Class 12 Chemistry Notes, Chapter 3: Chemical Kinetics

1. Why Do We Study Chemical Kinetics?

When a chemical reaction takes place, three important questions arise:

1. Can the reaction occur?
   • Answered by Thermodynamics.
2. How much reaction will occur?
   • Answered by Chemical Equilibrium.
3. How fast will the reaction occur?
   • Answered by Chemical Kinetics.

Example

Reaction	Speed
Explosion of crackers	Very Fast
Burning of LPG	Fast
Digestion of food	Moderate
Rusting of iron	Slow
Formation of diamond into graphite	Extremely Slow

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2. Reaction Rate

Definition

The rate of a chemical reaction is the change in concentration of reactants or products per unit time.

Simply,

Rate = Change in concentration ÷ Time taken

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Two Ways to Express Rate

(A) Rate of Disappearance of Reactant

Since the concentration of reactants decreases with time,$$\boxed{\text{Rate}=-\frac{\Delta[R]}{\Delta t}}$$Rate=−ΔtΔ[R]​​

The negative sign makes the rate positive.

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(B) Rate of Formation of Product

Since product concentration increases,$$\boxed{\text{Rate}=+\frac{\Delta[P]}{\Delta t}}$$Rate=+ΔtΔ[P]​​

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Important Point

Reactant concentration decreases.

Product concentration increases.

Therefore,

• Reactant → Negative sign
• Product → Positive sign

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3. Units of Reaction Rate

If concentration is measured in mol L⁻¹ and time in seconds,$$\boxed{\text{Unit}=mol\,L^{-1}s^{-1}}$$Unit=molL−1s−1​

For gaseous reactions expressed using pressure,$$\boxed{\text{Unit}=atm\,s^{-1}}$$Unit=atms−1​

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4. Average Rate of Reaction

Average rate tells us how much concentration changes during a certain time interval.

Formula:$$\boxed{\text{Average Rate}=-\frac{\Delta[R]}{\Delta t}}$$Average Rate=−ΔtΔ[R]​​

or$$\boxed{\text{Average Rate}=+\frac{\Delta[P]}{\Delta t}}$$Average Rate=+ΔtΔ[P]​​

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Characteristics

✔ Calculated over a finite time interval.

✔ Easy to calculate.

✔ Does not give the exact rate at one particular instant.

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5. Instantaneous Rate of Reaction

Sometimes we want to know the reaction rate at one exact moment.

This is called Instantaneous Rate.

Mathematically,$$\boxed{\text{Rate}=-\frac{d[R]}{dt}}$$Rate=−dtd[R]​​

or$$\boxed{\text{Rate}=+\frac{d[P]}{dt}}$$Rate=+dtd[P]​​

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Graphical Meaning

• Draw a tangent to the concentration–time curve.
• The slope of the tangent gives the instantaneous rate.

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Difference Between Average and Instantaneous Rate

Average Rate	Instantaneous Rate
Calculated over a time interval	Calculated at one instant
Uses Δ	Uses d (very small change)
Easier to calculate	Found from the tangent to the graph
Less accurate for a specific moment	Gives the exact rate at that instant

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6. Rate for Different Stoichiometric Coefficients

For a reaction:$$aA+bB \rightarrow cC+dD$$aA+bB→cC+dD

Rate is written as$$\boxed{ -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt} }$$−a1​dtd[A]​=−b1​dtd[B]​=c1​dtd[C]​=d1​dtd[D]​​

The stoichiometric coefficients are included so that the calculated rate is the same regardless of which reactant or product is used.

Example

For$$2HI \rightarrow H_2 + I_2$$2HI→H2​+I2​ $$\boxed{ \text{Rate} = -\frac{1}{2}\frac{d[HI]}{dt} = \frac{d[H_2]}{dt} = \frac{d[I_2]}{dt} }$$Rate=−21​dtd[HI]​=dtd[H2​]​=dtd[I2​]​​

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7. Factors Affecting Reaction Rate

The rate of a reaction depends on:

• Concentration of reactants
• Temperature
• Pressure (for gases)
• Catalyst
• Nature of reactants
• Surface area (for solids)

The chapter focuses mainly on concentration, temperature, pressure (for gases), and catalysts.

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8. Rate Law (Rate Equation)

Experiments show that reaction rate depends on reactant concentrations in a specific way.

General form:$$\boxed{\text{Rate}=k[A]^x[B]^y}$$Rate=k[A]x[B]y​

Where:

• k = Rate constant
• [A], [B] = Concentrations
• x, y = Experimentally determined powers

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Important Facts About Rate Law

• It is determined experimentally.
• It cannot usually be predicted from the balanced chemical equation.
• The exponents may or may not equal the stoichiometric coefficients.

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Example

Reaction:$$2NO+O_2\rightarrow2NO_2$$2NO+O2​→2NO2​

Experimentally,$$\boxed{\text{Rate}=k[NO]^2[O_2]}$$Rate=k[NO]2[O2​]​

Here,

• Order with respect to NO = 2
• Order with respect to O₂ = 1
• Overall order = 3

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9. Rate Constant (k)

The proportionality constant in the rate equation is called the rate constant.

Characteristics

• Constant for a given reaction at a fixed temperature.
• Independent of reactant concentration.
• Changes if temperature changes.
• Higher k means a faster reaction under the same conditions.

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Quick Formula Box

Reaction Rate

$$-\frac{\Delta[R]}{\Delta t}$$−ΔtΔ[R]​ $$+\frac{\Delta[P]}{\Delta t}$$+ΔtΔ[P]​

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Instantaneous Rate

$$-\frac{d[R]}{dt}$$−dtd[R]​

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General Rate Law

$$\boxed{Rate=k[A]^x[B]^y}$$Rate=k[A]x[B]y​

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One-Minute Revision

• Chemical kinetics studies the speed of reactions.
• Rate = change in concentration ÷ time.
• Reactants decrease → use a negative sign.
• Products increase → use a positive sign.
• Average rate is calculated over a time interval.
• Instantaneous rate is measured at a particular instant.
• Rate law is determined experimentally.
• General rate equation: Rate = k[A]^x[B]^y.
• Rate constant k is constant only at a fixed temperature.

10. Order of Reaction

Definition

The order of a reaction is the sum of the powers (exponents) of concentration terms present in the rate law expression.

For a reaction:$$\text{Rate}=k[A]^x[B]^y$$Rate=k[A]x[B]y

Order of reaction:$$\boxed{\text{Order}=x+y}$$Order=x+y​

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Examples

Example 1

$$Rate=k[A]^2[B]$$Rate=k[A]2[B]

Order:$$=2+1=3$$=2+1=3

So, it is a third-order reaction.

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Example 2

$$Rate=k[A]^{1/2}[B]^{3/2}$$Rate=k[A]1/2[B]3/2

Order:$$=\frac12+\frac32=2$$=21​+23​=2

Reaction order = 2

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Important Points About Order

• Order is found experimentally.
• It may be:
  • Zero
  • One
  • Two
  • Fractional
  • Even negative in some cases
• It is not always equal to the stoichiometric coefficients of the balanced equation.

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11. Molecularity of Reaction

Definition

The number of reacting species (atoms, ions, or molecules) that collide simultaneously in an elementary reaction is called molecularity.

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Types of Molecularity

1. Unimolecular Reaction

Only one molecule participates.

Example:$$NH_4NO_2 \rightarrow N_2+2H_2O$$NH4​NO2​→N2​+2H2​O

Molecularity = 1

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2. Bimolecular Reaction

Two reacting species participate.

Example:$$2HI\rightarrow H_2+I_2$$2HI→H2​+I2​

Molecularity = 2

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3. Termolecular Reaction

Three reacting species participate.

Example:$$2NO+O_2\rightarrow2NO_2$$2NO+O2​→2NO2​

Molecularity = 3

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Difference Between Order and Molecularity

Order of Reaction	Molecularity
Determined experimentally	Based on reaction mechanism
Applies to elementary and complex reactions	Applies only to elementary reactions
Can be zero or fractional	Always a whole number
Can be negative	Never negative
Gives information about rate dependence	Gives number of molecules involved in elementary step

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12. Rate Constant and Its Units

For:$$Rate=k[A]^x[B]^y$$Rate=k[A]x[B]y

where:$$x+y=n$$x+y=n

(n = order of reaction)

The units of rate constant depend on the order of reaction.

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Zero Order Reaction

Rate:$$Rate=k$$Rate=k

Units of k:$$\boxed{mol\,L^{-1}s^{-1}}$$molL−1s−1​

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First Order Reaction

Rate:$$Rate=k[A]$$Rate=k[A]

Units of k:$$\boxed{s^{-1}}$$s−1​

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Second Order Reaction

Rate:$$Rate=k[A]^2$$Rate=k[A]2

Units of k:$$\boxed{L\,mol^{-1}s^{-1}}$$Lmol−1s−1​

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13. Elementary and Complex Reactions

Elementary Reaction

A reaction that occurs in a single step is called an elementary reaction.

Example:$$A+B\rightarrow C$$A+B→C

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Complex Reaction

A reaction that occurs through multiple elementary steps is called a complex reaction.

The complete sequence of steps is called the mechanism of reaction.

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14. Rate Determining Step

In a multi-step reaction, the slowest step controls the overall reaction rate.

This slow step is called the:$$\boxed{\text{Rate Determining Step}}$$Rate Determining Step​

Example:

A relay race depends on the slowest runner. Similarly, the reaction speed depends on the slowest reaction step.

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15. Integrated Rate Equations

The rate law gives the relationship between rate and concentration.

However, measuring instantaneous rate is difficult.

Therefore, the differential rate equation is converted into an integrated rate equation, which relates:

• Concentration
• Time
• Rate constant

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A. Zero Order Reaction

Definition

A reaction in which the rate is independent of reactant concentration is called a zero-order reaction.

General reaction:$$R\rightarrow P$$R→P

Rate law:$$\boxed{Rate=k[R]^0}$$Rate=k[R]0​

Since:$$[R]^0=1$$[R]0=1

Therefore:$$\boxed{Rate=k}$$Rate=k​

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Integrated Rate Equation

$$-\frac{d[R]}{dt}=k$$−dtd[R]​=k

After integration:$$\boxed{[R]=[R]_0-kt}$$[R]=[R]0​−kt​

Where:

• [R]₀ = initial concentration
• [R] = concentration after time t
• k = rate constant

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Rate Constant Formula

$$\boxed{k=\frac{[R]_0-[R]}{t}}$$k=t[R]0​−[R]​​

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Graph

For a zero-order reaction:

Graph of:$$[R] \text{ vs } t$$[R] vs t

is a straight line.

• Slope = -k
• Intercept = [R]₀

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Examples of Zero Order Reactions

1. Decomposition of ammonia on platinum surface:

$$2NH_3\rightarrow N_2+3H_2$$2NH3​→N2​+3H2​

2. Reactions occurring on catalyst surfaces

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B. First Order Reaction

Definition

A reaction whose rate depends on the first power of reactant concentration is called a first-order reaction.

General reaction:$$R\rightarrow P$$R→P

Rate law:$$\boxed{Rate=k[R]}$$Rate=k[R]​

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Integrated Rate Equation

$$\boxed{ \ln\frac{[R]_0}{[R]}=kt }$$ln[R][R]0​​=kt​

or$$\boxed{ k=\frac{2.303}{t}\log\frac{[R]_0}{[R]} }$$k=t2.303​log[R][R]0​​​

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Exponential Form

$$\boxed{[R]=[R]_0e^{-kt}}$$[R]=[R]0​e−kt​

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Graph

For first-order reaction:

Graph between:$$\ln[R]\text{ and }t$$ln[R] and t

gives a straight line.

• Slope = -k
• Intercept = ln[R]₀

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Examples of First Order Reactions

1. Decomposition of N₂O₅

$$N_2O_5\rightarrow2NO_2+\frac12O_2$$N2​O5​→2NO2​+21​O2​

2. Radioactive decay

$$^{226}Ra\rightarrow^{222}Rn+He$$226Ra→222Rn+He

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Quick Revision Table

Feature	Zero Order	First Order
Rate law	Rate = k	Rate = k[R]
Order	0	1
Integrated equation	[R]=[R]₀−kt	ln([R]₀/[R])=kt
Unit of k	mol L⁻¹ s⁻¹	s⁻¹
Graph	[R] vs t	ln[R] vs t
Half-life	Depends on initial concentration	Independent of initial concentration

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Exam Remember Points ⭐

1. Order comes from rate law, not balanced equation.
2. Molecularity applies only to elementary reactions.
3. Zero-order rate is independent of concentration.
4. First-order reactions have constant half-life.
5. Unit of rate constant depends on reaction order.

16. Half-Life of a Reaction (t½)

Definition

The half-life of a reaction is the time required for the concentration of a reactant to become half of its initial value.

It is represented by:$$\boxed{t_{1/2}}$$t1/2​​

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A. Half-Life of Zero Order Reaction

For a zero-order reaction:$$[R]=[R]_0-kt$$[R]=[R]0​−kt

At half-life:$$[R]=\frac{[R]_0}{2}$$[R]=2[R]0​​

Substituting:$$\frac{[R]_0}{2}=[R]_0-kt_{1/2}$$2[R]0​​=[R]0​−kt1/2​

Therefore:$$\boxed{ t_{1/2}=\frac{[R]_0}{2k} }$$t1/2​=2k[R]0​​​

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Important Point

For zero-order reactions:$$\boxed{t_{1/2}\propto [R]_0}$$t1/2​∝[R]0​​

The half-life increases when the initial concentration increases.

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B. Half-Life of First Order Reaction

For first-order reaction:$$k=\frac{2.303}{t}\log\frac{[R]_0}{[R]}$$k=t2.303​log[R][R]0​​

At half-life:$$[R]=\frac{[R]_0}{2}$$[R]=2[R]0​​

Therefore:$$\boxed{ t_{1/2}=\frac{0.693}{k} }$$t1/2​=k0.693​​

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Important Point

For first-order reactions:$$\boxed{t_{1/2}\text{ is independent of initial concentration}}$$t1/2​ is independent of initial concentration​

This means every half-life interval takes the same amount of time.

Example:

If a reaction takes 10 minutes for half completion:

• First half → 10 min
• Next half → 10 min
• Next half → 10 min

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Comparison of Half-Life

Reaction Order	Half-Life Formula	Dependence
Zero Order	$\frac{[R]_0}{2k}$2k[R]0​​	Depends on initial concentration
First Order	$\frac{0.693}{k}$k0.693​	Independent of initial concentration

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17. Pseudo First Order Reaction

Definition

Some reactions are actually of higher order but behave like first-order reactions because one reactant is present in a very large amount.

Such reactions are called pseudo first-order reactions.

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Example: Hydrolysis of Ethyl Acetate

Actual reaction:$$CH_3COOC_2H_5+H_2O \rightarrow CH_3COOH+C_2H_5OH$$CH3​COOC2​H5​+H2​O→CH3​COOH+C2​H5​OH

Rate law:$$Rate=k[CH_3COOC_2H_5][H_2O]$$Rate=k[CH3​COOC2​H5​][H2​O]

It is a second-order reaction.

But water is taken in very large excess, so its concentration remains almost constant.

Therefore:$$Rate=k'[CH_3COOC_2H_5]$$Rate=k′[CH3​COOC2​H5​]

Hence, it behaves as a first-order reaction.

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Other Example

Inversion of cane sugar:$$C_{12}H_{22}O_{11}+H_2O \rightarrow Glucose+Fructose$$C12​H22​O11​+H2​O→Glucose+Fructose

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18. Temperature Dependence of Reaction Rate

Most reactions become faster when temperature increases.

Reason:

• Molecules gain more kinetic energy.
• More molecules can cross the activation energy barrier.
• More successful collisions occur.

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Arrhenius Equation

The effect of temperature on reaction rate is explained by the Arrhenius equation.$$\boxed{k=Ae^{-E_a/RT}}$$k=Ae−Ea​/RT​

Where:

Symbol	Meaning
k	Rate constant
A	Frequency factor
Ea	Activation energy
R	Gas constant
T	Temperature in Kelvin

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19. Activation Energy (Ea)

Definition

The minimum energy required by reactant molecules to form products is called activation energy.

It is the energy needed to form an activated complex.

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Activated Complex

During a reaction:

Reactants

↓

Activated Complex

↓

Products

The activated complex is an unstable intermediate state having high energy.

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Effect of Temperature on Rate

When temperature increases:

• Molecular kinetic energy increases.
• More molecules have energy greater than activation energy.
• Number of successful collisions increases.
• Reaction rate increases.

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Logarithmic Form of Arrhenius Equation

Taking natural logarithm:$$\boxed{ \ln k=-\frac{E_a}{RT}+\ln A }$$lnk=−RTEa​​+lnA​

This equation represents a straight line:$$y=mx+c$$y=mx+c

For graph:$$\ln k \text{ vs } \frac{1}{T}$$lnk vs T1​

• Slope:

$$\boxed{-\frac{E_a}{R}}$$−REa​​​

• Intercept:

$$\boxed{\ln A}$$lnA​

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20. Collision Theory of Reaction Rates

According to collision theory:

A reaction occurs only when reactant molecules collide effectively.

For a successful collision:

Conditions:

1. Molecules must collide.
2. Collision must have sufficient energy.
3. Molecules must have proper orientation.

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Effective Collision

An effective collision produces an activated complex and forms products.

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21. Catalyst

Definition

A catalyst is a substance that increases the rate of a reaction without undergoing permanent chemical change.

Example:

Manganese dioxide catalyses decomposition of potassium chlorate:$$2KClO_3 \rightarrow 2KCl+3O_2$$2KClO3​→2KCl+3O2​

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How Does a Catalyst Work?

A catalyst:

• Provides an alternative reaction pathway.
• Lowers activation energy.
• Increases reaction rate.

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Important Properties of Catalyst

✔ Does not get permanently consumed.

✔ Does not change Gibbs energy (ΔG) of reaction.

✔ Helps only feasible reactions.

✔ A small amount can catalyse a large amount of reactants.

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Complete Chapter Quick Revision Sheet

Chemical Kinetics

Study of:

• Rate of reaction
• Reaction mechanism

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Rate

$$Rate=\frac{\text{Change in concentration}}{\text{Time}}$$Rate=TimeChange in concentration​

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Rate Law

$$Rate=k[A]^x[B]^y$$Rate=k[A]x[B]y

Order:$$x+y$$x+y

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Molecularity

Number of species colliding in an elementary step.

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Zero Order

$$[R]=[R]_0-kt$$[R]=[R]0​−kt $$t_{1/2}=\frac{[R]_0}{2k}$$t1/2​=2k[R]0​​

Unit of k:$$mol\,L^{-1}s^{-1}$$molL−1s−1

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First Order

$$\ln\frac{[R]_0}{[R]}=kt$$ln[R][R]0​​=kt $$t_{1/2}=\frac{0.693}{k}$$t1/2​=k0.693​

Unit of k:$$s^{-1}$$s−1

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Arrhenius Equation

$$k=Ae^{-E_a/RT}$$k=Ae−Ea​/RT

Higher temperature → Faster reaction

Lower activation energy → Faster reaction

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Most Important Board Exam Points ⭐

1. Difference between order and molecularity.
2. Derivation of zero-order integrated rate equation.
3. Derivation of first-order integrated rate equation.
4. Half-life formulas.
5. Arrhenius equation and activation energy.
6. Effect of catalyst on activation energy.
7. Pseudo first-order reactions.
8. Units of rate constant.

A. Complete Formula Sheet

1. Rate of Reaction

Average Rate

$$\boxed{Rate=-\frac{\Delta[R]}{\Delta t}}$$Rate=−ΔtΔ[R]​​

or$$\boxed{Rate=\frac{\Delta[P]}{\Delta t}}$$Rate=ΔtΔ[P]​​

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Instantaneous Rate

$$\boxed{Rate=-\frac{d[R]}{dt}}$$Rate=−dtd[R]​​

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2. Rate Law

For:$$aA+bB\rightarrow Products$$aA+bB→Products $$\boxed{Rate=k[A]^x[B]^y}$$Rate=k[A]x[B]y​

Where:

• k = rate constant
• x = order with respect to A
• y = order with respect to B

Overall order:$$\boxed{Order=x+y}$$Order=x+y​

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3. Units of Rate Constant

Order	Unit of k
Zero order	mol L⁻¹ s⁻¹
First order	s⁻¹
Second order	L mol⁻¹ s⁻¹

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4. Zero Order Reaction

Rate Equation

$$Rate=k$$Rate=k

Integrated Rate Equation

$$\boxed{[R]=[R]_0-kt}$$[R]=[R]0​−kt​

Rate Constant

$$\boxed{k=\frac{[R]_0-[R]}{t}}$$k=t[R]0​−[R]​​

Half-Life

$$\boxed{t_{1/2}=\frac{[R]_0}{2k}}$$t1/2​=2k[R]0​​​

Graph

$$[R]\text{ vs }t$$[R] vs t

Straight line:

• Slope = -k
• Intercept = [R]₀

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5. First Order Reaction

Rate Equation

$$Rate=k[R]$$Rate=k[R]

Integrated Rate Equation

$$\boxed{ k=\frac{2.303}{t}\log\frac{[R]_0}{[R]} }$$k=t2.303​log[R][R]0​​​

Half-Life

$$\boxed{ t_{1/2}=\frac{0.693}{k} }$$t1/2​=k0.693​​

Concentration-Time Relation

$$\boxed{ [R]=[R]_0e^{-kt} }$$[R]=[R]0​e−kt​

Graph

$$\ln[R]\text{ vs }t$$ln[R] vs t

Straight line:

• Slope = -k
• Intercept = ln[R]₀

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6. Arrhenius Equation

$$\boxed{k=Ae^{-E_a/RT}}$$k=Ae−Ea​/RT​

Log form:$$\boxed{ \ln k=-\frac{E_a}{RT}+\ln A }$$lnk=−RTEa​​+lnA​

Two-temperature equation:$$\boxed{ \log\frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{T_2-T_1}{T_1T_2} \right) }$$logk1​k2​​=2.303REa​​(T1​T2​T2​−T1​​)​

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B. Important Numerical Types

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Type 1: Average Rate Calculation

Question

For reaction:$$R\rightarrow P$$R→P

Concentration of R changes from 0.03 M to 0.02 M in 25 minutes.

Find average rate.

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Solution

Formula:$$Rate=-\frac{\Delta[R]}{\Delta t}$$Rate=−ΔtΔ[R]​ $$=-\frac{0.02-0.03}{25}$$=−250.02−0.03​ $$=\frac{0.01}{25}$$=250.01​ $$\boxed{4\times10^{-4}M\,min^{-1}}$$4×10−4Mmin−1​

In seconds:$$25\times60=1500s$$25×60=1500s $$Rate=\frac{0.01}{1500}$$Rate=15000.01​ $$\boxed{6.67\times10^{-6}M\,s^{-1}}$$6.67×10−6Ms−1​

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Type 2: Finding Order from Rate Law

Given:$$Rate=k[A]^{1/2}[B]^{3/2}$$Rate=k[A]1/2[B]3/2

Order:$$=\frac12+\frac32$$=21​+23​ $$\boxed{Order=2}$$Order=2​

Reaction is second order.

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Type 3: Zero Order Calculation

Question

A zero-order reaction has:$$[R]_0=0.8M$$[R]0​=0.8M

After 20 seconds:$$[R]=0.4M$$[R]=0.4M

Find rate constant.

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Solution

Formula:$$k=\frac{[R]_0-[R]}{t}$$k=t[R]0​−[R]​ $$k=\frac{0.8-0.4}{20}$$k=200.8−0.4​ $$k=0.02$$k=0.02 $$\boxed{k=0.02\,mol\,L^{-1}s^{-1}}$$k=0.02molL−1s−1​

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Type 4: First Order Rate Constant

Question

A first-order reaction has:

Initial concentration:$$[R]_0=1M$$[R]0​=1M

After 100 seconds:$$[R]=0.25M$$[R]=0.25M

Find k.

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Solution

Formula:$$k=\frac{2.303}{t} \log\frac{[R]_0}{[R]}$$k=t2.303​log[R][R]0​​

Substitute:$$k=\frac{2.303}{100} \log\frac{1}{0.25}$$k=1002.303​log0.251​ $$=\frac{2.303}{100}\log4$$=1002.303​log4 $$\log4=0.602$$log4=0.602 $$k=0.0138$$k=0.0138 $$\boxed{k=1.38\times10^{-2}s^{-1}}$$k=1.38×10−2s−1​

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Type 5: Half-Life Calculation

First Order Reaction

Given:$$k=0.005s^{-1}$$k=0.005s−1

Formula:$$t_{1/2}=\frac{0.693}{k}$$t1/2​=k0.693​ $$=\frac{0.693}{0.005}$$=0.0050.693​ $$\boxed{138.6s}$$138.6s​

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Type 6: Arrhenius Equation Numerical

Given:$$k_1=0.02s^{-1}$$k1​=0.02s−1 $$k_2=0.07s^{-1}$$k2​=0.07s−1

Temperatures:$$T_1=500K$$T1​=500K $$T_2=700K$$T2​=700K

Use:$$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{T_2-T_1}{T_1T_2} \right)$$logk1​k2​​=2.303REa​​(T1​T2​T2​−T1​​)

Find activation energy.

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C. Important Concept Questions

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Q1. Why does reaction rate increase with temperature?

Answer:

Increasing temperature increases the kinetic energy of molecules. More molecules acquire energy greater than activation energy, producing more successful collisions and increasing reaction rate.

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Q2. Why is molecularity never fractional?

Answer:

Molecularity represents the number of molecules participating in an elementary step. Since molecules cannot be present in fractions, molecularity cannot be fractional.

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Q3. Why is order determined experimentally?

Answer:

The balanced chemical equation does not always represent the actual reaction mechanism. Therefore, rate law and order must be found experimentally.

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Q4. Why does a catalyst increase reaction rate?

Answer:

A catalyst provides an alternative pathway with lower activation energy, making the reaction faster.

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D. Board Exam Important Differences

Order vs Molecularity

Order	Molecularity
Experimental quantity	Theoretical concept
Applies to all reactions	Only elementary reactions
Can be zero/fractional	Always whole number
Obtained from rate law	Obtained from mechanism

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Average Rate vs Instantaneous Rate

Average Rate	Instantaneous Rate
Over a time interval	At a specific time
Uses Δ	Uses differential form
Less accurate	More accurate

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E. Chapter Mind Map

                 Chemical Kinetics
                        |
        --------------------------------
        |              |               |
      Rate          Rate Law       Mechanism
        |              |               |
 Average/Instant   Order          Molecularity
        |              |
 Integrated Rate Equations
        |
 --------------------------
 |                        |
Zero Order          First Order
 |                        |
[R]=[R]0-kt      ln[R]0/[R]=kt
 |                        |
t1/2=[R]0/2k     t1/2=0.693/k