Chemical Kinetics — Class 12 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Chemical Kinetics" — 7 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Chemical Kinetics" — 7 important questions with detailed answers for CBSE board exam…
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NCERT Solutions for Class 12 Chemistry — Chemical Kinetics. Step-by-step answers to all 7 textbook questions from this chapter, written for CBSE board preparation and free to use.
Chemical Kinetics — All 7 Questions Solved
Q1. Define rate of reaction. What factors affect the rate of chemical reaction?
Rate of Reaction:
The speed at which reactants are consumed or products are formed per unit time.
Mathematical expression:
Rate = Δ[concentration] / Δt (change in concentration per unit time)
Units: mol/(L·s) or mol·L⁻¹·s⁻¹
Can be expressed for reactants or products:
For reaction: aA + bB → cC + dD
Rate = -(1/a) × d[A]/dt = -(1/b) × d[B]/dt = (1/c) × d[C]/dt = (1/d) × d[D]/dt
Factors affecting rate of reaction:
1. Concentration (Pressure for gases):
Increasing concentration increases collision frequency between molecules, increasing reaction rate.
Rate ∝ [Concentration]^n
2. Temperature:
Increasing temperature increases kinetic energy of molecules.
For every 10°C increase, reaction rate approximately doubles (rule of thumb).
Exact relationship given by Arrhenius equation: k = Ae^(-E_a/RT)
3. Nature of reactants:
Different substances react at different rates due to differences in bond strength and reaction mechanism.
4. Presence of catalyst:
Catalyst speeds up reaction by lowering activation energy without being consumed.
Lowers E_a, increases reaction rate significantly.
5. Surface area (for heterogeneous reactions):
Increasing surface area increases contact between reactants, increasing reaction rate.
More finely divided solid reacts faster.
6. Presence of light:
Some reactions are photochemically initiated (e.g., photosynthesis, decomposition of AgCl)
Q2. Define order of reaction. How is it determined experimentally?
Order of Reaction:
The sum of powers of concentration terms in the rate equation.
For reaction: aA + bB → products
Rate = k[A]^m[B]^n
Order of reaction = m + n
where k = rate constant, m and n are experimentally determined powers (not necessarily equal to stoichiometric coefficients a and b)
Types of order:
1. Zero order: Order = 0, Rate = k, independent of concentration
2. First order: Order = 1, Rate = k[A], rate depends linearly on concentration
3. Second order: Order = 2, Rate = k[A][B] or k[A]²
4. Fractional order: Order = 0.5, 1.5, etc. (rare)
Experimental methods to determine order:
1. Method of Initial Rates:
Vary concentration of one reactant, keep others constant.
Measure initial rate for each concentration.
Use: Rate₁/Rate₂ = ([A₁]/[A₂])^m
Solve for m (order with respect to A)
Repeat for other reactants.
Example:
Experiment [A] Rate
1 0.1 M 1.0 mol/L·s
2 0.2 M 2.0 mol/L·s
Ratio: 2.0/1.0 = (0.2/0.1)^m = 2^m
Therefore m = 1 (first order in A)
2. Half-life Method:
Measure time taken for concentration to reduce to half.
If half-life is constant → zero order
If half-life increases as concentration decreases → first order
If half-life decreases as concentration decreases → second order
3. Integrated Rate Law Method:
Plot concentration vs time data:
Zero order: [A] vs t is linear
First order: ln[A] vs t is linear
Second order: 1/[A] vs t is linear
4. Differential Method:
Take differential of experimental data to directly apply rate law
Q3. A reaction has rate law: Rate = k[A]²[B]. Calculate the order of reaction and rate constant unit.
Given: Rate = k[A]²[B]
Step 1: Determine order of reaction
Order = sum of powers of concentration terms
Order = 2 (from [A]²) + 1 (from [B]) = 3
Order of reaction = 3 (third order)
Step 2: Determine unit of rate constant k
From rate equation: Rate = k[A]²[B]
k = Rate / ([A]²[B])
Substitute units:
k = [mol·L⁻¹·s⁻¹] / [mol·L⁻¹]² × [mol·L⁻¹]
k = [mol·L⁻¹·s⁻¹] / [mol³·L⁻³]
k = [mol·L⁻¹·s⁻¹] × [L³·mol⁻³]
k = L²·mol⁻²·s⁻¹
or
k = mol⁻²·L²·s⁻¹
Alternatively expressed as: M⁻²·s⁻¹ (where M = mol/L)
General pattern:
For nth order reaction:
Unit of k = (concentration)^(1-n) × (time)⁻¹
For 3rd order: k = (concentration)⁻² × (time)⁻¹ = L²·mol⁻²·s⁻¹
Q4. Write and explain the Arrhenius equation. What is activation energy?
Arrhenius Equation:
k = Ae^(-E_a/RT)
or in logarithmic form:
ln(k) = ln(A) - E_a/RT
or relating two rate constants at different temperatures:
ln(k₂/k₁) = (E_a/R) × (T₂ - T₁)/(T₁ × T₂)
where:
k = rate constant
A = pre-exponential factor (frequency factor, units same as k)
E_a = activation energy (J/mol or kcal/mol)
R = universal gas constant = 8.314 J/(mol·K)
T = absolute temperature (K)
e = Euler's number (2.718)
Activation Energy (E_a):
The minimum energy required for reactants to collide effectively and form products.
Characteristics:
1. Always positive value
2. Measured relative to energy of reactants
3. Independent of temperature
4. Lower E_a → faster reaction
5. Catalyst lowers E_a without being consumed
Significance of Arrhenius Equation:
1. Explains temperature effect on reaction rate: Small temperature increase significantly increases k
2. Shows exponential relationship between rate and temperature
3. Provides quantitative prediction of rate at different temperatures
4. E_a can be determined experimentally from slope of ln(k) vs 1/T plot
5. Pre-exponential factor A related to frequency and orientation of collisions
Example:
For reaction with E_a = 50 kJ/mol at 300 K:
Doubling temperature to 600 K increases k by factor of e^(50000/(8.314) × (1/600 - 1/300)) ≈ 10⁵
Effect of catalyst:
Catalyst lowers E_a to E'_a
New rate constant k' = Ae^(-E'_a/RT) >> k
Reaction rate increases dramatically without changing E_a itself
Q5. For a first-order reaction with k = 0.01 s⁻¹, calculate the half-life.
Given: First-order reaction, k = 0.01 s⁻¹
For first-order reaction:
Integrated rate law: ln[A] = ln[A]₀ - kt
or: [A] = [A]₀e^(-kt)
Half-life (t₁/₂) definition:
Time required for concentration to reduce to half of initial value.
At t = t₁/₂, [A] = [A]₀/2
Step 1: Substitute into integrated rate law
ln([A]₀/2) = ln[A]₀ - kt₁/₂
ln[A]₀ - ln(2) = ln[A]₀ - kt₁/₂
-ln(2) = -kt₁/₂
ln(2) = kt₁/₂
t₁/₂ = ln(2)/k
Step 2: Calculate numerical value
ln(2) = 0.693
t₁/₂ = 0.693 / 0.01
t₁/₂ = 69.3 s
Key insight for first-order reactions:
1. Half-life is independent of initial concentration
2. Each successive half-life is equal:
- After t₁/₂: 50% remains
- After 2t₁/₂: 25% remains
- After 3t₁/₂: 12.5% remains
3. Half-life depends only on rate constant
Answer: Half-life = 69.3 s (or approximately 69 s)
Comparison with other orders:
Zero order: t₁/₂ = [A]₀/(2k) - depends on [A]₀
Second order: t₁/₂ = 1/(k[A]₀) - depends on [A]₀
Q6. Explain the concept of a catalyst. How does it affect activation energy and reaction mechanism?
Catalyst:
A substance that speeds up a reaction by lowering the activation energy without being consumed in the reaction.
Characteristics of catalysts:
1. Increases reaction rate (forward and reverse)
2. Not consumed overall (regenerated)
3. Does not change equilibrium position
4. Does not change ΔG (Gibbs free energy)
5. Works by providing alternative reaction mechanism with lower E_a
6. Small amount can catalyze large amount of reactants
7. Specific to particular reactions
Effect on Activation Energy:
Uncatalyzed reaction:
k_uncatalyzed = Ae^(-E_a/RT)
Catalyzed reaction:
k_catalyzed = Ae^(-E'_a/RT)
where E'_a < E_a (lower activation energy)
k_catalyzed / k_uncatalyzed = e^((E_a - E'_a)/RT)
Small decrease in E_a causes exponential increase in rate.
Example: Lowering E_a by 50 kJ/mol increases rate ~10⁶ fold at room temperature.
Effect on Reaction Mechanism:
Uncatalyzed:
Reactants → Products (direct, high E_a)
Catalyzed:
Reactants + C → Intermediate₁
Intermediate₁ → Intermediate₂
Intermediate₂ → Products + C
Alternative mechanism with intermediate steps, each with lower E_a.
Overall: No change in reactants or products, just different pathway.
Energy diagram:
- Uncatalyzed: One large energy barrier
- Catalyzed: Multiple smaller barriers
- Catalyst enters and exits unchanged
Examples:
1. MnO₂ catalyzes decomposition of H₂O₂
2. Enzymes catalyze biological reactions
3. Pt in catalytic converters oxidize pollutants
4. Concentrated H₂SO₄ catalyzes esterification
Limitations:
1. Cannot make unfavorable reaction become favorable (ΔG remains same)
2. Cannot increase yield at equilibrium
3. Speeds up both forward and reverse reactions equally
Q7. A reaction shows the following data. Determine the order of reaction using initial rate method. Experiment [A] (M) [B] (M) Initial Rate (mol/L·s) 1 0.1 0.1 1.0 × 10⁻⁴ 2 0.2 0.1 4.0 × 10⁻⁴ 3 0.1 0.2 2.0 × 10⁻⁴
Given reaction rate data for determining order using initial rate method.
Assume rate law: Rate = k[A]^m[B]^n
Step 1: Find order with respect to A
Compare Experiment 1 and 2 (B is constant):
Rate₁/Rate₂ = [A₁]^m / [A₂]^m
(1.0 × 10⁻⁴) / (4.0 × 10⁻⁴) = [0.1]^m / [0.2]^m
0.25 = (0.1/0.2)^m
0.25 = (0.5)^m
0.25 = (1/2)^m
Note: (1/2)² = 0.25
Therefore: m = 2
Order with respect to A = 2 (second order in A)
Step 2: Find order with respect to B
Compare Experiment 1 and 3 (A is constant):
Rate₁/Rate₃ = [B₁]^n / [B₃]^n
(1.0 × 10⁻⁴) / (2.0 × 10⁻⁴) = [0.1]^n / [0.2]^n
0.5 = (0.1/0.2)^n
0.5 = (0.5)^n
0.5 = (0.5)¹
Therefore: n = 1
Order with respect to B = 1 (first order in B)
Step 3: Write rate law
Rate = k[A]²[B]¹
or
Rate = k[A]²[B]
Overall order = 2 + 1 = 3 (third order reaction)
Step 4: Calculate rate constant k
Using Experiment 1:
1.0 × 10⁻⁴ = k × (0.1)² × (0.1)
1.0 × 10⁻⁴ = k × 0.01 × 0.1
1.0 × 10⁻⁴ = k × 1.0 × 10⁻³
k = (1.0 × 10⁻⁴) / (1.0 × 10⁻³)
k = 0.1 L²·mol⁻²·s⁻¹
Answer:
Rate law: Rate = k[A]²[B]
Order of reaction = 3
Rate constant k = 0.1 L²·mol⁻²·s⁻¹
Chemical Kinetics — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the unit of the rate of a chemical reaction?
- A. mol L⁻¹ s⁻¹
- B. L mol⁻¹ s⁻¹
- C. mol s⁻¹
- D. L s⁻¹
Answer: A. mol L⁻¹ s⁻¹ — The rate of reaction is defined as the change in concentration per unit time. Concentration is typically measured in mol L⁻¹ and time in s, so the unit is mol L⁻¹ s⁻¹.
MCQ 2. For a reaction, the rate constant k has units of L mol⁻¹ s⁻¹. What is the order of the reaction?
- A. Zero order
- B. First order
- C. Second order
- D. Third order
Answer: C. Second order — The general unit for the rate constant k for an n-th order reaction is (mol L⁻¹)¹⁻ⁿ s⁻¹. For second order (n=2), the unit is (mol L⁻¹)¹⁻² s⁻¹ = (mol L⁻¹)⁻¹ s⁻¹ = L mol⁻¹ s⁻¹.
MCQ 3. For a complex reaction, what is true about its molecularity?
- A. It is always equal to the order of the reaction.
- B. It can be fractional.
- C. It is determined for each elementary step and refers to the number of reacting species in that step.
- D. It is an experimentally determined value.
Answer: C. It is determined for each elementary step and refers to the number of reacting species in that step. — Molecularity is defined for an elementary reaction and is the number of reactant molecules taking part in that step. It cannot be fractional and is always an integer.
MCQ 4. A first-order reaction is 50% complete in 20 minutes. How long will it take for the reaction to be 75% complete?
- A. 30 minutes
- B. 40 minutes
- C. 60 minutes
- D. 80 minutes
Answer: B. 40 minutes — For a first-order reaction, t₅₀% (half-life) = 20 minutes. To be 75% complete means 25% remains, which corresponds to two half-lives (100% → 50% → 25%). Therefore, time taken = 2 × 20 minutes = 40 minutes.
MCQ 5. The activation energy of a chemical reaction can be reduced by:
- A. Increasing the temperature
- B. Increasing the concentration of reactants
- C. Adding a catalyst
- D. Increasing the pressure
Answer: C. Adding a catalyst — A catalyst provides an alternative reaction pathway with a lower activation energy, thereby increasing the rate of reaction without being consumed. Temperature, concentration, and pressure affect the rate but not the activation energy itself.
MCQ 6. For the reaction 2A + B → 3C, if the rate law is given as Rate = k[A]¹[B]², what is the overall order of the reaction?
- A. 1
- B. 2
- C. 3
- D. Cannot be determined from the stoichiometry.
Answer: C. 3 — The overall order of reaction is the sum of the exponents of the concentration terms in the rate law. Here, 1 (for A) + 2 (for B) = 3.
MCQ 7. A first-order reaction has a rate constant of 1.15 x 10⁻³ s⁻¹. How long will it take for 5 g of this reactant to reduce to 3 g? (Given: ln 5 = 1.609, ln 3 = 1.098)
- A. 444 s
- B. 198 s
- C. 388 s
- D. 222 s
Answer: A. 444 s — For a first-order reaction, the integrated rate law is t = (1/k) ln([A]₀/[A]). Substituting k = 1.15 × 10⁻³ s⁻¹, [A]₀ = 5 g, and [A] = 3 g, we get t = (1 / 1.15 × 10⁻³) × (ln(5/3)) = (1 / 1.15 × 10⁻³) × (1.609 - 1.098) = (1 / 1.15 × 10⁻³) × 0.511 ≈ 444 s.
MCQ 8. Which of the following is an example of a pseudo first-order reaction?
- A. Hydrolysis of ester in acidic medium
- B. Decomposition of N₂O₅
- C. H₂ + I₂ → 2HI
- D. 2NO₂ → 2NO + O₂
Answer: A. Hydrolysis of ester in acidic medium — In the hydrolysis of an ester (e.g., ethyl acetate) in acidic medium, water is present in a large excess, so its concentration remains effectively constant. Thus, the reaction appears to be first order with respect to the ester only, making it a pseudo first-order reaction.
MCQ 9. According to collision theory, not all collisions between reactant molecules lead to product formation. Which of the following is a necessary condition for an effective collision?
- A. Molecules must collide with sufficient energy (activation energy).
- B. Molecules must collide with the correct orientation.
- C. Molecules must collide at high frequency.
- D. Both a) and b).
Answer: D. Both a) and b). — For a collision to be effective and lead to product formation, reactant molecules must possess energy equal to or greater than the activation energy (energy barrier) and must collide in the proper orientation.
MCQ 10. In an Arrhenius plot (ln k vs 1/T), what does the slope represent?
- A. -Ea/R
- B. Ea/R
- C. -R/Ea
- D. R/Ea
Answer: A. -Ea/R — The Arrhenius equation is k = A e^(-Ea/RT). Taking the natural logarithm gives ln k = ln A - Ea/RT. This is in the form of a straight line y = c + mx, where y = ln k, x = 1/T, and the slope m = -Ea/R.
MCQ 11. The sum of the powers of the concentration of the reactants in the rate law expression is called the:
- A. Molecularity
- B. Order of reaction
- C. Rate constant
- D. Activation energy
Answer: B. Order of reaction — The order of reaction is defined as the sum of the powers of the concentration terms of the reactants in the experimentally determined rate law expression.
MCQ 12. For a zero-order reaction, which statement is correct?
- A. The rate of reaction increases with time.
- B. The rate of reaction is independent of reactant concentration.
- C. The half-life is independent of the initial concentration.
- D. The unit of rate constant is s⁻¹.
Answer: B. The rate of reaction is independent of reactant concentration. — For a zero-order reaction, Rate = k[A]⁰ = k. This means the rate of reaction is constant and does not depend on the concentration of the reactant.
MCQ 13. If the half-life of a reaction is inversely proportional to the initial concentration of the reactant, what is the order of the reaction?
- A. Zero order
- B. First order
- C. Second order
- D. Third order
Answer: C. Second order — For a zero-order reaction, t½ ∝ [A]₀. For a first-order reaction, t½ is independent of [A]₀. For a second-order reaction, t½ ∝ 1/[A]₀, meaning it is inversely proportional to the initial concentration.
MCQ 14. What is the primary role of a catalyst in a chemical reaction?
- A. To shift the equilibrium position
- B. To increase the temperature of the reaction
- C. To provide an alternative reaction pathway with lower activation energy
- D. To increase the concentration of reactants
Answer: C. To provide an alternative reaction pathway with lower activation energy — A catalyst speeds up a reaction by providing a new reaction mechanism that has a lower activation energy barrier, making it easier for reactants to form products. It does not change the equilibrium position.
MCQ 15. For a first-order reaction, a plot of ln[A] vs time (t) yields a straight line with a negative slope. What does this indicate?
- A. The reaction is zero-order.
- B. The rate constant is positive.
- C. The concentration of reactant decreases exponentially with time.
- D. The activation energy is very high.
Answer: C. The concentration of reactant decreases exponentially with time. — For a first-order reaction, the integrated rate law is ln[A] = -kt + ln[A]₀. A straight line plot of ln[A] vs t with a negative slope (-k) is characteristic of an exponential decrease in reactant concentration over time.
MCQ 16. The rate of a reaction doubles when its temperature changes from 298 K to 308 K. The activation energy of the reaction is approximately: (Given: R = 8.314 J K⁻¹ mol⁻¹, ln 2 = 0.693)
- A. 52.8 kJ mol⁻¹
- B. 26.4 kJ mol⁻¹
- C. 105.6 kJ mol⁻¹
- D. 13.2 kJ mol⁻¹
Answer: A. 52.8 kJ mol⁻¹ — Using the Arrhenius equation: ln(k₂/k₁) = (Ea/R) * (1/T₁ - 1/T₂). Substituting k₂/k₁ = 2, T₁ = 298 K, T₂ = 308 K, and R = 8.314 J/Kmol gives ln 2 = (Ea/8.314) * (1/298 - 1/308). Solving for Ea yields approximately 52.8 kJ mol⁻¹.
MCQ 17. In a multi-step reaction, the slowest step is known as the:
- A. Elementary step
- B. Rate-determining step
- C. Overall step
- D. Activation step
Answer: B. Rate-determining step — The slowest step in a reaction mechanism controls the overall rate of the reaction, as the reaction cannot proceed faster than this step. Hence, it is called the rate-determining step.
MCQ 18. For an elementary reaction 2A → products, what can be said about its molecularity and order?
- A. Molecularity = 1, Order = 2
- B. Molecularity = 2, Order = 1
- C. Molecularity = 2, Order = 2
- D. Molecularity = 1, Order = 1
Answer: C. Molecularity = 2, Order = 2 — For an elementary reaction, the molecularity is equal to the number of reacting species, which is 2. Also, for an elementary reaction, the order of reaction is equal to its molecularity, so the order is also 2.
MCQ 19. Which of the following does NOT affect the activation energy of a reaction?
- A. Presence of a catalyst
- B. Nature of reactants
- C. Temperature
- D. Reaction pathway
Answer: C. Temperature — Activation energy is an intrinsic property of a particular reaction and depends on the nature of reactants and the reaction pathway. A catalyst changes the reaction pathway, thus changing activation energy. Temperature affects the kinetic energy of molecules but not the activation energy barrier itself.
MCQ 20. The half-life of a first-order reaction is 10 minutes. If the initial concentration is 0.1 M, what is the concentration after 30 minutes?
- A. 0.05 M
- B. 0.025 M
- C. 0.0125 M
- D. 0.00625 M
Answer: C. 0.0125 M — For a first-order reaction, each half-life reduces the concentration by half. After 10 min (1 t½), concentration = 0.1/2 = 0.05 M. After 20 min (2 t½), concentration = 0.05/2 = 0.025 M. After 30 min (3 t½), concentration = 0.025/2 = 0.0125 M.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.