Consider a complex reaction taking place in three steps with rate constants k1 , k2 and k3 respectively. The overall…

Year	2026	2025	2024	Total
Questions	14	20	8	42

Q jee_main_2025_24_jan_morning First Order Reactions and Pressure dependence

For a reaction,

N₂O5(g) arrow 2NO2(g) + (1)/(2)O2(g)

in a constant volume container, no products were present initially. The final pressure of the system when

50%

of reaction gets completed is

A. $7 / 2$ $7 / 2$ times of initial pressure
B. 5 times of initial pressure
C. $5 / 2$ $5 / 2$ times of initial pressure
D. $7 / 4$ $7 / 4$ times of initial pressure

Solution

Core Logic

Let the initial pressure of the reactant $N₂O₅$ $\mathrm{N}_2\mathrm{O}_{5}$ be $P₀$ $P_0$.

Setting up the stoichiometric reaction table:

$arraylcccc & N₂O5(g) & arrow & 2NO2(g) & + & (1)/(2)O2(g) Initially (t = 0): & P₀ & & 0 & & 0 At time t: & P₀ - x & & 2x & & (x)/(2) array$

The total pressure of the gaseous mixture at any time $t$ $t$ is given by:

Ptotal = (P₀ - x) + 2x + (x)/(2) = P₀ + (3x)/(2)

When $50%$ $50\%$ of the reaction is completed, the change in the reactant's pressure is:

x = 0.5 P₀ = (P₀)/(2)

Substituting the value of $x$ $x$ into the total pressure expression:

Ptotal = P₀ + (3)/(2)((P₀)/(2)) = P₀ + (3P₀)/(4) = (7)/(4)P₀

Pattern Recognition

Track the change in the total pressure carefully using stoichiometric coefficients. For a $50%$ $50\%$ completion step, substitute the fractional equivalent ( $x = 0.5 P₀$ $x = 0.5 P_0$) directly into your total pressure expression.

Q26 jee_main_2025_28_jan_evening Order and Rate of Reaction

consider the elementary reaction

A(g) + B(g) arrow C(g) + D(g)

If the volume of reaction mixture is suddenly reduced to

(1)/(3)

of its initial volume, the reaction rate will become 'x' times of the original reaction rate. The value of x is:

A. $(1)/(9)$ $\frac{1}{9}$
B. $9$ $9$
C. $(1)/(3)$ $\frac{1}{3}$
D. $3$ $3$

Solution

Related Formula

For an elementary reaction, the rate law corresponds directly to its stoichiometry:

$R = K[A]¹[B]¹$ $R = K[A]^1[B]^1$

Concentration ( $C$ $C$) is inversely proportional to volume ( $V$ $V$):

C = (n)/(V)

Core Logic

Initial rate expression:

R₁ = K[(nA)/(V)]¹[(nB)/(V)]¹

When volume is reduced to $(1)/(3)V$ $\frac{1}{3}V$, the new concentration becomes $3$ $3$ times the initial concentration:

R₂ = K[(3nA)/(V)]¹[(3nB)/(V)]¹ = 9 \cdot K[(nA)/(V)]¹[(nB)/(V)]¹

Step 1: Calculating the Value of x

Comparing the two rates:

$R₂ = 9R₁$ $R_2 = 9R_1$

Therefore, the value of $x$ $x$ is $9$ $9$.

Pattern Recognition

For a second-order overall elementary reaction ( $1+1=2$ $1+1=2$), reducing the volume by a factor of $n$ $n$ increases the rate by a factor of $n²$ $n^2$. Here $n=3$ $n=3$, so the rate increases by $3² = 9$ $3^2 = 9$ times.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q40 jee_main_2025_28_jan_evening First Order Kinetics

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?

A. (1)
B. (2)
C. (3)
D. (4)

Solution

Related Formula

Exponential growth equation model:

$N = N₀ eKt$ $N = N_0 e^{Kt}$

Normalized configuration formula:

(N)/(N₀) = eKt

Core Logic

Radioactive decay follows a decreasing exponential path ( $N = N₀ e-λ t$ $N = N_0 e^{-\lambda t}$).

Conversely, cell culture growth functions via an increasing exponential pattern because the rate of growth is directly proportional to the current population size ( $dN/dt = KN$ $dN/dt = KN$). This results in an exponential curve that starts at $(N)/(N₀) = 1$ $\frac{N}{N_0} = 1$ when $t = 0$ $t = 0$ and curves sharply upward over time.

Step 1: Finding the Matching Curve

Plotting $(N)/(N₀)$ $\frac{N}{N_0}$ against time shows an upward-clinging exponential profile starting from $1$ $1$, which perfectly matches the curve in option (4).

Exponential growth profile plot for Q40

Pattern Recognition

The expression $eKt$ $e^{Kt}$ dictates an exponential increase. Ensure the curve starts from a non-zero value ( $1$ $1$) at $t=0$ $t=0$, as $(N₀)/(N₀) = 1$ $\frac{N_0}{N_0} = 1$, rather than starting from the origin ( $0$ $0$).

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q jee_main_2025_29_jan_morning Reaction Mechanism and Rate Law

The reaction

A₂ + B₂ arrow 2 AB

follows the mechanism:

A₂ k₋₁ k₁leftharpoons A + A (fast)

A + B₂ k₂ AB + B (slow)

A + B arrow AB (fast)

The overall order of the reaction is :

A. 1.5
B. 3
C. 2.5
D. 2

Solution

Related Formula

Rate = k \cdot [Reactants]order

Core Logic

The slowest elementary step controls the net kinetic pathway rate law :

Rate = k₂[A][B2] Equation (1)

Since $[A]$ $[\mathrm{A}]$ behaves as a transient intermediate species, replace it using the prior fast equilibrium step :

k₁k-1 = [A]²[A2] [A]² = ( k₁k-1) [A2]

[A] = k₁k-1 \cdot [A2]1/2

Substitute $[A]$ $[\mathrm{A}]$ back into Equation (1) :

Rate = k₂ k₁k-1 \cdot [A2]1/2[B2]

Sum of powers determining overall order:

Order = (1)/(2) + 1 = 1.5

Hence, Option (1) is correct.

Pattern Recognition

Whenever a fast initial step dissociates a molecule into matching independent halves, it always injects a fractional order component of $0.5$ $0.5$ relative to that parent species.

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

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Chemical Kinetics Previous-Year Questions — Page 7

Solution

Core Logic

Pattern Recognition

Solution

Related Formula

Core Logic

Step 1: Calculating the Value of x

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding the Matching Curve

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Solution

Related Formula

Core Logic

Step 1: Calculate Half-lives

Step 2: Calculate Age

Pattern Recognition

Chapter Mix

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