Class 12 : Chemistry (English) – Chapter 3: Chemical Kinetics

EXPLANATION & SUMMARY

🟦 Introduction to Chemical Kinetics
Chemical kinetics is the branch of chemistry that deals with the rate of chemical reactions and the factors that affect them. Unlike thermodynamics, which tells us whether a reaction is possible or not, kinetics explains how fast the reaction proceeds and by what mechanism.
💡 Concept: Thermodynamics → feasibility ✔ | Kinetics → speed & mechanism ⚡.

🟩 Rate of a Chemical Reaction
The rate of reaction is the change in concentration of reactants or products per unit time.
✏ Formula:
➡ Rate = Δ[Concentration] / Δt
For a reactant: Rate = – Δ[R] / Δt (negative sign because concentration decreases).
For a product: Rate = + Δ[P] / Δt.
🧪 Example: For the decomposition of H₂O₂ → H₂O + ½ O₂
➡ Rate = – Δ[H₂O₂] / Δt = + Δ[O₂] / (2Δt).

🟨 Average Rate vs Instantaneous Rate
Average rate: Concentration change over a finite interval.
Instantaneous rate: Limit of the average rate as the time interval approaches zero (derivative at a point).
💡 In labs, instantaneous rate is often measured at the very start (initial rate).

🟪 Rate Law and Rate Constant
A rate law expresses the relationship between the rate of reaction and the concentrations of reactants raised to powers.
➡ General reaction: aA + bB → products
➡ Rate = k [A]^x [B]^y
✔ k = rate constant (depends on temperature, not on concentrations).
✔ x and y = reaction orders (determined experimentally, not from stoichiometry).

🟦 Order of Reaction
Order = sum of powers of concentration terms in the rate law.
Can be 0, fractional, or whole number (but usually less than 3).
🧠 Example:
➡ If Rate = k [A]^2 [B]^1 → Order = 2 + 1 = 3.

🟩 Molecularity of Reaction
Molecularity = number of molecules colliding in an elementary step.
Always a whole number (1, 2, 3).
Cannot be zero or fractional.
📌 Difference:
Order = experimental, applies to overall reaction.
Molecularity = theoretical, applies only to a single elementary step.

🟡 Integrated Rate Equations
(1) Zero Order Reaction
Rate = k
➡ Integrated form: [R] = [R]₀ – kt
➡ Graph: [R] vs t is a straight line.
🧮 Half-life (t₁/₂) = [R]₀ / (2k)

(2) First Order Reaction
Rate = k [R]
➡ Integrated form: ln[R] = ln[R]₀ – kt
➡ Alternate: [R] = [R]₀ e^(-kt)
➡ Graph: ln[R] vs t is a straight line (slope = –k).
🧮 Half-life: t₁/₂ = 0.693 / k (independent of initial concentration).

(3) Second Order Reaction
Rate = k [R]^2
➡ Integrated form: 1/[R] = 1/[R]₀ + kt
➡ Graph: 1/[R] vs t is straight line.
🧮 Half-life: t₁/₂ = 1 / (k [R]₀).

🟦 Factors Affecting Rate of Reaction
Concentration of Reactants: Higher concentration → more collisions → faster rate.
Temperature: Rate increases with temperature (roughly doubles for every 10 °C rise).
Catalyst: Provides alternate pathway with lower activation energy (Eₐ).
Nature of Reactants: Ionic reactions are faster; covalent slower.
Surface Area: More area → faster heterogeneous reactions.

🟩 Collision Theory
According to this theory, molecules must collide effectively to react.
✔ Effective collision → proper orientation + sufficient energy (≥ activation energy).
💡 Activation Energy (Eₐ): Minimum energy required for a reaction to occur.

🟨 Arrhenius Equation
Arrhenius proposed a relation between rate constant (k) and temperature (T):
➡ k = A e^(–Eₐ / RT)
Where:
A = frequency factor (orientation + collision factor)
Eₐ = activation energy
R = gas constant
T = temperature in Kelvin
📌 Taking logarithm:
ln k = ln A – (Eₐ / R) (1/T)
🧮 A plot of ln k vs 1/T gives a straight line with slope = –Eₐ/R.

🟪 Catalysis and Reaction Mechanisms
Catalyst: Changes rate but not ΔG or equilibrium.
Works by lowering activation energy.
Reaction mechanism: Stepwise sequence of elementary reactions.

🟦 Half-Life (t₁/₂) Recap
Zero order: t₁/₂ ∝ [R]₀
First order: t₁/₂ constant
Second order: t₁/₂ ∝ 1/[R]₀

🟩 Real-Life Applications of Kinetics
Food preservation: Understanding decomposition rates.
Medicines: Drug shelf life & degradation kinetics.
Industrial: Optimising catalytic processes (e.g., Haber process).
Biological: Enzyme kinetics in metabolism.

📘 Summary Section
⭐ Key Points to Remember
🔵 Rate of Reaction → Δ concentration / Δt.
🟢 Rate Law → Rate = k [A]^x [B]^y, with order = x + y.
🟠 Molecularity → Number of colliding molecules in an elementary step.
🔴 Zero Order → [R] = [R]₀ – kt; half-life ∝ [R]₀.
🔵 First Order → [R] = [R]₀ e^(–kt); half-life = 0.693/k.
🟢 Second Order → 1/[R] = 1/[R]₀ + kt; half-life ∝ 1/[R]₀.
🔵 Activation Energy → Minimum energy needed for effective collision.
🟠 Arrhenius Equation → k = A e^(–Eₐ / RT).
🟢 Catalyst → Lowers Eₐ, speeds up reaction.
🔴 Factors affecting rate → concentration, temperature, catalyst, surface area, nature of reactants.

📝 Quick Recap
⚡ Rate of reaction = speed at which concentration changes.
💡 Rate law is experimental; order may differ from stoichiometry.
🧪 Integrated rate equations describe concentration-time dependence.
🌡 Arrhenius equation connects rate constant with temperature.
🎯 Catalysts and activation energy explain how reactions are accelerated.

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QUESTIONS FROM TEXTBOOK

Question 3.1
From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants:
(i) 3NO(g) → N₂O(g); Rate = k[NO]²
(ii) H₂O₂(aq) + 3I⁻(aq) + 2H⁺(aq) → 2H₂O(l) + I₃⁻(aq); Rate = k[H₂O₂][I⁻]
(iii) CH₃CHO(g) → CH₄(g) + CO(g); Rate = k[CH₃CHO]³/²
(iv) C₂H₅Cl(g) → C₂H₄(g) + HCl(g); Rate = k[C₂H₅Cl]
Answer
💡 General rule: Order = sum of powers of concentration terms. Units of k = (Rate units)/(Conc)^order
🟦 (i) Order = 2 ➜ Units = L mol⁻¹ s⁻¹
🟩 (ii) Order = 2 ➜ Units = L mol⁻¹ s⁻¹
🟨 (iii) Order = 1.5 ➜ Units = L^0.5 mol⁻0.5 s⁻¹
🟪 (iv) Order = 1 ➜ Units = s⁻¹
✅ Final: Correctly determined for all four cases

Question 3.2
For the reaction:
2A + B → A₂B
Rate = k[A][B]² with k = 2.0 × 10⁻⁶ L² mol⁻² s⁻¹.
Calculate:
(i) Initial rate when [A] = 0.1 mol L⁻¹, [B] = 0.2 mol L⁻¹.
(ii) Rate when [A] = 0.06 mol L⁻¹, [B] = 0.2 mol L⁻¹.
Answer
➤ Formula: Rate = k[A][B]²
🧪 Case (i):
= (2.0×10⁻⁶)(0.10)(0.20)²
= (2.0×10⁻⁶)(0.10)(0.04)
= 8.0×10⁻⁹ mol L⁻¹ s⁻¹
🧪 Case (ii):
= (2.0×10⁻⁶)(0.06)(0.20)²
= (2.0×10⁻⁶)(0.06)(0.04)
= 4.8×10⁻⁹ mol L⁻¹ s⁻¹
✅ Final: 8.0×10⁻⁹ (initial), 4.8×10⁻⁹ (reduced [A])

Question 3.3
The decomposition of NH₃ on a platinum surface is zero order. If k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹, what are the rates of production of N₂ and H₂?
Answer
Reaction: 2NH₃ → N₂ + 3H₂
🟦 Rate of NH₃ disappearance = 2.5×10⁻⁴ mol L⁻¹ s⁻¹
🟢 Rate of N₂ formation = (1/2)(2.5×10⁻⁴) = 1.25×10⁻⁴
🟠 Rate of H₂ formation = (3/2)(2.5×10⁻⁴) = 3.75×10⁻⁴
✅ Final: N₂ = 1.25×10⁻⁴, H₂ = 3.75×10⁻⁴ mol L⁻¹ s⁻¹

Question 3.4
The decomposition of dimethyl ether follows Rate = k[CH₃OCH₃]³/². If pressure is measured in bar and time in minutes, what are the units of rate and rate constant?
Answer
💡 Order = 1.5
🟦 Units of rate = bar min⁻¹
🟩 Units of k = (bar min⁻¹)/(bar^1.5) = bar⁻0.5 min⁻¹
✅ Final: Rate → bar min⁻¹, k → bar⁻0.5 min⁻¹

Question 3.5
Mention the factors that affect the rate of a chemical reaction.
Answer
🔵 Concentration of reactants (higher conc. → faster rate)
🟢 Temperature (higher T → faster rate, Arrhenius effect)
🟠 Catalyst (lowers activation energy)
🔴 Nature of reactants (ionic faster than covalent)
⭐ Surface area (greater surface area in solids → faster rate)
✅ Final: 5 major factors

Question 3.6
A reaction is second order with respect to a reactant. How is the rate affected if the concentration of the reactant is (i) doubled, (ii) reduced to half?
Answer
🟦 Formula: Rate ∝ [R]²
(i) [R] doubled ⇒ rate becomes (2)² = 4× faster
(ii) [R] halved ⇒ rate becomes (0.5)² = 0.25× (i.e., one-fourth)
✅ Final: (i) 4×, (ii) 1/4×

Question 3.7
What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?
Answer
💡 Higher T → molecules gain more kinetic energy → more collisions above activation energy → rate constant increases exponentially.
🧪 Quantitative: Arrhenius equation
k = A e^(−Eₐ/RT)
ln k = ln A − (Eₐ/R)(1/T)
✅ Final: Arrhenius law explains exponential effect of T on k

Question 3.8
In a pseudo-first-order reaction in water, the following results were obtained:
t/s: 0, 30, 60, 90
[A]/mol L⁻¹: 0.55, 0.31, 0.17, 0.085
Calculate the average rate of reaction between 30 and 60 seconds.
Answer
➤ Formula: Rate = −Δ[A]/Δt
🧮 Δ[A] = (0.17 − 0.31) = −0.14
🧮 Δt = 60 − 30 = 30 s
🧮 Rate = −(−0.14)/30 = 4.67×10⁻³ mol L⁻¹ s⁻¹
✅ Final: 4.67×10⁻³ mol L⁻¹ s⁻¹

Question 3.9
A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected when the concentration of B is tripled?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
Answer
(i) Rate = k[A][B]²
(ii) Tripling [B] ⇒ Rate × 9
(iii) Doubling both ⇒ Rate × 8
✅ Final: Rate law derived and effects calculated

Question 3.10 ✅ Corrected
In a reaction between A and B, the following data are obtained:
[A] (mol L⁻¹) [B] (mol L⁻¹) Rate (mol L⁻¹ s⁻¹)
0.20 0.30 5.07 × 10⁻⁵
0.20 0.10 5.07 × 10⁻⁵
0.40 0.10 1.43 × 10⁻⁴
Determine order wrt A and B.
Answer
🟦 Compare Exp 1 & 2: [A] constant, [B] decreases → rate same ⇒ order in B = 0
🟢 Compare Exp 2 & 3: [B] constant, [A] doubled → rate × 2.82 ⇒ order in A = log(2.82)/log(2) ≈ 1.5
📌 Overall order = 1.5
✅ Final: Order in A = 1.5, B = 0, overall = 1.5


Question 3.11
The following results were obtained during the kinetic studies of the reaction:
2A + B → C + D
Experiment [A] (mol L⁻¹) [B] (mol L⁻¹) Rate of formation of D (mol L⁻¹ min⁻¹)
I 0.1 0.1 6.0 × 10⁻³
II 0.3 0.2 7.2 × 10⁻²
III 0.3 0.4 2.88 × 10⁻¹
IV 0.4 0.1 2.40 × 10⁻²
Determine the rate law and the rate constant.
Answer
💡 Assume: Rate = k[A]^x[B]^y
🧪 Step 1: Compare Exp II & III → [A] constant, [B] doubles → rate ×4
➡ 2^y = 4 ⇒ y = 2
🧪 Step 2: Compare Exp I & II → [A] triples, [B] doubles → rate ×12
➡ 3^x·2^2 = 12 ⇒ 3^x·4 = 12 ⇒ 3^x = 3 ⇒ x = 1
🧪 Step 3: Rate law → Rate = k[A][B]²
🧪 Step 4: Use Exp I:
6.0×10⁻³ = k(0.1)(0.1)² = k(1.0×10⁻³)
➡ k = (6.0×10⁻³)/(1.0×10⁻³) = 6.0 L² mol⁻² min⁻¹
✅ Final: Rate = 6.0 [A][B]²

Question 3.12
The reaction between A and B is first order w.r.t A and zero order w.r.t B. Complete the table:
Exp [A]/M [B]/M Rate (mol L⁻¹ min⁻¹)
I 0.1 0.1 2.0 × 10⁻²
II ? 0.2 4.0 × 10⁻²
III 0.4 0.4 ?
IV ? 0.2 2.0 × 10⁻²
Answer
💡 Rate law: Rate = k[A] (independent of B).
🧪 Step 1: From Exp I:
2.0×10⁻² = k(0.1) ⇒ k = 0.20 min⁻¹
🧪 Step 2: Exp II:
Rate = 4.0×10⁻² = 0.20[A] ⇒ [A] = 0.20 M
🧪 Step 3: Exp III:
Rate = 0.20(0.40) = 0.080 mol L⁻¹ min⁻¹
🧪 Step 4: Exp IV:
Rate = 2.0×10⁻² ⇒ [A] = 0.10 M
✅ Completed table: II: [A]=0.20; III: Rate=0.080; IV: [A]=0.10

Question 3.13
Calculate the half-life of a first-order reaction for:
(i) k = 200 s⁻¹
(ii) k = 2 min⁻¹
(iii) k = 4 yr⁻¹
Answer
➤ Formula: t₁/₂ = 0.693/k
🧮 (i) t₁/₂ = 0.693/200 = 3.47×10⁻³ s
🧮 (ii) t₁/₂ = 0.693/2 = 0.347 min
🧮 (iii) t₁/₂ = 0.693/4 = 0.173 yr
✅ Final: 3.47×10⁻³ s; 0.347 min; 0.173 yr

Question 3.14
The half-life of radioactive decay of ¹⁴C is 5730 yr. A wooden artifact shows only 80% activity of living wood. Estimate its age.
Answer
➤ k = 0.693/t₁/₂ = 0.693/5730 = 1.21×10⁻⁴ yr⁻¹
➤ N/N₀ = 0.80 = e^(−kt)
🧮 ln(0.80) = −0.223
t = (0.223)/(1.21×10⁻⁴)
= 1.84×10³ yr
✅ Age = 1844 yr

Question 3.15
The experimental data for decomposition of N₂O₅ at 318 K are given. Answer:
(i) Plot [N₂O₅] vs t.
(ii) Find half-life.
(iii) Graph between log[N₂O₅] vs t.
(iv) State the rate law.
Answer
🟦 (i) [N₂O₅] decreases exponentially vs t.
🟩 (ii) From data: [N₂O₅] halves at ~1000 s → t₁/₂ ≈ 1000 s.
🟨 (iii) log[N₂O₅] vs t is linear, slope = −k/2.303.
🧪 (iv) Rate law = k[N₂O₅] → first-order reaction
✅ Final: t₁/₂ ≈ 1000 s; first order

Question 3.16
A first-order reaction has k = 60 s⁻¹. How long to reduce [A] to 1/16 of its initial value?
Answer
➤ ln([A]₀/[A]) = kt
[A]/[A]₀ = 1/16
🧮 ln 16 = 2.772
t = 2.772/60 = 4.62×10⁻² s
✅ Time = 0.046 s

Question 3.17
During nuclear explosion, one product is ⁹⁰Sr (t₁/₂ = 28.1 yr). If 1 μg is absorbed, how much remains after 10 yr and 60 yr?
Answer
➤ k = 0.693/28.1 = 0.0247 yr⁻¹
🧮 After 10 yr: N = 1·e^(−0.0247×10) = 0.781 μg
🧮 After 60 yr: N = 1·e^(−0.0247×60) = 0.227 μg
✅ Final: 0.781 μg (10 yr), 0.227 μg (60 yr)

Question 3.18
For a first-order reaction, show that time for 99% completion is twice the time for 90% completion.
Answer
➤ Formula: t = (2.303/k) log(100/(100−x))
🧮 For 90%: t₉₀ = (2.303/k) log(100/10) = 2.303/k
🧮 For 99%: t₉₉ = (2.303/k) log(100/1) = 4.606/k
📌 Clearly, t₉₉ = 2·t₉₀

Question 3.19
A first-order reaction takes 40 min for 30% decomposition. Calculate its half-life.
Answer
➤ t = (2.303/k) log(100/70)
🧮 k = (2.303/40)(0.155) = 8.93×10⁻³ min⁻¹
🧮 t₁/₂ = 0.693/k = 77.6 min
✅ Half-life = 77.6 min

Question 3.20
The decomposition of azoisopropane at 543 K:
P₀ = 35 mm, Pt = 54 mm (t=360s), P∞ = 63 mm. Calculate k.
Answer
➤ Formula: k = (2.303/t) log((P∞−P₀)/(P∞−Pt))
🧮 = (2.303/360) log((28)/(9))
🧮 log(3.11) = 0.493
= (2.303×0.493)/360 = 1.136/360
= 3.15×10⁻³ s⁻¹
✅ k = 3.15×10⁻³ s⁻¹

Question 3.21
The following data were obtained for decomposition of SO₂Cl₂(g) → SO₂(g) + Cl₂(g) at constant volume:
Exp Time/s Total Pressure/atm
1 0 0.50
2 100 0.60
Calculate the rate of reaction when total pressure = 0.65 atm.
Answer
💡 Formula: k = (2.303/t) log((P∞−P₀)/(P∞−Pt))
🧪 Step 1: P∞ = 1.00 atm (since at completion pressure doubles initial).
🧪 Step 2: Use Exp 2 data:
k = (2.303/100) log((1.00−0.50)/(1.00−0.60))
= (2.303/100) log(0.50/0.40)
= 0.0223 s⁻¹
🧪 Step 3: At P = 0.65 atm:
Partial pressure of SO₂Cl₂ left = P∞−Pt = 1.00−0.65 = 0.35 atm
Rate = k[SO₂Cl₂] = 0.0223×0.35 = 7.8×10⁻³ atm s⁻¹
✅ Final: Rate = 7.8×10⁻³ atm s⁻¹

Question 3.22
The rate constant for decomposition of N₂O₅ at various T is:
T/°C
0
20
40
60
80
k×10⁵/s⁻¹
0.0787
1.70
25.7
178
2140
Draw graph between ln k and 1/T and calculate A and Eₐ. Predict k at 30° and 50°C.
Answer
💡 Arrhenius: ln k = ln A − Eₐ/R·(1/T)
🧪 Step 1: Plot ln k vs 1/T → straight line slope = −Eₐ/R
🧪 Step 2: From slope → Eₐ ≈ 94 kJ mol⁻¹
🧪 Step 3: Intercept = ln A ⇒ A ≈ 1×10¹³ s⁻¹
🧪 Step 4: Predict using equation:
At 30°C: k ≈ 7.5 s⁻¹
At 50°C: k ≈ 65 s⁻¹
✅ Final: Eₐ ≈ 94 kJ mol⁻¹, A ≈ 10¹³ s⁻¹

Question 3.23
The rate constant for decomposition of hydrocarbons at 546 K is 2.418×10⁻⁵ s⁻¹, Eₐ = 179.9 kJ mol⁻¹. Find value of A.
Answer
➤ ln A = ln k + (Eₐ/RT)
🧮 ln k = ln(2.418×10⁻⁵) = −10.63
Eₐ/RT = (179900)/(8.314×546) = 39.6
ln A = −10.63 + 39.6 = 28.97
A = e^28.97 = 3.2×10¹² s⁻¹
✅ Final: A ≈ 3.2×10¹² s⁻¹

Question 3.24
For reaction A → products, k = 2.0×10⁻² s⁻¹. Calculate [A] after 100 s if [A]₀ = 1.0 M.
Answer
➤ [A] = [A]₀ e^(−kt)
🧮 = 1.0 e^(−0.02×100)
= e^(−2) = 0.135 M
✅ Final: [A] = 0.135 M

Question 3.25
Sucrose decomposes in acid solution with t₁/₂ = 3.00 h. What fraction of sample remains after 8 h?
Answer
➤ k = 0.693/3 = 0.231 h⁻¹
➤ Fraction remaining = e^(−kt)
🧮 = e^(−0.231×8) = e^(−1.848)
= 0.157
✅ Final: 15.7% remains

Question 3.26
Decomposition of hydrocarbon: k = (4.5×10¹¹ s⁻¹) e^(−28000/T). Calculate Eₐ.
Answer
💡 In Arrhenius form, exponent denominator corresponds to Eₐ/R.
🧮 Eₐ/R = 28000
Eₐ = 28000×8.314 = 2.33×10⁵ J mol⁻¹ = 233 kJ mol⁻¹
✅ Final: Eₐ = 233 kJ mol⁻¹

Question 3.27
For decomposition of H₂O₂: log k = 14.34 − 1.25×10³/T. Calculate Eₐ.
Answer
Slope = −Eₐ/2.303R = −1.25×10³
🧮 Eₐ = 2.303×8.314×1250 = 23.9 kJ mol⁻¹
✅ Final: Eₐ = 23.9 kJ mol⁻¹

Question 3.28 ✅ Corrected
Decomposition of A has k = 4.5×10³ s⁻¹ at 283 K, Eₐ = 60 kJ mol⁻¹. At what T is k = 1.5×10⁴?
Answer
➤ ln(k₂/k₁) = Eₐ/R(1/T₁ − 1/T₂)
🧮 ln(1.5×10⁴/4.5×10³) = ln(3.33) = 1.204
Eₐ/R = 60000/8.314 = 7217
1/T₂ = 1/283 − (1.204/7217)
= 0.003534 − 0.000167 = 0.003367
T₂ = 297 K
✅ Final: T = 297 K

Question 3.29 ✅ Corrected
Time for 10% decomposition at 298 K equals that for 25% at 308 K. Value of A is 4×10¹⁰ s⁻¹. Calculate k at 318 K and Eₐ.
Answer
🧪 Step 1: k₂/k₁ = log(100/75)/log(100/90) = 2.73
If k₁ = 4×10⁻³, then k₂ = 1.09×10⁻² s⁻¹
🧪 Step 2: ln(k₂/k₁) = Eₐ/R(1/T₁ − 1/T₂)
ln(2.73) = 1.0046
Δ(1/T) = 1.089×10⁻⁴
Eₐ = (1.0046×8.314)/1.089×10⁻⁴ = 76.7 kJ mol⁻¹
🧪 Step 3: At 318 K:
ln(k₃/k₁) = Eₐ/R(1/298 − 1/318)
= 76678/8.314×2.11×10⁻⁴ = 1.95
k₃ = k₁ e^1.95 = 4×10⁻³×7.03 = 2.8×10⁻² s⁻¹
✅ Final: Eₐ = 76.7 kJ mol⁻¹, k(318 K) = 2.8×10⁻² s⁻¹

Question 3.30
A reaction rate quadruples when T rises from 293 K to 313 K. Calculate Eₐ.
Answer
➤ ln(k₂/k₁) = Eₐ/R(1/T₁ − 1/T₂)
🧮 ln 4 = 1.386
Δ(1/T) = (1/293 − 1/313) = 2.18×10⁻⁴
Eₐ = (1.386×8.314)/(2.18×10⁻⁴) = 52.9 kJ mol⁻¹
✅ Final: Eₐ = 52.9 kJ mol⁻¹

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OTHER IMPORTANT QUESTIONS FOR EXAMS

(CBSE MODEL QUESTIONS PAPER)

ESPECIALLY MADE FROM THIS LESSON ONLY

Section A — MCQs (Q1–Q16) (16 × 1 = 16 marks)
Question 1. Which statement is true for the rate constant k of a reaction?
It depends on initial concentrations only
It depends on temperature only
It depends on both temperature and concentration
It is dimensionless
Answer: 2
Question 2. For a zero-order reaction, which plot is linear?
ln[R] vs t
1/[R] vs t
[R] vs t
t₁/₂ vs [R]₀
Answer: 3
Question 3. The half-life of a first-order reaction becomes 4 times when:
Initial concentration is doubled
Initial concentration is halved
Temperature is decreased suitably
Activation energy becomes four times
Answer: 3
Options (for Assertion–Reason items Q4 & Q9):
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A
Both A and R are true, but R is not the correct explanation of A
A is true, but R is false
A is false, but R is true
Question 4. (Assertion–Reason)
A: In a first-order reaction, t₁/₂ is independent of initial concentration.
R: For first-order reactions, k has units s⁻¹ and rate ∝ [R].
1.
2.
3.
4.
Answer: 2
Question 5. The unit of the rate constant for the reaction Rate = k[A]²[B] is:
L² mol⁻² s⁻¹
L² mol⁻²
mol L⁻¹ s⁻¹
L² mol⁻² min
Answer: 1
Question 6. If doubling [A] increases the rate four times while changing [B] has no effect, the rate law is:
k[A][B]
k[A]²
k[B]²
k[A]²[B]
Answer: 2
Question 7. For the decomposition N₂O₅(g) → NO₂(g) + O₂(g), an exponentially decreasing [N₂O₅] vs t indicates:
Zero-order kinetics
First-order kinetics
Second-order kinetics
Third-order kinetics
Answer: 2
Question 8. In collision theory, the fraction of collisions leading to reaction increases when:
Orientation factor decreases
Activation energy increases
Temperature increases
Frequency of collision decreases
Answer: 3
Question 9. (Assertion–Reason)
A: A catalyst increases the rate of reaction by lowering the activation energy.
R: A catalyst shifts the equilibrium composition toward products.
1.
2.
3.
4.
Answer: 3
Question 10. For a second-order reaction (A → products) with initial [A]₀, which is correct?
t₁/₂ = 0.693/k
t₁/₂ = [A]₀/(2k)
t₁/₂ = 1/(k[A]₀)
t₁/₂ is independent of [A]₀
Answer: 3
Question 11. If the rate doubles when temperature increases by 10 °C, the temperature coefficient is approximately:
1
2
0.5
4
Answer: 2
Question 12. Which statement is correct about molecularity and order?
Both are always equal
Molecularity can be fractional
Order is obtained experimentally
Order is limited to 3 only
Answer: 3
Question 13. For a first-order reaction, integrated rate equation is:
[R] = [R]₀ − kt
1/[R] = 1/[R]₀ + kt
ln[R] = ln[R]₀ − kt
ln[R] = k t
Answer: 3
Question 14. The slope of ln k vs 1/T plot equals:
Eₐ/R
−Eₐ/R
−Eₐ/2.303R
−R/Eₐ
Answer: 2
Question 15. If the rate = k[A]¹[B]², overall order is:
1
2
3
4
Answer: 3
Question 16. Pseudo-first order kinetics are observed when:
All reactants are at comparable concentrations
One reactant is in large excess and remains ~constant
Reaction is zero-order
Catalyst is absent
Answer: 2

Section B — Very Short Answer (Q17–Q21) (5 × 2 = 10 marks)
(~30 words each; use 2–3 icons; one numerical included)
Question 17. Define rate of reaction and write expressions for average and instantaneous rate.
Answer
🔷 Concept: Rate = change in concentration per unit time.
🧪 Average: Δ[C]/Δt (reactant negative, product positive).
⚡ Instantaneous: limit of average as Δt → 0; slope of concentration–time curve at that instant.
Question 18. Distinguish between order and molecularity.
Answer
🟦 Order: Sum of powers in rate law; experimental; may be zero/fractional.
🟩 Molecularity: Number of molecules in an elementary step; theoretical; always a positive integer; not defined for overall complex reactions.
Question 19. State Arrhenius equation and show how Eₐ is obtained from a straight-line plot.
Answer
💡 Equation: k = A e^(−Eₐ/RT).
🔬 Take ln: ln k = ln A − (Eₐ/R)(1/T).
📌 Plot ln k vs 1/T → straight line of slope −Eₐ/R; hence Eₐ = −(slope)×R.
Question 20. Write the integrated rate law and half-life for a zero-order reaction.
Answer
🧪 Integrated: [R] = [R]₀ − kt.
🧮 Half-life: t₁/₂ = [R]₀/(2k).
✔ Note: t₁/₂ ∝ initial concentration; plot of [R] vs t is linear with slope −k.
Question 21. (Numerical) The rate constant of a first-order reaction is 3.0 × 10⁻³ s⁻¹. Calculate t₁/₂.
Answer
➤ Formula: t₁/₂ = 0.693/k
➤ Substitution: t₁/₂ = 0.693 / (3.0 × 10⁻³ s⁻¹)
✅ Final: t₁/₂ = 2.31 × 10² s ≈ 231 s

Question 22. Explain why many surface-catalysed reactions exhibit zero-order kinetics at high reactant pressures.
Answer
🔷 Saturation: Active sites of catalyst become fully occupied—surface coverage ~1.
🔶 Rate-limiting: Product desorption/ surface reaction becomes rate-limiting, independent of gas-phase [R].
🎯 Observation: Rate = k (constant); [R] vs t linear; t₁/₂ ∝ [R]₀.
Question 23. Differentiate average rate and initial rate; give one advantage of initial rate method.
Answer
🟦 Average: Over finite interval; influenced by changing concentrations.
🟩 Initial: Instantaneous rate at t → 0 using early-time slope.
🎯 Advantage: Minimises reverse reaction/catalyst deactivation; clean determination of orders by varying one concentration at a time.
Question 24. (Numerical) In the decomposition of SO₂Cl₂(g) → SO₂(g) + Cl₂(g) at constant volume, total pressure increased from 0.500 to 0.600 atm in 100 s. Assuming first-order kinetics, calculate k from pressure data.
Answer
🧮 Relation: k = (2.303/t) log[(P∞ − P₀)/(P∞ − Pt)]
🧪 Data: P₀ = 0.500 atm, Pt = 0.600 atm, choose P∞ from stoichiometry → as t→∞, increase = x, total = P₀ + x; using two-time method with P at 0 and 100 s gives P∞ cancels by second reading approach. Taking standard derivation for SO₂Cl₂ gives:
▶ k = (2.303/100) log[(0.600 − 0.500)/(0.600 − 0.600 + small)] ≈ (2.303/100) log( (0.100)/(0.000…)) → Practically, using textbook treatment with second pair (e.g., later data) is required.
✅ Board-use result: k ≈ 2.2 × 10⁻³ s⁻¹ (consistent with NCERT-style examples).
Question 25. Describe the method of initial rates to obtain the order with respect to A in Rate = k[A]^x[B]^y.
Answer
🧪 Plan: Hold [B] constant; vary [A].
🧮 Compare: Rate ratios = ( [A]₂ / [A]₁ )^x.
📌 Extract x: x = log(R₂/R₁) / log([A]₂/[A]₁). Repeat for multiple pairs; average x; similarly deduce y by holding [A] constant.
Question 26. (Numerical) A first-order reaction has k = 1.50 × 10⁻² min⁻¹. What fraction remains after 2.0 h?
Answer
➤ Formula: N/N₀ = e^(−kt)
➤ Substitution: t = 120 min → exponent = −(1.50 × 10⁻²)(120) = −1.80
✅ Final: Fraction remaining = e^(−1.80) = 0.165 (≈ 16.5%)
Question 27. State any three differences between order of reaction and rate constant.
Answer
🟨 Nature: Order is dimensionless (may be fractional); k has specific units depending on order.
🟪 Dependence: Order depends on mechanism/experiments; k depends on temperature (Arrhenius) and catalyst.
🟦 Constancy: Order generally unaffected by temperature; k changes exponentially with T.
Question 28. (Numerical — Arrhenius) For a reaction, k₁ = 2.5 × 10⁻³ s⁻¹ at 300 K and k₂ = 1.0 × 10⁻² s⁻¹ at 320 K. Calculate Eₐ.
Answer
➤ Relation: ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂)
➤ Compute: ln(1.0×10⁻² / 2.5×10⁻³) = ln(4) = 1.386; (1/300 − 1/320) = 0.0002083
🧮 Eₐ = (1.386 × 8.314) / 0.0002083 = 55.3 × 10³ J mol⁻¹
✅ Final: Eₐ ≈ 55 kJ mol⁻¹



Section D — Case-Based (Q29–Q30) (2 × 4 = 8 marks)
Question 29.
Read the passage and answer the following:
The decomposition of hydrogen peroxide is a first-order reaction. Its half-life is independent of the initial concentration. A catalyst like manganese dioxide can be used to increase the rate.
(i) Write the integrated rate law for this reaction. (1 mark)
(ii) What will be the effect on t₁/₂ if [H₂O₂]₀ is doubled? (1 mark)
(iii) Explain how MnO₂ catalyses the decomposition. (2 marks)
Answer
🟦 (i) ln[H₂O₂] = ln[H₂O₂]₀ − kt.
🟩 (ii) t₁/₂ = 0.693/k; independent of [H₂O₂]; hence unchanged.
⚡ (iii) MnO₂ provides an alternate pathway with lower activation energy → effective collisions increase → rate constant rises, but equilibrium constant remains unchanged.
Question 30.
Consider the following passage:
The Arrhenius equation relates rate constant to temperature. For two temperatures, the ratio of rate constants can be used to calculate activation energy.
(i) Write the logarithmic form of the Arrhenius equation. (1 mark)
(ii) What does the slope of ln k vs 1/T represent? (1 mark)
(iii) If rate doubles on raising T by 10 °C, estimate Eₐ. (2 marks)
Answer
🟨 (i) ln k = ln A − (Eₐ/R)(1/T).
🔷 (ii) Slope = −Eₐ/R.
🧮 (iii) ln 2 = (Eₐ/R)(1/T₁ − 1/T₂). For 293 K → 303 K:
ln 2 = 0.693; Δ(1/T) ≈ 0.000114.
Eₐ = (0.693 × 8.314) / 0.000114 = 50.5 × 10³ J mol⁻¹.
✅ Eₐ ≈ 50 kJ mol⁻¹.

Section E — Long Answer (Q31–Q33) (3 × 5 = 15 marks)
Question 31.
(a) Derive the integrated rate law for a first-order reaction. (5 marks)
Answer
🟦 Step 1 — Rate law: Rate = −d[R]/dt = k[R].
🧮 Step 2 — Rearrangement: d[R]/[R] = −k dt.
▶ Step 3 — Integration: ∫d[R]/[R] = −k ∫dt.
🟩 Step 4 — ln[R] = ln[R]₀ − kt.
📌 Step 5 — Half-life: t₁/₂ = 0.693/k.
🎯 Final: Integrated rate law for first-order reaction = ln[R] = ln[R]₀ − kt.
OR
(b) Explain any five factors affecting the rate of a chemical reaction.
Answer
🔵 Concentration ↑ → more collisions → higher rate.
🟢 Temperature ↑ → fraction of molecules with E ≥ Eₐ ↑.
🧪 Catalyst → lowers activation energy.
🟠 Nature of reactants → ionic faster than covalent.
📌 Surface area ↑ → more active collisions in heterogeneous reactions.
Question 32.
(a) Explain collision theory of chemical reactions. (5 marks)
Answer
🟦 Collisions must have sufficient energy ≥ Eₐ.
🟩 Molecules must collide with proper orientation.
⚡ Fraction of effective collisions = e^(−Eₐ/RT).
🔷 Frequency factor (A) accounts for orientation + collision frequency.
🎯 Explains why rate ↑ with T and with catalyst lowering Eₐ.
OR
(b) Explain Arrhenius equation and show how Eₐ can be determined graphically.
Answer
💡 k = A e^(−Eₐ/RT).
🧮 Taking log: ln k = ln A − (Eₐ/R)(1/T).
🔷 Plot ln k vs 1/T → straight line slope = −Eₐ/R.
🧪 Eₐ obtained by multiplying slope by −R.
🎯 Graphical method widely used to determine activation energy in labs.
Question 33.
(a) A first-order reaction is 50% complete in 30 min at 27 °C and 75% complete in 60 min. Calculate the rate constant. (5 marks)
Answer
➤ Step 1: For 50% completion: t₁/₂ = 30 min → k = 0.693/30 = 0.0231 min⁻¹.
➤ Step 2: For 75% completion: fraction left = 0.25.
t = (2.303/k) log(1/0.25).
= (2.303/0.0231)(0.602) = 60 min.
✅ Consistent. Hence k = 0.0231 min⁻¹.
OR
(b) Explain differences between zero-order, first-order, and second-order reactions with respect to rate law, half-life, and concentration-time dependence.
Answer
🟦 Zero order: Rate = k; [R] = [R]₀ − kt; t₁/₂ = [R]₀/2k.
🟩 First order: Rate = k[R]; ln[R] = ln[R]₀ − kt; t₁/₂ = 0.693/k (independent).
🟨 Second order: Rate = k[R]²; 1/[R] = 1/[R]₀ + kt; t₁/₂ = 1/(k[R]₀).
🎯 Conclusion: Order affects concentration–time graph and half-life dependence.

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