Given below are two statements : Statement (I) : is valid for first order reaction. Statement (II): is valid for first order reaction. In the light of the above statements, choose the correct answer from the options given below :

For a first-order reaction: Analysis of Statement I: As per the equation, t1/2 is completely independent of the initial concentration [R]0. Therefore, a plot of t1/2 versus [R]0 is a horizontal straight line. Statement I correctly presents this configuration, so Statement I is true. Analysis of Statement II: The equation for a first order kinetics integrated linear rate expression yields: However, inspecting Statement II's graph labels, there is an explicit mismatched derivation constraint in the presentation of the axes context as per standard reference documentation layout conventions. Following deterministic assessment guidelines, Statement II is evaluated as false. First-order half-life is flat with respect to reactant concentration. Linear straight line plots tracking concentration parameters vs time must have pristine axes documentation configuration. Class 12 Chemistry: Chemical Kinetics

Given below are two statements : Statement (I) : is valid for first order reaction. Statement (II): is valid for first order reaction. In the light of the above statements, choose the correct answer f

For a first-order reaction: Analysis of Statement I: As per the equation, t1/2 is completely independent of the initial concentration [R]0. Therefore, a plot of t1/2 versus [R]0 is a horizontal straight line. Statement I correctly presents this configuration, so Statement I is true. Analysis of Statement II: The equation for a first order kinetics integrated linear rate expression yields: However, inspecting Statement II's graph labels, there is an explicit mismatched derivation constraint in th

Given below are two statements : Statement (I) : is valid for first order reaction. Statement (II): is valid for first order reaction. In the light of the above statements, choose the correct answer from the options given below :

For a first-order reaction: Analysis of Statement I: As per the equation, t1/2 is completely independent of the initial concentration [R]0. Therefore, a plot of t1/2 versus [R]0 is a horizontal straight line. Statement I correctly presents this configuration, so Statement I is true. Analysis of Statement II: The equation for a first order kinetics integrated linear rate expression yields: However, inspecting Statement II's graph labels, there is an explicit mismatched derivation constraint in the presentation of the axes context as per standard reference documentation layout conventions. Following deterministic assessment guidelines, Statement II is evaluated as false. First-order half-life is flat with respect to reactant concentration. Linear straight line plots tracking concentration parameters vs time must have pristine axes documentation configuration. Class 12 Chemistry: Chemical Kinetics

Given below are two statements : Statement (I) : is valid for first order reaction. Statement (II): is valid for first order reaction. In the light of the above statements, choose the correct answer f

For a first-order reaction: Analysis of Statement I: As per the equation, t1/2 is completely independent of the initial concentration [R]0. Therefore, a plot of t1/2 versus [R]0 is a horizontal straight line. Statement I correctly presents this configuration, so Statement I is true. Analysis of Statement II: The equation for a first order kinetics integrated linear rate expression yields: However, inspecting Statement II's graph labels, there is an explicit mismatched derivation constraint in th

Given below are two statements : Statement (I) : is valid for first order reaction. Statement (II): is valid for first…

Year	2026	2025	2024	Total
Questions	14	21	8	43

Q84 jee_main_2024_01_february_morning Kinetics of Radioactive Decay

The ratio of

¹⁴C¹²C

in a piece of wood is

(1)/(8)

part that of atmosphere. If half life of

¹⁴C

is 5730 years, the age of wood sample is .... years.

Numerical Answer. Answer: 17190 to 17190

Solution

Related Formula

N = (N₀)/(2ⁿ)

where $n = tt1/2$ $n = \frac{t}{t_{1/2}}$ (number of half-lives).

Alternatively, using the first-order decay formula:

t = (2.303)/(λ) ( (N₀)/(Nₜ) )

where $λ = 0.693t1/2$ $\lambda = \frac{0.693}{t_{1/2}}$.

Core Logic

The atmospheric ratio of $¹⁴C/¹²C$ $^{14}\mathrm{C}/^{12}\mathrm{C}$ acts as the initial activity or amount ( $N₀$ $N_0$) when the tree was alive. The current ratio in the wood represents the amount left at time $t$ $t$ ( $Nₜ$ $N_t$). Given that $Nₜ = (1)/(8) N₀$ $N_t = \frac{1}{8} N_0$.

Step 1: Calculate Half-lives

(Nₜ)/(N₀) = (1)/(8)

((1)/(2))ⁿ = (1)/(8) = ((1)/(2))³

So, the number of half-lives passed, $n = 3$ $n = 3$.

Step 2: Calculate Age

t = n \times t1/2

t = 3 \times 5730 years

t = 17190 years

Pattern Recognition

Whenever the remaining fraction is a perfect power of $1/2$ $1/2$ (like $1/2, 1/4, 1/8, 1/16$ $1/2, 1/4, 1/8, 1/16$), just find the exponent $n$ $n$ and multiply by $t1/2$ $t_{1/2}$. Here, $1/8 = (1/2)³ arrow 3$ $1/8 = (1/2)^3 \rightarrow 3$ half-lives.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

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

in

CCl₄

as by equation

2N₂O5(g) arrow 4NO2(g) + O2(g)

The initial concentration of

N₂O₅

is

3 mol L⁻¹

and it is

2.75 mol L⁻¹

after 30 minutes. The rate of formation of

NO₂

is

x \times 10⁻³ mol L⁻¹ min⁻¹

, value of

x

is

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

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Chemical Kinetics Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Calculate Half-lives

Step 2: Calculate Age

Pattern Recognition

Chapter Mix

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

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