For a given reaction Rarrow P,t1 / 2 is related to [A]0 as given in table : [A]0 / mol,L-1 t1/2 / min 0.100 200 0.025…

Year	2026	2025	2024	Total
Questions	14	20	8	42

Year

2026

2025

2024

Total

Questions

$[A]₀ / mol L⁻¹$ $[A]_0 / \mathrm{mol\,L^{-1}}$	$t1/2 / min$ $t_{1/2} / \mathrm{min}$
0.100	200
0.025	100

[A]₀ / mol L⁻¹

t1/2 / min

0.100

200

0.025

100

Solution & Explanation

Related Formula

The dependence of half-life on initial concentration is given by:

t1/2 \propto 1[A]₀ⁿ⁻¹

Step 1: Finding the reaction order (n)

Using the values provided:

(t1/2)₁(t1/2)₂ = ( [A]0,2[A]0,1 )ⁿ⁻¹

(200)/(100) = ( (0.025)/(0.100) )ⁿ⁻¹ \Rightarrow 2 = ( (1)/(4) )ⁿ⁻¹

2 = 2-2(n-1) \Rightarrow 1 = -2n + 2 \Rightarrow n = (1)/(2)

Hence, statement A is correct.

Step 2: Checking half-life at other concentrations

Since $n = (1)/(2)$ $n = \frac{1}{2}$, $t1/2 \propto \sqrt([A]₀)$ $t_{1/2} \propto \sqrt{[A]_0}$.

For $[A]₀ = 1 M$ $[A]_0 = 1\,\mathrm{M}$:

200t1/2 = \sqrt((0.1)/(1)) \Rightarrow t1/2 = 200\sqrt(10) min

Hence, statement B is correct.

For $[A]₀ = 1.6 M$ $[A]_0 = 1.6\,\mathrm{M}$:

200t1/2 = \sqrt((0.1)/(1.6)) = \sqrt((1)/(16)) = (1)/(4) \Rightarrow t1/2 = 800 min

Hence, statement D is correct.

Pattern Recognition

Sees: Half-life reducing as initial concentration decreases. Trap: Assuming all reactions are first or zero order without calculations. Shortcut: Reduction of $[A]₀$ $[A]_0$ by 4 causes reduction of $t1/2$ $t_{1/2}$ by 2 $arrow$ $\rightarrow$ indicates a square root dependence ( $t1/2 \propto \sqrt(A₀)$ $t_{1/2} \propto \sqrt{A_0}$), which implies $n = 0.5$ $n = 0.5$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

More Chemical Kinetics Previous-Year Questions — Page 8

Q jee_main_2024_29_january_evening First Order Kinetics and Half Life

The half-life of radioisotopic bromine - 82 is 36 hours. The fraction which remains after one day is ________

\times 10⁻²

. (Given antilog 0.2006 = 1.587)

Numerical Answer. Answer: 63 to 63

Solution

Related Formula

k = 0.693t1/2 and t = (2.303)/(k) ₁₀ ((a)/(a-x))

Core Logic

Given $t1/2 = 36 hours$ $t_{1/2} = 36\text{ hours}$, calculate the decay constant ( $k$ $k$):

k = (0.693)/(36) = 0.01925 hr⁻¹

We want to find the fraction remaining after $1 day = 24 hours$ $1\text{ day} = 24\text{ hours}$:

₁₀ ((a)/(a-x)) = (k \times t)/(2.303) = (0.01925 \times 24)/(2.303) = 0.2006

Step 1: Antilog Application

Taking the antilog on both sides:

(a)/(a-x) = 1.587 Fraction remaining ((a-x)/(a)) = (1)/(1.587) \approx 0.6301

Expressing the remaining fraction in the requested format:

0.6301 = 63 \times 10⁻²

Thus, the required integer value is 63.

Pattern Recognition

Ensure all time variables are in matching units (hours) before substituting values into first-order kinetic equations.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q82 jee_main_2024_27_jan_morning Determination of Order of Reaction

Consider the following data for the given reaction:

2HI(g) arrow H2(g) + I2(g)

Experiment	$[HI] (mol L⁻¹)$	Rate $(mol L⁻¹s⁻¹)$
1	0.005	$7.5 \times 10⁻⁴$ $7.5 \times 10^{-4}$
2	0.01	$3.0 \times 10⁻³$ $3.0 \times 10^{-3}$
3	0.02	$1.2 \times 10⁻²$ $1.2 \times 10^{-2}$

The order of the reaction is

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

Rate law relation expression:

R = k[HI]ⁿ

where $n$ $n$ represents the overall reaction order indicator.

Step 1: Set up ratios using data subsets

Comparing data from experiment 1 and experiment 2:

(R₂)/(R₁) = 3.0 \times 10⁻³7.5 \times 10⁻⁴ = ((0.01)/(0.005))ⁿ

$4 = (2)ⁿ$ $4 = (2)^n$

2² = 2ⁿ n = 2

Pattern Recognition

Doubling concentration ( $0.005 arrow 0.01$ $0.005 \rightarrow 0.01$) increases the reaction rate by 4 times ( $7.5 \times 10⁻⁴ arrow 3.0 \times 10⁻³$ ). Hence, it is a clear second-order ( $2² = 4$ $2^2 = 4$) dynamic pattern.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q83 jee_main_2024_29_jan_morning Arrhenius Equation and Activation Energy

For a reaction taking place in three steps at same temperature, overall rate constant

K = K₁K₂K₃

. If

Ea₁

Ea₂

and

Ea₃

are 40, 50 and 60 kJ/mol respectively, the overall Ea is ______ kJ/mol.

Numerical Answer. Answer: 30 to 30

Solution

Related Formula

K = A e-Eₐ/RT

Core Logic

Given the relationship between the rate constants:

K = (K₁ \cdot K₂)/(K₃)

Substituting the Arrhenius equation for each rate constant:

A \cdot e-Eₐ/RT = A₁ \cdot e^-Eₐ₁/RT \cdot A₂ \cdot e^-Eₐ₂/RTA₃ \cdot e^-Eₐ₃/RT

Combining the exponential terms using rules of exponents:

A \cdot e-Eₐ/RT = ((A₁ \cdot A₂)/(A₃)) \cdot e^ -(Eₐ₁ + Eₐ₂ - Eₐ₃)RT

Step 1: Equating Activation Energies

By comparing the powers of $e$ $e$ on both sides, the overall activation energy $Eₐ$ $E_a$ is related to the individual steps as follows:

Eₐ = Eₐ₁ + Eₐ₂ - Eₐ₃

Substitute the given values ( $Eₐ₁ = 40$ $E_{a1} = 40$, $Eₐ₂ = 50$ $E_{a2} = 50$, $Eₐ₃ = 60$ $E_{a3} = 60$ kJ/mol):

Eₐ = 40 + 50 - 60

$Eₐ = 90 - 60$ $E_a = 90 - 60$

Eₐ = 30 kJ/mol

Pattern Recognition

When rate constants are multiplied or divided ( $K = K₁^a K₂^b / K₃^c$ $K = K_1^a K_2^b / K_3^c$), the corresponding overall activation energy follows the linear combination of the exponents: $Eₐ = a Eₐ₁ + b Eₐ₂ - c Eₐ₃$ $E_a = a E_{a1} + b E_{a2} - c E_{a3}$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q82 jee_main_2024_30_january_evening Rate of Chemical Reaction

NO₂

required for a reaction is produced by decomposition of

N₂O₅

CCl₄

as by equation

2N₂O5(g) arrow 4NO2(g) + O2(g)

The initial concentration of

N₂O₅

3 mol L⁻¹

and it is

2.75 mol L⁻¹

after 30 minutes. The rate of formation of

NO₂

x \times 10⁻³ mol L⁻¹ min⁻¹

, value of

x

Numerical Answer. Answer: 17 to 17

Solution

Related Formula

Rate of Reaction (ROR) = -(1)/(2) Δ [N₂O₅]Δ t = (1)/(4) Δ [NO₂]Δ t

Core Logic

First, find the rate of disappearance of $N₂O₅$ $\mathrm{N}_2\mathrm{O}_5$.

- Δ [N₂O₅]Δ t = - ((2.75 - 3))/(30) = (0.25)/(30) mol L⁻¹ min⁻¹

Now, equate it to the general Rate of Reaction:

ROR = -(1)/(2) Δ [N₂O₅]Δ t = (1)/(2) ((0.25)/(30)) = (0.125)/(30) = (1)/(240) mol L⁻¹ min⁻¹

Step 1: Calculate the Rate of Formation of $NO₂$ $\mathrm{NO}_2$

Rate of formation of $NO₂ = Δ [NO₂]Δ t = 4 \times ROR$

= 4 \times (1)/(240) = (1)/(60) mol L⁻¹ min⁻¹

Convert this to scientific notation to find $x$ $x$:

(1)/(60) \approx 0.01666 = 16.66 \times 10⁻³ mol L⁻¹ min⁻¹

Rounding to the nearest integer, we get $x = 17$ $x = 17$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q81 jee_main_2024_30_jan_morning First Order Reactions

The rate of first order reaction is

0.04 mol L⁻¹s⁻¹

at 10 minutes and

0.03 mol L⁻¹s⁻¹

at 20 minutes after initiation. Half life of the reaction is ________ minutes. (Given

2=0.3010

3=0.4771

)

Numerical Answer. Answer: 24 to 24.1

Solution

Related Formula

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

[A] = [A]₀ e-kt

t1/2 = (ln 2)/(k)

Core Logic

For a first order reaction, rate is directly proportional to concentration.

R₁ = k[A]₁₀ = k[A]₀ e-k(10 \times 60)

R₂ = k[A]₂₀ = k[A]₀ e-k(20 \times 60)

Step 1: Setting up equations

0.04 = k[A]₀ e-600k (1)

0.03 = k[A]₀ e-1200k (2)

Step 2: Solving for k

Dividing equation (1) by (2):

(0.04)/(0.03) = e-600ke-1200k

(4)/(3) = e600k

Take natural log on both sides:

ln((4)/(3)) = 600k

k = (ln(4/3))/(600) s⁻¹

Step 3: Calculating half life

t1/2 = (ln 2)/(k) = (ln 2)/((ln(4/3))/(600)) = (600 ln 2)/(ln 4 - ln 3) seconds

Convert to minutes by dividing by 60:

t1/2 = (10 ln 2)/(ln 4 - ln 3) minutes

Substitute log values (since $ln x = 2.303 x$ $\ln x = 2.303 \log x$, the 2.303 cancels out):

t1/2 = 10 \times ( 2)/( 4 - 3) minutes

t1/2 = 10 \times (0.3010)/(2(0.3010) - 0.4771)

t1/2 = 10 \times (0.3010)/(0.6020 - 0.4771) = 10 \times (0.3010)/(0.1249)

t1/2 = 24.099 \approx 24 minutes

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Solution & Explanation

Related Formula

Step 1: Finding the reaction order (n)

Step 2: Checking half-life at other concentrations

Pattern Recognition

Chapter Mix

Reference Study Guides

More Chemical Kinetics Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Antilog Application

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Set up ratios using data subsets

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Equating Activation Energies

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the Rate of Formation of NO₂$\mathrm{NO}_2$

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setting up equations

Step 2: Solving for k

Step 3: Calculating half life

Chapter Mix

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Step 1: Calculate the Rate of Formation of $NO₂$ $\mathrm{NO}_2$