Chemical Kinetics Notes for NEET: Class 12 NCERT Chemistry

Introduction

Chemical Kinetics is about how fast a reaction goes and what controls that speed. Thermodynamics tells you whether a reaction is possible. Kinetics tells you how long it will actually take.

This chapter is consistently tested in NEET with 3 to 5 questions per year. The most frequent topics are half-life calculations for first-order reactions, the Arrhenius equation, and determining order from experimental data. Master these and you can score full marks from this chapter.

By the end of this chapter you will understand why some reactions are over in microseconds while others take years, and how temperature, concentration, and catalysts control reaction speed.

Rate of Reaction and Rate Law

Defining Rate

The rate of a reaction measures how quickly reactants are consumed or products are formed. For a reaction A → products:

Rate = - \frac{Δ [ A ]}{Δ t}

The negative sign makes the rate positive (since [A] decreases over time). For a reaction aA + bB → cC + dD, all the individual rates are related by the stoichiometric coefficients:

Rate = - \frac{1}{a} \frac{d [ A ]}{d t} = - \frac{1}{b} \frac{d [ B ]}{d t} = + \frac{1}{c} \frac{d [ C ]}{d t} = + \frac{1}{d} \frac{d [ D ]}{d t}

Average Rate vs Instantaneous Rate

Average rate: the change in concentration divided by the time interval. Depends on which time interval you choose.
Instantaneous rate: the rate at a specific instant, equal to the slope of the tangent drawn to the concentration-time curve at that point (mathematically, the derivative d[A]/dt).

In NEET, instantaneous rate is more important. When a question says "rate of the reaction at time t," it means instantaneous rate.

Rate Law Expression

The rate law (rate equation) expresses the rate as a function of concentration. For aA + bB → products:

rate = k [A]^{m} [B]^{n}

Here k is the rate constant, m is the order with respect to A, and n is the order with respect to B. The exponents m and n are not the stoichiometric coefficients a and b. They are determined experimentally.

Units of the Rate Constant k

The units of k depend on the overall order of the reaction. For an nth-order reaction:

units of k = (concentration)^{1 - n} \times time^{- 1}

Order	Units of k (mol/L, seconds)
Zero (n = 0)	mol L⁻¹ s⁻¹
First (n = 1)	s⁻¹
Second (n = 2)	L mol⁻¹ s⁻¹
Third (n = 3)	L² mol⁻² s⁻¹

A quick shortcut: if the units of k contain mol/L with a power, the order equals (1 minus that power).

Explore how concentration changes over time for zero, first, and second-order reactions. Adjust the initial concentration and rate constant, then query any time point.

Zero-order

First-order

Second-order

Integrated rate law

[A] = [A]₀ × e^(−k·t)

Initial concentration [A]₀: 1.00 mol/L

Rate constant k: 0.05 s⁻¹

Half-life calculation

t½ = 0.693 / k = 0.693 / 0.05 = 13.86 s

Time to reach 25% of [A]₀: 27.73 s

Concentration at each half-life interval

Time (s)

[A] (mol/L)

% remaining

0.00

1.0000

100.0%

13.86

0.5001

50.0%

27.72

0.2501

25.0%

41.58

0.1251

12.5%

55.44

0.0625

6.3%

Query concentration at any time

t = 10 s

[A] at t = 10s

0.6065 mol/L

(60.7% of [A]₀)

Order and Molecularity

Order of Reaction

The order of a reaction is the sum of all the exponents in the rate law. It is determined experimentally, not from the balanced equation.

Zero-order: rate = k. The reaction proceeds at a constant rate regardless of concentration.
First-order: rate = k[A]. The rate is proportional to concentration.
Second-order: rate = k[A]², or rate = k[A][B]. The rate depends on the square of one concentration, or the product of two concentrations.
Fractional order: possible in complex reactions (e.g., rate = k[A]^(1/2)).

Molecularity

Molecularity applies to a single elementary step in a reaction mechanism. It is the number of molecules (or atoms or ions) that collide in that step.

Unimolecular (molecularity = 1): one molecule rearranges or decomposes.
Bimolecular (molecularity = 2): two molecules collide.
Termolecular (molecularity = 3): three molecules collide simultaneously (rare).

Property	Order	Molecularity
Determined by	Experiment	Theory / mechanism
Applies to	Overall reaction or any step	Elementary step only
Can be fractional?	Yes	No (always 1, 2, or 3)
Can be zero?	Yes	No

Determining Order from Initial Rate Data

In NEET, you are often given a table of initial concentrations and initial rates. The method is to compare two experiments where one concentration changes while the other stays constant:

m = \frac{lo g ( r _{2} / r _{1} )}{lo g ([ A ] _{2} / [ A ] _{1} )}

If doubling [A] doubles the rate, m = 1. If it quadruples the rate, m = 2. If the rate does not change, m = 0.

Practice rate law calculations

Our AI tutor walks you through every step of rate law problems. Try it free with no time limit.

Integrated Rate Laws and Half-life

Integrated rate laws give you the concentration of a reactant as a function of time. They are the most useful formulas for NEET numerical problems.

Zero-order Reaction

[A] = [A]_{0} - k t

A graph of [A] vs time gives a straight line with slope = -k and intercept = [A]₀. The reaction proceeds at a constant rate until the reactant is exhausted.

t_{1/2} = \frac{[ A ] _{0}}{2 k}

Note: t½ depends on [A]₀ for zero-order. Each successive half-life is shorter (because [A]₀ for the next half-life is half as large).

First-order Reaction

[A] = [A]_{0} e^{- k t}

Taking the natural log:

ln [A] = ln [A]_{0} - k t

A graph of ln[A] vs time gives a straight line with slope = -k. This is the test for first-order kinetics in NEET graphical questions.

t_{1/2} = \frac{0.693}{k} (independent of [A]_{0})

The constant half-life is the defining feature of first-order kinetics. Radioactive decay follows first-order kinetics.

To find the time for [A] to reach any fraction f of [A]₀:

t = \frac{1}{k} ln \frac{[ A ] _{0}}{[ A ]} = \frac{t _{1/2}}{0.693} ln \frac{[ A ] _{0}}{[ A ]}

Common fractions you should memorize:

Fraction remaining	Number of half-lives
1/2 (50%)	1
1/4 (25%)	2
1/8 (12.5%)	3
1/10 (10%)	ln(10)/0.693 ≈ 3.32
1/16 (6.25%)	4

Second-order Reaction

\frac{1}{[ A ]} = \frac{1}{[ A ] _{0}} + k t

A graph of 1/[A] vs time gives a straight line with slope = k. The half-life is:

t_{1/2} = \frac{1}{k [ A ] _{0}}

Note: for second-order, t½ depends on [A]₀ and increases as the reaction proceeds (successive half-lives get longer).

Summary: How to Identify Order from Graphs

Order	Rate law	Integrated law	Straight-line plot	Half-life
Zero	r = k	[A] = [A]₀ - kt	[A] vs t (slope = -k)	[A]₀ / 2k
First	r = k[A]	ln[A] = ln[A]₀ - kt	ln[A] vs t (slope = -k)	0.693/k
Second	r = k[A]²	1/[A] = 1/[A]₀ + kt	1/[A] vs t (slope = k)	1/(k[A]₀)

Calculate half-life and time to reach any fraction of [A]₀ for zero, first, or second-order reactions. For first-order you can enter either k or t½ directly.

Zero-order

First-order

Second-order

t½ = 0.693 / k

Half-life is independent of [A]₀ for first-order reactions.

k (s⁻¹)

Half-life result

t½ = 0.693 / k = 0.693 / 0.0500 = 13.8600 s

t½ = 13.8600 s

Time for [A] to reach a fraction of [A]₀

½ (50%)

¼ (25%)

⅛ (12.5%)

1/16 (6.25%)

1/10 (10%)

1/100 (1%)

Custom fraction (0–1)

e.g. 0.33 for 33%

Calculation

t = ln([A]₀/[A]) / k = ln(1/0.25) / 0.0500 = 1.3863 / 0.0500 = 27.7259 s

Time required

27.7259 s

In half-lives

= 2.00 half-lives

Temperature Dependence and Arrhenius Equation

Increasing temperature always increases the rate of a reaction (for elementary reactions). The Arrhenius equation quantifies this relationship:

k = A e^{- E_{a} / R T}

where:

k is the rate constant
A is the pre-exponential factor (frequency factor, same units as k)
E_a is the activation energy in J/mol
R = 8.314 J mol⁻¹ K⁻¹
T is temperature in Kelvin

Linearised Form (Arrhenius Plot)

Taking the natural log:

ln k = ln A - \frac{E _{a}}{R T}

Plotting ln k (y-axis) vs 1/T (x-axis) gives a straight line with slope = -E_a/R and y-intercept = ln A.

Two-Temperature Form (Most Used in NEET)

When you have k at two temperatures, subtract the linearised forms to eliminate ln A:

ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

Or using log base 10:

lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})

Always convert temperature to Kelvin before using these formulas. Add 273 to any Celsius temperature.

Effect of Temperature on the Arrhenius Exponential

The term e^(-Ea/RT) is the fraction of molecules with energy at least equal to E_a. As T increases, this fraction increases, so more collisions are effective. A small increase in temperature can cause a large increase in rate because the exponential is sensitive to T. As a rough rule, a 10 °C rise approximately doubles the rate for many reactions (though this depends on E_a).

Use the Arrhenius equation in its two-temperature form: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂). Switch between finding the rate constant ratio or the activation energy.

Ea (kJ/mol)

T₁ (K)

Lower temperature

T₂ (K)

Higher temperature

Step-by-step working

Ea = 55 kJ/mol = 55000 J/mol
1/T₁ - 1/T₂ = 1/300 - 1/320 = 20.8333 × 10⁻⁵ K⁻¹
ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂)
= (55000/8.314) × 2.083e-4
= 1.3782

k₂/k₁ = 3.968

The rate increases by a factor of 3.97 when temperature rises from 300 K to 320 K.

Load a preset reaction

Acid hydrolysis (55 kJ/mol)

Enzyme reaction (40 kJ/mol)

High Ea reaction (250 kJ/mol)

Low Ea reaction (20 kJ/mol)

Remember: Always use T in Kelvin. Convert from Celsius by adding 273 (or 273.15). R = 8.314 J mol⁻¹ K⁻¹. Ea is in J/mol when using R in these units. Divide by 1000 to get kJ/mol.

Arrhenius equation made easy

Our AI tutor explains every step of Arrhenius calculation problems with clear working and units. Start learning free today.

Collision Theory and Reaction Mechanisms

Collision Theory

Collision theory says that molecules must collide for a reaction to happen, but not all collisions lead to a reaction. Two conditions must be met for a collision to be effective:

Energy condition: the colliding molecules must have kinetic energy at least equal to the activation energy E_a (called the threshold energy condition).
Orientation condition: the molecules must be oriented correctly relative to each other at the moment of collision (the steric requirement).

The fraction of molecules with energy above E_a at temperature T is e^(-Ea/RT). The pre-exponential factor A accounts for the frequency of collisions with proper orientation.

Activation Energy and Energy Profile Diagrams

An energy profile diagram (reaction coordinate diagram) plots energy on the y-axis and reaction progress on the x-axis. The key features are:

Reactants are at the left (their energy level).
Transition state (activated complex) is at the top of the energy hill, which is the energy maximum. It cannot be isolated; it is an unstable arrangement of atoms midway between reactants and products.
Products are at the right.
E_a (forward) = energy difference between reactants and the transition state.
ΔH = energy difference between reactants and products. For an exothermic reaction, products are lower in energy than reactants.
E_a (reverse) = E_a(forward) − ΔH.

A catalyst provides an alternative pathway with a lower activation energy. It lowers the energy hill without changing the energy of reactants or products (ΔH stays the same).

Reaction Mechanisms

Many reactions do not occur in a single step. A reaction mechanism is a sequence of elementary steps that adds up to the overall balanced equation.

The rate-determining step (RDS) is the slowest step. It controls the overall rate, just as the slowest vehicle in a convoy controls the speed of the whole group. The rate law for the overall reaction is the rate law of the RDS.

Pseudo-First-Order Reactions

When one reactant is present in such a large excess that its concentration barely changes during the reaction, it is treated as a constant. The reaction appears to be first-order even if it is not.

Example: hydrolysis of methyl acetate in excess water.

CH_{3} COOCH_{3} + H_{2} O \to CH_{3} COOH + CH_{3} OH

True rate law: rate = k[CH₃COOCH₃][H₂O]. Since water is the solvent, [H₂O] ≈ constant. Observed rate law: rate = k'[CH₃COOCH₃], where k' = k[H₂O]. This is called a pseudo-first-order rate constant.

Worked NEET Problems

NEET-style problem · Determining order from initial rate data

Question

For the reaction A + B → products, the following data were collected:
Expt 1: [A] = 0.10 M, [B] = 0.10 M, rate = 2.0 × 10⁻⁴ M/s
Expt 2: [A] = 0.20 M, [B] = 0.10 M, rate = 4.0 × 10⁻⁴ M/s
Expt 3: [A] = 0.10 M, [B] = 0.30 M, rate = 1.8 × 10⁻³ M/s
Find the rate law and calculate k.

Solution

Step 1: Find order in A. Compare Expt 1 and 2 (only [A] changes):

\frac{r _{2}}{r _{1}} = \frac{k ( 0.20 ) ^{m} ( 0.10 ) ^{n}}{k ( 0.10 ) ^{m} ( 0.10 ) ^{n}} = 2^{m} = \frac{4.0 \times 1 0 ^{- 4}}{2.0 \times 1 0 ^{- 4}} = 2

So 2^m = 2, therefore m = 1. First-order in A.

Step 2: Find order in B. Compare Expt 1 and 3 (only [B] changes):

\frac{r _{3}}{r _{1}} = 3^{n} = \frac{1.8 \times 1 0 ^{- 3}}{2.0 \times 1 0 ^{- 4}} = 9 = 3^{2}

So n = 2. Second-order in B.

Step 3: Rate law = k[A][B]². Overall order = 1 + 2 = 3 (third-order).

Step 4: Find k using Expt 1:

k = \frac{r}{[ A ] [ B ] ^{2}} = \frac{2.0 \times 1 0 ^{- 4}}{( 0.10 ) ( 0.10 ) ^{2}} = \frac{2.0 \times 1 0 ^{- 4}}{1.0 \times 1 0 ^{- 3}} = 0.20 L^{2} mol^{- 2} s^{- 1}

NEET-style problem · First-order half-life and fraction remaining

Question

A first-order reaction has k = 1.386 × 10⁻² min⁻¹. (a) What is the half-life? (b) What fraction remains after 200 min? (c) How long for the concentration to fall to 6.25% of its initial value?

Solution

(a) Half-life: t½ = 0.693/k = 0.693 / (1.386 × 10⁻²) = 50 min.

(b) Fraction after 200 min. Number of half-lives = 200/50 = 4. Fraction = (1/2)⁴ = 1/16 = 0.0625 = 6.25%.

t = \frac{0.693}{k} \times n = 50 \times 4 = 200 min

NEET-style problem · Arrhenius equation: finding Ea

Question

The rate constant for a reaction is 4.0 × 10⁻³ s⁻¹ at 300 K and 2.0 × 10⁻² s⁻¹ at 350 K. Calculate the activation energy. (R = 8.314 J mol⁻¹ K⁻¹)

Solution

Use the two-temperature Arrhenius equation:

ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

ln \frac{2.0 \times 1 0 ^{- 2}}{4.0 \times 1 0 ^{- 3}} = ln 5 = 1.609

\frac{1}{300} - \frac{1}{350} = \frac{350 - 300}{300 \times 350} = \frac{50}{105000} = 4.762 \times 1 0^{- 4} K^{- 1}

E_{a} = \frac{1.609 \times 8.314}{4.762 \times 1 0 ^{- 4}} = \frac{13.38}{4.762 \times 1 0 ^{- 4}} \approx 28, 100 J/mol = 28.1 kJ/mol

NEET-style problem · Rate law from mechanism

Question

A proposed mechanism for the decomposition of H₂O₂ in the presence of I⁻ is:
Step 1 (slow): H₂O₂ + I⁻ → H₂O + IO⁻
Step 2 (fast): H₂O₂ + IO⁻ → H₂O + O₂ + I⁻
What is the predicted rate law?

Solution

The rate-determining step (slow step) is Step 1. The rate law is based on the stoichiometry of Step 1:

rate = k [H_{2} O_{2}] [I^{-}]

This is second-order overall (first-order in H₂O₂ and first-order in I⁻). The intermediate IO⁻ is formed in Step 1 and consumed in Step 2, so it does not appear in the overall rate law. This mechanism is consistent with the experimentally observed rate law for catalytic decomposition of H₂O₂.

NEET-style problem · Energy profile and catalyst

Question

For a reaction A → B, E_a(forward) = 120 kJ/mol and ΔH = -50 kJ/mol. (a) What is E_a(reverse)? (b) If a catalyst reduces E_a(forward) to 80 kJ/mol, what is the new E_a(reverse)?

Solution

(a) E_a(reverse) = E_a(forward) − ΔH = 120 − (−50) = 120 + 50 = 170 kJ/mol.

The reverse reaction is endothermic (going uphill), so its activation energy is higher than the forward reaction's.

(b) The catalyst lowers both E_a(forward) by 40 kJ/mol and E_a(reverse) by the same amount (since ΔH does not change):

E_{a} (reverse, catalysed) = 170 - 40 = 130 kJ/mol

Or: E_a(reverse) = 80 − (−50) = 130 kJ/mol. Same result. The catalyst lowers both barriers by the same amount so that ΔH is unchanged.

Summary Cheat Sheet

Concept	Formula / Key Point
Rate law	rate = k[A]^m[B]^n; m, n from experiment
Units of k (nth order)	(concentration)^(1-n) × time⁻¹; first-order = s⁻¹
Zero-order integrated	[A] = [A]₀ - kt; t½ = [A]₀/2k
First-order integrated	ln[A] = ln[A]₀ - kt; t½ = 0.693/k (independent of [A]₀)
Second-order integrated	1/[A] = 1/[A]₀ + kt; t½ = 1/(k[A]₀)
Graph test (order)	[A] vs t linear = zero; ln[A] vs t linear = first; 1/[A] vs t linear = second
Arrhenius equation	k = A × e^(-Ea/RT)
Two-temperature Arrhenius	ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂); T in Kelvin
Arrhenius plot slope	Slope of ln k vs 1/T = -Ea/R
Ea(reverse)	Ea(reverse) = Ea(forward) - ΔH
Catalyst effect	Lowers Ea (both forward and reverse by same amount); does not change ΔH or K
Order vs molecularity	Order: experimental, can be fractional; Molecularity: theoretical, only 1/2/3
Rate-determining step	Slowest step in mechanism; its rate law = overall rate law
Pseudo-first-order	One reactant in large excess: absorbed into k' = k × [excess reactant]
Effective collision	Needs: (1) energy >= Ea, (2) correct orientation

Frequently asked questions

How many questions come from Chemical Kinetics in NEET?

You can expect 3 to 5 questions from Chemical Kinetics in NEET every year. The most tested topics are half-life calculations for first-order reactions and the Arrhenius equation. Questions on determining order from experimental data and rate law expressions also appear regularly. This chapter is manageable because the formulas are small in number but frequently applied.

What is the difference between order of reaction and molecularity?

Order of reaction is determined experimentally from the rate law: rate = k[A]^m[B]^n, where order = m + n. It can be zero, a fraction, or a whole number. Molecularity is a theoretical concept that applies only to an elementary step in a reaction mechanism. It is the number of molecules that collide in that single step and is always a small positive integer (1, 2, or 3). For a complex (multi-step) reaction, order is measurable but molecularity is not defined for the overall reaction.

Why is the half-life of a first-order reaction independent of initial concentration?

For a first-order reaction, the integrated rate law gives t½ = 0.693/k. The rate constant k depends only on temperature, not on concentration. So t½ comes out the same no matter what [A]₀ is. Physically, this happens because the rate at which the remaining reactant falls to half always takes the same time when the reaction rate is proportional to the current amount. In contrast, for zero-order reactions t½ = [A]₀/2k, which does depend on initial concentration.

How do you use the Arrhenius equation to find activation energy?

When you have rate constants k₁ and k₂ at temperatures T₁ and T₂, use the two-temperature form: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂), where R = 8.314 J/mol/K and temperatures are in Kelvin. Rearrange to get Ea = R × ln(k₂/k₁) / (1/T₁ - 1/T₂). Make sure to convert temperature to Kelvin (add 273 to Celsius) before plugging in. The result gives Ea in J/mol; divide by 1000 to convert to kJ/mol.

What is a pseudo-first-order reaction? Can you give an example?

A pseudo-first-order reaction is one that is actually second-order (or higher) but behaves as first-order because one reactant is present in such a large excess that its concentration barely changes. The classic example is the hydrolysis of sucrose: sucrose + H₂O → glucose + fructose. The rate law is rate = k[sucrose][H₂O], which is second-order overall. But water is the solvent and is present in huge excess, so [H₂O] is essentially constant. The reaction then follows rate = k'[sucrose], where k' = k[H₂O]. This is called the pseudo-first-order rate constant.

How do you determine the order of reaction from experimental data?

The most common method in NEET problems is the initial rates method. You compare two experiments where one reactant concentration is doubled while the other is kept constant. If doubling [A] doubles the rate, the order with respect to A is 1. If it quadruples the rate, the order is 2. If the rate does not change, the order is 0. Mathematically: m = log(r₂/r₁) / log([A]₂/[A]₁). You can also use concentration-time graphs: a linear [A] vs t graph means zero-order, a linear ln[A] vs t graph means first-order, and a linear 1/[A] vs t graph means second-order.

What does activation energy mean physically?

Activation energy (Ea) is the minimum energy that colliding molecules must have for a reaction to occur. Even if two molecules collide, the collision is productive only if the combined kinetic energy at the moment of collision is at least equal to Ea. You can picture it as an energy barrier on the reaction profile diagram. Reactants climb this barrier to reach the transition state (activated complex) and then fall down to form products. A higher Ea means fewer molecules in the system have enough energy to react at a given temperature, so the rate is slower.

What is the difference between rate constant k and rate of reaction r?

The rate constant k is a proportionality constant that depends only on temperature (and not on concentration). It is fixed for a given reaction at a given temperature. The rate of reaction r is the actual speed at which the reaction occurs at a specific moment and depends on both k and the current concentrations: r = k[A]^m[B]^n. As the reaction proceeds and concentrations fall, r decreases even though k stays the same. Think of k as a property of the reaction itself, and r as the instantaneous speed that changes as concentrations change.

Continue with the next chapter notes

Stay in NCERT order — the next chapter's notes are one click away.

Electrochemistry

The d- and f-Block Elements

Open notes

Track Your NEET Score Across All 90 Chapters

Free 14-day trial. AI tutor, full mock tests and chapter analytics — built for NEET 2027.

Free 14-day trial · No credit card required

Chemical KineticsNEET Chemistry · Class 12 · NCERT Chapter 3

Introduction

Rate of Reaction and Rate Law

Defining Rate

Average Rate vs Instantaneous Rate

Rate Law Expression

Units of the Rate Constant k

Order and Molecularity

Order of Reaction

Molecularity

Determining Order from Initial Rate Data

Integrated Rate Laws and Half-life

Zero-order Reaction

First-order Reaction

Second-order Reaction

Summary: How to Identify Order from Graphs

Temperature Dependence and Arrhenius Equation

Linearised Form (Arrhenius Plot)

Two-Temperature Form (Most Used in NEET)

Effect of Temperature on the Arrhenius Exponential

Collision Theory and Reaction Mechanisms

Collision Theory

Activation Energy and Energy Profile Diagrams

Reaction Mechanisms

Pseudo-First-Order Reactions

Worked NEET Problems

Summary Cheat Sheet

Frequently asked questions

How many questions come from Chemical Kinetics in NEET?

What is the difference between order of reaction and molecularity?

Why is the half-life of a first-order reaction independent of initial concentration?

How do you use the Arrhenius equation to find activation energy?

What is a pseudo-first-order reaction? Can you give an example?

How do you determine the order of reaction from experimental data?

What does activation energy mean physically?

What is the difference between rate constant k and rate of reaction r?

Continue with the next chapter notes

Track Your NEET Score Across All 90 Chapters