu²⁺ is 3d⁹) and exhibit coloured compounds and variable oxidation states.
Q2. Explain why Zn and Cd are normally not considered transition metals.
Answer:
In their common +2 oxidation state, Zn (3d¹⁰) and Cd (4d¹⁰) have completely filled d-orbitals. They do not show variable oxidation states or form coloured ions; hence they are not true transition metals.
Q3. Zirconium and Hafnium exhibit similar chemical properties. Justify your answer.
Answer:
Due to lanthanide contraction, atomic and ionic radii of Zr and Hf are almost identical, leading to very similar chemical and physical properties such as oxidation states and coordination behaviour.
Q4. Write the chemical formula of manganate ion and dichromate ion.
Answer:
Manganate ion: MnO₄²⁻
Dichromate ion: Cr₂O₇²⁻
Q5. What is Misch metal? State its composition and one use.
Answer:
Misch metal is an alloy of rare-earth metals containing about 70% cerium, 25% lanthanum, and small amounts of neodymium and praseodymium.
Use: In flints for gas lighters and in magnesium alloys.
Q6. List any three characteristic properties of transition elements.
Answer:
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Variable oxidation states.
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Formation of coloured ions/complexes.
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Paramagnetic behaviour and catalytic activity due to unpaired d-electrons.
Q7. Explain why copper is considered a transition metal.
Answer:
Copper is considered a transition metal because in its +2 oxidation state (Cu²⁺) it has an incomplete 3d⁹ configuration and forms coloured ions and complexes.
Q8. Zn²⁺ ion is colorless whereas Cu²⁺ ion is colored. Explain with reason.
Answer:
Zn²⁺ (3d¹⁰) has no unpaired d-electrons, so no d–d transitions occur → colourless.
Cu²⁺ (3d⁹) has unpaired electrons → d–d transitions → coloured compounds.
Q9. Why are Fe²⁺ compounds more stable than Fe³⁺ compounds towards oxidation to +3 state?
Answer:
Fe³⁺ has a stable half-filled d⁵ configuration, so Fe²⁺ tends to get oxidised to Fe³⁺. However, in reducing environments or complex formation, Fe²⁺ can be stabilised due to hydration energy differences.
Q10. Why are lanthanoids called inner transition metals?
Answer:
Because their differentiating electrons enter the 4f inner orbitals (penultimate shell) while outer shells remain unchanged. Hence, they are called inner transition elements.
Q11. Transition metals have high melting and boiling points. Give reason.
Answer:
They possess strong metallic bonding due to delocalised s and d electrons, resulting in high enthalpies of atomisation and consequently high melting and boiling points.
Q12. Why do transition metals exhibit variable oxidation states?
Answer:
The energy difference between (n−1)d and ns orbitals is small, allowing variable numbers of d and s electrons to participate in bonding → multiple oxidation states.
Q13. The +2 oxidation state of lead is more stable than that of tin. Explain.
Answer:
Due to the inert pair effect, the 6s² electrons in lead are less available for bonding. Hence, Pb²⁺ is more stable than Pb⁴⁺, while Sn⁴⁺ remains relatively more stable than Sn²⁺.
Q14. Why do Zn, Cd, and Hg have comparatively low melting and boiling points?
Answer:
They have completely filled d¹⁰ orbitals, leading to weaker metallic bonding. In Hg, relativistic effects further weaken bonding, making it liquid at room temperature.
