Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction…

Year	2026	2025	2024	Total
Questions	14	21	8	43

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

ln k = ln A - (Eₐ)/(RT)

Core Logic

For Reaction 1: $ln k₁ = ln A - (E₁)/(RT)$ $\ln k_1 = \ln A - \frac{E_1}{RT}$ For Reaction 2: $ln k₂ = ln A - (E₂)/(RT)$ $\ln k_2 = \ln A - \frac{E_2}{RT}$ (Pre-exponential factor $A$ $A$ is the same).

Subtracting the first from the second:

ln k₂ - ln k₁ = -(E₂)/(RT) - (-(E₁)/(RT))

ln ((k₂)/(k₁)) = (E₁ - E₂)/(RT)

Given that $E₁$ $E_1$ exceeds $E₂$ $E_2$ by $20 kJ mol⁻¹$ $20\text{ kJ mol}^{-1}$, $E₁ - E₂ = 20000 J mol⁻¹$ $E_1 - E_2 = 20000\text{ J mol}^{-1}$. $T = 300 K$ $T = 300\text{ K}$, $R = 8.3 J K⁻¹ mol⁻¹$ .

ln ((k₂)/(k₁)) = (20000)/(8.3 \times 300) = (200)/(8.3 \times 3) = (200)/(24.9)

ln ((k₂)/(k₁)) = 8.032

Rounding off to nearest integer gives 8.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

More Chemical Kinetics Previous-Year Questions — Page 9

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

Q82 jee_main_2024_31_jan_evening First Order Kinetics

r = k[A]

for a reaction,

50%

of A is decomposed in

120 minutes

. The time taken for

90%

decomposition of A is ________ minutes.

Numerical Answer. Answer: 398.5 to 399.5

Solution

Related Formula

k = 0.693t1/2

t = (2.303)/(k) ( (a)/(a - x) )

Core Logic

Since $r = k[A]$ $r = k[A]$, the reaction follows first-order kinetics. The half-life ( $50%$ $50\%$ decomposition) is $t1/2 = 120 minutes$ $t_{1/2} = 120\text{ minutes}$.

Step 1: Calculating for 90% decomposition

For $90%$ $90\%$ completion of the reaction, $[A]₀ = 100$ $[A]_0 = 100$ and $[A]ₜ = 100 - 90 = 10$ $[A]_t = 100 - 90 = 10$.

t = (2.303)/(k) (100)/(10)

t = 2.303( 0.693t1/2 ) (10)

t = (2.303 \times 120)/(0.693) \times 1

t = 398.78 minutes

Rounding off to the nearest integer, we get $399 minutes$ $399\text{ minutes}$.

Pattern Recognition

For a first order reaction, $t90% \approx 3.32 \times t50%$ $t_{90\%} \approx 3.32 \times t_{50\%}$. $120 \times 3.32 = 398.4$ $120 \times 3.32 = 398.4$, so roughly $399$ $399$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q73 jee_main_2024_31_jan_morning First Order Reactions

Integrated rate law equation for a first order gas phase reaction is given by (where

Pᵢ

is initial pressure and

Pₜ

is total pressure at time t)

A. $k = (2.303)/(t) \times (Pᵢ)/((2Pᵢ - Pₜ))$
B. $k = (2.303)/(t) \times (2Pᵢ)/((2Pᵢ - Pₜ))$
C. $k = (2.303)/(t) \times ((2Pᵢ - Pₜ))/(Pᵢ)$
D. $k = (2.303)/(t) \times (Pᵢ)/((2Pᵢ - Pₜ))$

Solution

Core Logic

Consider a general gas phase reaction: $A arrow B + C$ $A \rightarrow B + C$

Initial ( $t=0$ $t=0$): $Pᵢ 0 0$ $\quad P_i \quad\quad 0 \quad\quad 0$ At time $t$ $t$: $Pᵢ - x x x$

Total pressure at time $t$ $t$:

Pₜ = (Pᵢ - x) + x + x = Pᵢ + x

$x = Pₜ - Pᵢ$ $x = P_t - P_i$

Partial pressure of A at time $t$ $t$ ( $PA$ $P_A$): $PA = Pᵢ - x$ $P_A = P_i - x$

PA = Pᵢ - (Pₜ - Pᵢ) = 2Pᵢ - Pₜ

For a first-order reaction:

k = (2.303)/(t) (P₀)/(Pₜ)

Here, $P₀ = Pᵢ$ $P_0 = P_i$ and the pressure of the reactant at time $t$ $t$ is $PA$ $P_A$.

k = (2.303)/(t) (Pᵢ)/(2Pᵢ - Pₜ)

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Solution & Explanation

Related Formula

Core Logic

Chapter Mix

Reference Study Guides

More Chemical Kinetics Previous-Year Questions — Page 9

Solution

Related Formula

Core Logic

Step 1: Setting up equations

Step 2: Solving for k

Step 3: Calculating half life

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating for 90% decomposition

Pattern Recognition

Chapter Mix

Solution

Core Logic

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Solution & Explanation

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Reference Study Guides

More Chemical Kinetics Previous-Year Questions — Page 9

Solution

Related Formula

Core Logic

Step 1: Setting up equations

Step 2: Solving for k

Step 3: Calculating half life

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating for 90% decomposition

Pattern Recognition

Chapter Mix

Solution

Core Logic

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