Chemical kinetics is the branch of chemistry that studies the speed of reactions and the factors that control them. It explains why some reactions finish instantly, while others take minutes, years, or even longer.
A reaction may be possible, yet still move so slowly that no visible change appears for years. Important Questions Class 12 Chemistry Chapter 3 help students understand this difference through rate, rate law, order, molecularity, integrated equations, half-life, temperature effect, catalyst action, and collision theory. CBSE 2026 questions from Chemical Kinetics often test definitions with numerical reasoning. This chapter is scoring when students know the correct formula, unit, graph, and condition behind each result.
Key Takeaways
- Reaction Rate: Rate measures change in concentration of a reactant or product per unit time.
- Rate Law: Rate law must be determined experimentally and cannot be predicted from the balanced equation.
- First-order Reaction: Its half-life is constant and independent of initial concentration.
- Arrhenius Equation: Higher temperature increases the fraction of molecules crossing activation energy.
Important Questions Class 12 Chemistry Chapter 3 Structure 2026
| Concept |
Formula |
Key Variable |
| Rate Of Reaction |
Rate = change in concentration / time |
Concentration, time |
| First-order Reaction |
k = 2.303/t log([R]₀/[R]) |
Rate constant, time |
| Arrhenius Equation |
k = A e^(-Ea/RT) |
Activation energy, temperature |
Important Questions Class 12 Chemistry Chapter 3 Chemical Kinetics Overview
Chemical Kinetics Class 12 starts with the idea that reaction speed depends on measurable conditions. These questions build the base before formula-based numericals.
Q1. What Is Chemical Kinetics?
Chemical kinetics is the branch of chemistry that studies reaction rates and reaction mechanisms. It also studies how concentration, temperature, pressure, and catalysts affect reaction speed.
Thermodynamics tells whether a reaction is feasible. Chemical kinetics tells how fast a feasible reaction occurs.
Example: Diamond can convert into graphite thermodynamically, but the rate is extremely slow.
Q2. Why Is Chemical Kinetics Important?
Chemical kinetics is important because it explains how fast reactions occur and how their rates can change. It helps chemists control reactions in food storage, medicines, engines, and industry.
For example, it helps explain food spoilage. It also helps design fast-setting dental materials. It shows why fuel burns faster under some conditions.
Q3. What Is Rate Of A Chemical Reaction?
Rate of reaction Class 12 means change in concentration of reactant or product per unit time. It can be written using disappearance of reactant or appearance of product.
For R → P:
Rate of disappearance of R:
Rate = -Δ[R] / Δt
Rate of appearance of P:
Rate = +Δ[P] / Δt
Unit:
mol L⁻¹ s⁻¹
For gaseous reactions, rate can also be expressed in atm s⁻¹.
Chemical Kinetics Class 12 Important Questions On Rate Law And Order
Rate law and order decide how concentration affects speed. These Chemical Kinetics Class 12 Important Questions are frequent in board-style answers.
Q4. What Is Average Rate And Instantaneous Rate?
Average rate is calculated over a time interval, while instantaneous rate is calculated at a particular moment. Instantaneous rate gives the rate at a specific time.
Average rate:
rₐᵥ = -Δ[R] / Δt
Instantaneous rate:
rᵢₙₛₜ = -d[R] / dt
Graph method:
Draw a tangent to the concentration-time curve at that instant. The slope of the tangent gives instantaneous rate.
Q5. Calculate Average Rate When Concentration Changes From 0.03 M To 0.02 M
The average rate is 4.0 × 10⁻⁴ mol L⁻¹ min⁻¹ or 6.67 × 10⁻⁶ mol L⁻¹ s⁻¹.
Given Data:
Initial concentration = 0.03 M
Final concentration = 0.02 M
Time = 25 min
Formula Used:
Rate = -Δ[R] / Δt
Calculation:
Δ[R] = 0.02 - 0.03
Δ[R] = -0.01 M
Rate = -(-0.01) / 25
Rate = 0.0004 mol L⁻¹ min⁻¹
In seconds:
25 min = 1500 s
Rate = 0.01 / 1500
Rate = 6.67 × 10⁻⁶ mol L⁻¹ s⁻¹
Final Result: Rate = 4.0 × 10⁻⁴ mol L⁻¹ min⁻¹
Q6. What Is Rate Law?
Rate law is the expression that relates reaction rate to molar concentration of reactants. The powers in rate law are found experimentally.
For a reaction:
aA + bB → products
Rate law may be:
Rate = k[A]ˣ[B]ʸ
Here, x and y may differ from a and b.
Rate law cannot be written only by seeing the balanced equation.
Q7. What Is Rate Constant?
Rate constant is the proportionality constant in the rate law. It equals the rate when concentration terms become unity.
For rate law:
Rate = k[A]ˣ[B]ʸ
k is the rate constant.
The unit of k depends on reaction order.
For first order:
Unit of k = s⁻¹
For second order:
Unit of k = L mol⁻¹ s⁻¹
Q8. What Is Order Of Reaction?
Order of reaction Class 12 means the sum of powers of concentration terms in the rate law. It is an experimental quantity.
For:
Rate = k[A]ˣ[B]ʸ
Order = x + y
Example:
Rate = k[A]¹ᐟ²[B]³ᐟ²
Order = 1/2 + 3/2
Order = 2
Final Result: The reaction is second order.
Q9. Find The Order For Rate = k[A]³ᐟ²[B]⁻¹
The order of the reaction is 1/2. Add the powers of concentration terms.
Given Rate Law:
Rate = k[A]³ᐟ²[B]⁻¹
Calculation:
Order = 3/2 + (-1)
Order = 3/2 - 1
Order = 1/2
Final Result: Overall order = 1/2
A reaction order can be fractional or zero.
Q10. What Is Molecularity Of A Reaction?
Molecularity is the number of reacting species that collide in one elementary step. It applies only to elementary reactions.
Unimolecular reaction involves one reacting species.
Example: NH₄NO₂ → N₂ + 2H₂O
Bimolecular reaction involves two reacting species.
Example: 2HI → H₂ + I₂
Trimolecular reaction involves three reacting species.
Example: 2NO + O₂ → 2NO₂
Molecularity cannot be zero or fractional.
Class 12 Chemistry Chapter 3 Questions And Answers On Integrated Rate Equations
Integrated rate equations connect concentration and time directly. These Class 12 Chemistry Chapter 3 questions and answers cover graphs, half-life, and rate constants.
Q11. What Is A Zero Order Reaction?
A zero order reaction has a rate independent of reactant concentration. Its rate stays constant under fixed conditions.
For R → P:
Rate law:
Rate = k[R]⁰
Since [R]⁰ = 1:
Rate = k
Integrated equation:
[R] = [R]₀ - kt
Rate constant:
k = ([R]₀ - [R]) / t
Graph:
A plot of [R] versus t gives a straight line.
Q12. What Is A First Order Reaction?
A first order reaction has a rate directly proportional to the first power of reactant concentration. Its rate decreases as reactant concentration decreases.
For R → P:
Rate law:
Rate = k[R]
Integrated equation:
k = 2.303/t log([R]₀/[R])
Exponential form:
[R] = [R]₀ e^(-kt)
Graph:
A plot of log[R] versus t gives a straight line.
Q13. What Is Half-life Of Reaction?
Half-life of reaction Class 12 means the time required for reactant concentration to become half of its initial value.
For zero order reaction:
t₁ᐟ₂ = [R]₀ / 2k
For first order reaction:
t₁ᐟ₂ = 0.693 / k
A first-order half-life does not depend on initial concentration. A zero-order half-life depends on initial concentration.
Q14. Calculate Half-life For k = 5.5 × 10⁻¹⁴ s⁻¹
The half-life is 1.26 × 10¹³ s. Since k has unit s⁻¹, the reaction is first order.
Given Data:
k = 5.5 × 10⁻¹⁴ s⁻¹
Formula Used:
t₁ᐟ₂ = 0.693 / k
Calculation:
t₁ᐟ₂ = 0.693 / 5.5 × 10⁻¹⁴
t₁ᐟ₂ = 1.26 × 10¹³ s
Final Result: t₁ᐟ₂ = 1.26 × 10¹³ s
Q15. How Is A Pseudo First Order Reaction Formed?
A pseudo first order reaction is actually higher order but behaves like first order. This happens when one reactant is present in large excess.
Example: Hydrolysis of ethyl acetate.
CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH
Water is present in large excess. Its concentration changes very little. Rate depends mainly on ethyl acetate concentration.
Q16. What Is The Difference Between Order And Molecularity?
Order is experimental, while molecularity is theoretical for an elementary step. Both can match only for elementary reactions.
| Basis |
Order |
Molecularity |
| Meaning |
Sum of powers in rate law |
Number of species colliding |
| Value |
Can be zero or fractional |
Always a positive integer |
| Applies to |
Elementary and complex reactions |
Elementary reactions only |
| Found by |
Experiment |
Mechanism |
Q17. What Is Rate Determining Step?
Rate determining step is the slowest step in a reaction mechanism. It controls the overall rate of a complex reaction.
Example:
H₂O₂ + I⁻ → H₂O + IO⁻ is slow.
H₂O₂ + IO⁻ → H₂O + I⁻ + O₂ is fast.
The slow step decides the rate law.
This is similar to the slowest runner deciding a relay team’s final time.
Q18. What Are The Units Of Rate Constant For Different Orders?
The unit of rate constant depends on the overall order of reaction. This helps identify order from given units.
| Reaction Order |
Unit Of k |
| Zero order |
mol L⁻¹ s⁻¹ |
| First order |
s⁻¹ |
| Second order |
L mol⁻¹ s⁻¹ |
| Third order |
L² mol⁻² s⁻¹ |
Example:
k = 3 × 10⁻⁴ s⁻¹ means first order.
Chemical Kinetics Class 12 Numericals With Step-by-step Solutions
Chemical Kinetics Class 12 numericals need clean substitution. These Class 12 Chemistry Chapter 3 extra questions and answers follow board-answer format.
Q19. Calculate Rate For 2A → Products
The rate of reaction is 0.005 mol L⁻¹ min⁻¹. The rate is divided by the stoichiometric coefficient of A.
Given Data:
[A] decreases from 0.5 mol L⁻¹ to 0.4 mol L⁻¹
Time = 10 min
Reaction = 2A → products
Formula Used:
Rate = -1/2 × Δ[A]/Δt
Calculation:
Δ[A] = 0.4 - 0.5
Δ[A] = -0.1 mol L⁻¹
Rate = -1/2 × (-0.1/10)
Rate = 0.005 mol L⁻¹ min⁻¹
Final Result: Rate = 0.005 mol L⁻¹ min⁻¹
Q20. Calculate Average Rate For Decomposition Of N₂O₅
The average rate is 6.79 × 10⁻⁴ mol L⁻¹ min⁻¹.
Given Data:
[N₂O₅] changes from 2.33 mol L⁻¹ to 2.08 mol L⁻¹
Time = 184 min
Reaction = 2N₂O₅ → 4NO₂ + O₂
Formula Used:
Rate = -1/2 × Δ[N₂O₅]/Δt
Calculation:
Δ[N₂O₅] = 2.08 - 2.33
Δ[N₂O₅] = -0.25 mol L⁻¹
Rate = -1/2 × (-0.25 / 184)
Rate = 0.25 / 368
Rate = 6.79 × 10⁻⁴ mol L⁻¹ min⁻¹
Final Result: Rate = 6.79 × 10⁻⁴ mol L⁻¹ min⁻¹
Q21. Find The Rate Of Formation Of NO₂
The rate of formation of NO₂ is 2.72 × 10⁻³ mol L⁻¹ min⁻¹.
Given Data:
Average reaction rate = 6.79 × 10⁻⁴ mol L⁻¹ min⁻¹
Reaction = 2N₂O₅ → 4NO₂ + O₂
Formula Used:
Rate = 1/4 × Δ[NO₂]/Δt
So:
Δ[NO₂]/Δt = 4 × Rate
Calculation:
Δ[NO₂]/Δt = 4 × 6.79 × 10⁻⁴
Δ[NO₂]/Δt = 2.716 × 10⁻³
Final Result: Rate of NO₂ formation = 2.72 × 10⁻³ mol L⁻¹ min⁻¹
Q22. Calculate Rate Constant For First Order N₂O₅ Decomposition
The rate constant is 0.0304 min⁻¹.
Given Data:
Initial concentration = 1.24 × 10⁻² mol L⁻¹
Final concentration = 0.20 × 10⁻² mol L⁻¹
Time = 60 min
Formula Used:
k = 2.303/t log([R]₀/[R])
Calculation:
k = 2.303/60 log(1.24 × 10⁻² / 0.20 × 10⁻²)
k = 2.303/60 log(6.2)
k = 2.303/60 × 0.792
k = 0.0304 min⁻¹
Final Result: k = 0.0304 min⁻¹
Q23. Find Time For 5 g Reactant To Reduce To 3 g
The time required is 444 s. The reaction follows first-order kinetics.
Given Data:
k = 1.15 × 10⁻³ s⁻¹
Initial amount = 5 g
Final amount = 3 g
Formula Used:
k = 2.303/t log([R]₀/[R])
Rearranged:
t = 2.303/k log([R]₀/[R])
Calculation:
t = 2.303 / 1.15 × 10⁻³ × log(5/3)
log(5/3) = 0.2218
t = 2.303 × 0.2218 / 1.15 × 10⁻³
t = 444 s
Final Result: t = 444 s
Q24. Calculate Rate Constant From Half-life Of 60 Minutes
The rate constant is 1.925 × 10⁻⁴ s⁻¹.
Given Data:
t₁ᐟ₂ = 60 min
t₁ᐟ₂ = 3600 s
Reaction order = first order
Formula Used:
t₁ᐟ₂ = 0.693 / k
Rearranged:
k = 0.693 / t₁ᐟ₂
Calculation:
k = 0.693 / 3600
k = 1.925 × 10⁻⁴ s⁻¹
Final Result: k = 1.925 × 10⁻⁴ s⁻¹
Q25. Find The Effect On Rate When B Is Tripled
The rate becomes nine times when B is tripled. The reaction is second order with respect to B.
Given Rate Law:
Rate = k[A][B]²
When [B] becomes 3[B]:
New rate = kA²
New rate = k[A] × 9[B]²
New rate = 9k[A][B]²
Final Result: Rate increases 9 times
Chemical Kinetics Class 12 Important Questions On Temperature And Collision Theory
Temperature and catalysts explain why reactions speed up. These Chemical Kinetics Class 12 questions with answers cover Arrhenius equation, activation energy, and collision theory.
Q26. What Is The Arrhenius Equation?
Arrhenius equation Class 12 shows the relation between rate constant, temperature, and activation energy.
Formula:
k = A e^(-Eₐ/RT)
Here:
k = rate constant
A = Arrhenius factor
Eₐ = activation energy
R = gas constant
T = temperature in kelvin
Log form:
ln k = -Eₐ/RT + ln A
A plot of ln k versus 1/T gives a straight line.
Q27. What Is Activation Energy?
Activation energy Class 12 means the minimum extra energy needed to form the activated complex. Reactant molecules must cross this energy barrier to form products.
Symbol:
Eₐ
Unit:
J mol⁻¹ or kJ mol⁻¹
Lower activation energy gives faster reaction. A catalyst increases rate by providing a path with lower activation energy.
Q28. Calculate Eₐ When Rate Doubles For 10 K Rise
The activation energy is 52.897 kJ mol⁻¹. This uses the Arrhenius temperature relation.
Given Data:
k₂/k₁ = 2
T₁ = 298 K
T₂ = 308 K
R = 8.314 J mol⁻¹ K⁻¹
Formula Used:
log(k₂/k₁) = Eₐ / 2.303R × [(T₂ - T₁) / (T₁T₂)]
Calculation:
log 2 = Eₐ / (2.303 × 8.314) × [10 / (298 × 308)]
0.3010 = Eₐ / 19.147 × 0.000109
Eₐ = 0.3010 × 19.147 / 0.000109
Eₐ = 52897 J mol⁻¹
Final Result: Eₐ = 52.897 kJ mol⁻¹
Q29. What Is The Effect Of Catalyst On Reaction Rate?
A catalyst increases reaction rate by lowering activation energy. It provides an alternate reaction pathway.
A catalyst does not change Gibbs energy. A catalyst does not change equilibrium constant. A catalyst speeds up forward and backward reactions equally.
Example:
MnO₂ catalyses decomposition of KClO₃.
2KClO₃ → 2KCl + 3O₂
Q30. What Is Collision Theory Of Chemical Reactions?
Collision theory states that reacting molecules must collide with sufficient energy and proper orientation. Only effective collisions form products.
For a bimolecular reaction:
A + B → products
Rate expression:
Rate = PZAB e^(-Eₐ/RT)
Here:
P = steric factor
ZAB = collision frequency
e^(-Eₐ/RT) = fraction of molecules with enough energy
A collision fails if molecules have poor orientation.
Class 12 Chemistry Important Links