A person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The wound was later treated with some antibacterial medicine and the rate of bacterial decay (r) was found to be proportional with the square of the existing number of bacteria at any instance. Which of the following set of graphs correctly represents the 'before' and 'after' situation of the application of the medicine? [Given: N = No. of bacteria, t = time, bacterial growth follows Ist order kinetics.]

### Related Formula ### Core Logic Let's analyze the kinetics for the two stages: 1. **Before applying medicine**: - Bacterial growth follows 1st order kinetics: (dN/dt) = k1 N. - Integrating this yields: N(t) = N0 ek1 t. - The graph of (N/N0) vs t is a rising exponential curve starting from 1 (since at t=0, (N/N0) = 1). 2. **After applying medicine**: - Reductive rate is proportional to the square of existing bacteria (2nd order decay): - Integrating this yields: - A plot of N vs t or (N/N0) vs t for this decay is a hyperbolic curve decreasing gradually. - Option B correctly matches the exponential growth before medicine and the hyperbolic decay after medicine. ### Pattern Recognition First-order growth is an exponential curve (N0 ekt), while second-order decay behaves as a rational hyperbolic relationship (1 / (1 + bt)). Option B shows the exact transition from an exponential rise to a hyperbolic decay curve. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

A person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The wound was later treated with some antibacterial medicine and the rate of bacterial decay

### Related Formula ### Core Logic Let's analyze the kinetics for the two stages: 1. **Before applying medicine**: - Bacterial growth follows 1st order kinetics: (dN/dt) = k1 N. - Integrating this yields: N(t) = N0 ek1 t. - The graph of (N/N0) vs t is a rising exponential curve starting from 1 (since at t=0, (N/N0) = 1). 2. **After applying medicine**: - Reductive rate is proportional to the square of existing bacteria (2nd order decay): - Integrating this yields: - A plot of N vs t or (N/N0) vs

A person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The wound was later treated with some antibacterial medicine and the rate of bacterial decay (r) was found to be proportional with the square of the existing number of bacteria at any instance. Which of the following set of graphs correctly represents the 'before' and 'after' situation of the application of the medicine? [Given: N = No. of bacteria, t = time, bacterial growth follows Ist order kinetics.]

### Related Formula ### Core Logic Let's analyze the kinetics for the two stages: 1. **Before applying medicine**: - Bacterial growth follows 1st order kinetics: (dN/dt) = k1 N. - Integrating this yields: N(t) = N0 ek1 t. - The graph of (N/N0) vs t is a rising exponential curve starting from 1 (since at t=0, (N/N0) = 1). 2. **After applying medicine**: - Reductive rate is proportional to the square of existing bacteria (2nd order decay): - Integrating this yields: - A plot of N vs t or (N/N0) vs t for this decay is a hyperbolic curve decreasing gradually. - Option B correctly matches the exponential growth before medicine and the hyperbolic decay after medicine. ### Pattern Recognition First-order growth is an exponential curve (N0 ekt), while second-order decay behaves as a rational hyperbolic relationship (1 / (1 + bt)). Option B shows the exact transition from an exponential rise to a hyperbolic decay curve. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

A person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The wound was later treated with some antibacterial medicine and the rate of bacterial decay

### Related Formula ### Core Logic Let's analyze the kinetics for the two stages: 1. **Before applying medicine**: - Bacterial growth follows 1st order kinetics: (dN/dt) = k1 N. - Integrating this yields: N(t) = N0 ek1 t. - The graph of (N/N0) vs t is a rising exponential curve starting from 1 (since at t=0, (N/N0) = 1). 2. **After applying medicine**: - Reductive rate is proportional to the square of existing bacteria (2nd order decay): - Integrating this yields: - A plot of N vs t or (N/N0) vs

A person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The wound…

Year	2026	2025	2024	Total
Questions	14	20	8	42

Q75 jee_main_2026_22_january_evening Activation Energy Calculation

Consider

A k₁ B

and

C k₂ D

are two reactions. If the rate constant (

k₁

) of the

A arrow B

reaction can be expressed by the following equation

₁₀k = 14.34 - 1.5 \times 10⁴T/K

and activation energy of

C arrow D

reaction (

Eₐ₂

) is

(1)/(5)th

of the

A arrow B

reaction (

Eₐ₁

), then the value of (

Eₐ₂

) is ____

kJ mol⁻¹

. (Nearest Integer)

Numerical Answer. Answer: 57 to 57

Solution

Related Formula

₁₀ k = ₁₀ A - (Eₐ)/(2.303 R T)

Eₐ₁2.303 R = 1.5 \times 10⁴

Core Logic

Step 1: Calculate $Eₐ₁$ $E_{a1}$ from the Arrhenius equation slope:

Eₐ₁ = 1.5 \times 10⁴ \times 2.303 \times 8.314 J mol⁻¹

Eₐ₁ = 287207 J mol⁻¹ = 287.207 kJ mol⁻¹

Step 2: Calculate $Eₐ₂$ $E_{a2}$:

Eₐ₂ = Eₐ₁5 = (287.207)/(5) = 57.4414 kJ mol⁻¹ \approx 57 kJ mol⁻¹

Pattern Recognition

Sees: Arrhenius log equation comparing slopes. Shortcut: $Eₐ₁ = 1.5 \times 10⁴ \times 2.303 \times 8.314 = 287.2 kJ$ ; divide by 5 gives $57.44 \approx 57 kJ/mol$ $57.44 \approx 57\text{ kJ/mol}$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q71 jee_main_2026_23_january_morning Zero Order Kinetics

For the thermal decomposition of reaction AB(g), the following is constructed.

Zero Order Kinetics diagram for Q71 - JEE Main 2026 Morning — The image shows a linear downward slope on a graph of concentration vs. time, indicating a zero-order reaction.

The half life of the reaction is '

x

' min.

x =

____ min. (Nearest integer)

Numerical Answer. Answer: 10 to 10

Solution

Related Formula

[AB]ₜ = [AB]₀ - kt

t1/2 = ([AB]₀)/(2k)

Core Logic

The graph of concentration vs time is a straight line with a negative slope, which is the hallmark of a zero-order reaction. The slope of the line equals $-k$ $-k$.

Step 1: Finding Rate Constant (k)

From the given graph coordinates: At $t = 0$ $t = 0$, $[AB]₀ = 0.60$ $[AB]_0 = 0.60$ M At $t = 100$ $t = 100$ s, $[AB]₁₀₀ = 0.55$ $[AB]_{100} = 0.55$ M

k = ([AB]₀ - [AB]ₜ)/(t)

k = (0.60 - 0.55)/(100) = (0.05)/(100) = 5 \times 10⁻⁴ M/s

Step 2: Calculating Half-Life

t1/2 = ([AB]₀)/(2k) = 0.602 \times 5 \times 10⁻⁴

t1/2 = 0.6010⁻³ = 600 seconds

Converting to minutes:

x = (600)/(60) = 10 min

Pattern Recognition

Linear $[A]$ $[A]$ vs $t$ $t$ plot $arrow$ $\rightarrow$ Zero order. Half-life for zero order is proportional to initial concentration. $t1/2 = [A]₀ / 2k$ $t_{1/2} = [A]_0 / 2k$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q51 jee_main_2026_23_january_evening Rate of Reaction

Rate of Reaction diagram for Q51 - JEE Main 2026 Evening — A graph showing concentration vs time for reactant A and product B over an 80 minute period.

Given above is the concentration vs time plot for a dissociation reaction :

A arrow nB

. Based on the data of the initial phase of the reaction (initial

10 min

), the value of n is ____.

A. $4$ $4$
B. $3$ $3$
C. $2$ $2$
D. $5$ $5$

Solution

Related Formula

Rate = -(Δ[A])/(Δ t) = (1)/(n)(Δ[B])/(Δ t)

Core Logic

For the reaction $A arrow nB$ $A \rightarrow nB$, the stoichiometry determines the relationship between the change in concentration of $A$ $A$ and $B$ $B$.

From the graph, at $t = 0 min$ $t = 0 \text{ min}$: $[A] = 0.05 M$ $[A] = 0.05 \text{ M}$ $[B] = 0 M$ $[B] = 0 \text{ M}$

At $t = 10 min$ $t = 10 \text{ min}$: $[A] = 0.04 M$ $[A] = 0.04 \text{ M}$ Change in $[A] = 0.05 - 0.04 = 0.01 M$ $[A] = 0.05 - 0.04 = 0.01 \text{ M}$

Since $1$ $1$ mole of $A$ $A$ dissociates to give $n$ $n$ moles of $B$ $B$, the concentration of $B$ $B$ formed will be $n \times change in [A]$ $n \times \text{change in } [A]$. $Δ[B] = 0.01 \times n$ $\Delta[B] = 0.01 \times n$

Step 1: Calculating n

From the graph, at $t = 10 min$ $t = 10 \text{ min}$, the concentration of $B$ $B$ is $0.03 M$ $0.03 \text{ M}$.

Equating this to our stoichiometric expression:

0.01 \times n = 0.03

$n = 3$ $n = 3$

Pattern Recognition

Visually compare the vertical drops/rises at the first time interval. A drops by 1 unit, B rises by 3 units. The ratio directly gives the stoichiometric coefficient $n = 3$ $n = 3$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q61 jee_main_2026_23_january_evening First Order Reactions

Observe the following reactions at T(K) I.

A arrow products

. II.

5Br⁻(aq) + BrO₃⁻(aq) + 6H⁺(aq) arrow 3Br₂(aq) + 3H₂O(l)

Both the reactions are started at 10.00 am. The rates of these reactions at 10.10 am are same. The value of

- Δ[Br⁻]Δ t

at 10.10 am is

2 \times 10⁻⁴ mol L⁻¹ min⁻¹

. The concentration of A at 10.10 am is

10⁻² mol L⁻¹

. What is the first order rate constant (in

min⁻¹

) of reaction I?

A. $2 \times 10⁻³$ $2 \times 10^{-3}$
B. $10⁻³$ $10^{-3}$
C. $10⁻²$ $10^{-2}$
D. $4 \times 10⁻³$ $4 \times 10^{-3}$

Solution

Related Formula

Rate of reaction = (-1)/(νᵢ) (d[Reactant])/(dt)

Rate of first order reaction = k[A]

Core Logic

At $t = 10$ $t = 10$ minutes ( $10:10 am$ $10:10 \text{ am}$): For reaction II: $5Br⁻ + BrO₃⁻ + 6H⁺ arrow 3Br₂ + 3H₂O$ The overall rate of reaction II is expressed by dividing the rate of disappearance of $Br^-$ $Br^-$ by its stoichiometric coefficient:

RateII = -(1)/(5) Δ[Br⁻]Δ t

Given $- Δ[Br⁻]Δ t = 2 \times 10⁻⁴ mol L⁻¹ min⁻¹$ , we have:

RateII = (1)/(5) \times (2 \times 10⁻⁴) = 4 \times 10⁻⁵ mol L⁻¹ min⁻¹

Step 1: Equating Rates

The problem states that the rates of both reactions are identical at this time. Therefore, $RateI = RateII = 4 \times 10⁻⁵ mol L⁻¹ min⁻¹$ .

For the first-order reaction I ( $A arrow products$ $A \rightarrow \text{products}$):

RateI = k[A]

Step 2: Calculating Rate Constant

Substitute the known values at $t = 10 min$ $t = 10 \text{ min}$:

4 \times 10⁻⁵ = k \times (10⁻²)

k = 4 \times 10⁻⁵10⁻² = 4 \times 10⁻³ min⁻¹

Pattern Recognition

Remember to divide the given rate of disappearance of a species by its stoichiometric coefficient to find the true, normalized "Rate of Reaction" before equating it to another reaction's rate.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q52 jee_main_2026_24_january_morning Reaction Profile and Energy Diagram

A arrow D

is an endothermic reaction occurring in three steps (elementary). (i)

A arrow B Δ Hᵢ = +ve

(ii)

B arrow C Δ Hᵢᵢ = -ve

(iii)

C arrow D Δ Hᵢᵢᵢ = -ve

Which of the following graphs between potential energy (y-axis) vs reaction coordinate (x-axis) correctly represents the reaction profile of

A arrow D

?

A. $Graph 1$ $\text{Graph 1}$
B. $Graph 2$ $\text{Graph 2}$
C. $Graph 3$ $\text{Graph 3}$
D. $Graph 4$ $\text{Graph 4}$

Solution

Core Logic

Given the overall reaction $A arrow D$ $\text{A} \rightarrow \text{D}$ is endothermic:

Δᵣ H = +ve ED > EA

Analyzing the mechanism steps: Step (i): $A arrow B$ $\text{A} \rightarrow \text{B}$, $Δᵣ H = +ve EB > EA$ Step (ii): $B arrow C$ $\text{B} \rightarrow \text{C}$, $Δᵣ H = -ve EC < EB$ Step (iii): $C arrow D$ $\text{C} \rightarrow \text{D}$, $Δᵣ H = -ve ED < EC$

Putting it all together, we must have a profile where the final energy $ED$ $E_D$ is higher than the initial energy $EA$ $E_A$, with intermediate dips corresponding to $B$ $B$ and $C$ $C$ as guided by the individual step enthalpies.

Reaction Profile for A to D

Step 1: Final Conclusion

Graph 3 matches all criteria: Overall endothermic ( $D > A$ $D > A$), first step endothermic ( $B > A$ $B > A$), followed by two exothermic steps ( $C < B$ $C < B$ and $D < C$ $D < C$).

Pattern Recognition

Endothermic overall means the tail end (products) sits higher than the origin (reactants). Exothermic intermediate steps mean the curve drops from the previous intermediate well.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics Class 11 Chemistry: Thermodynamics

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More Chemical Kinetics Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Rate Constant (k)

Step 2: Calculating Half-Life

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating n

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Equating Rates

Step 2: Calculating Rate Constant

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

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