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

More Chemical Kinetics Previous-Year Questions — Page 4

Q46 jee_main_2025_02_april_evening First-Order Reactions and Half-Life

For the reaction

A arrow B

the following graph was obtained. The time required (in seconds) for the concentration of A to reduce to

2.5~g~L⁻¹

(if the initial concentration of A was

50~g~L⁻¹

) is _______. (Nearest integer) Given:

2 = 0.3010

Concentration of A versus time graph for Q46 - JEE Main 2025 Evening — The graph plots the concentration of A in g/L against time in seconds, indicating marked coordinates at t=5s, t=10s, t=15s, t=20s, and t=25s.

Numerical Answer. Answer: 43 to 43

Solution

Related Formula

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

t = (1)/(k) ln((A₀)/(Aₜ))

Core Logic

Let's first analyze the order of the reaction using the concentration-time coordinates from the graph:

At $t=5~s$ $t=5~\mathrm{s}$, concentration $[A] = 40~ g~L⁻¹$ $[A] = 40~\mathrm{g~L^{-1}}$
At $t=15~s$ $t=15~\mathrm{s}$, concentration $[A] = 20~ g~L⁻¹$ $[A] = 20~\mathrm{g~L^{-1}}$

Notice that the concentration drops to exactly half of its value ( $40 arrow 20$ $40 \rightarrow 20$) over a time interval of $Δ t = 15 - 5 = 10~s$ $\Delta t = 15 - 5 = 10~\mathrm{s}$.

At $t=25~s$ $t=25~\mathrm{s}$, concentration $[A] = 10~ g~L⁻¹$ $[A] = 10~\mathrm{g~L^{-1}}$

Again, the concentration drops to half ( $20 arrow 10$ $20 \rightarrow 10$) in another interval of $Δ t = 25 - 15 = 10~s$ $\Delta t = 25 - 15 = 10~\mathrm{s}$.

Since the half-life ( $t1/2$ $t_{1/2}$) is constant and independent of the initial concentration, this reaction follows first-order kinetics.

Step 1: Calculate the Rate Constant (k)

The half-life of the reaction is $t1/2 = 10~s$ $t_{1/2} = 10~\mathrm{s}$.

k = (ln 2)/(10) = (2.303 2)/(10) = (2.303 \times 0.3010)/(10) \approx 0.0693~ s⁻¹

Step 2: Calculate the Time for Decay to 2.5 g/L

Given initial concentration $A₀ = 50~ g~L⁻¹$ $A_0 = 50~\mathrm{g~L^{-1}}$ and target concentration $Aₜ = 2.5~ g~L⁻¹$ $A_t = 2.5~\mathrm{g~L^{-1}}$:

t = (2.303)/(k) ((A₀)/(Aₜ))

t = (2.303)/((2.303 2)/(10)) ((50)/(2.5))

t = (10)/( 2) (20)

t = 10 \times ( (10) + (2))/( (2))

t = 10 \times ( (1 + 0.3010)/(0.3010) )

t = 10 \times ( (1.3010)/(0.3010) ) \approx 43.22~s

Rounding to the nearest integer gives 43 seconds.

Pattern Recognition

Shortcut trick: If $t1/2 = 10~s$ $t_{1/2} = 10~\mathrm{s}$, any concentration drop of $2ⁿ$ $2^n$ times takes $n \times t1/2$ $n \times t_{1/2}$ seconds. Here, $(50)/(2.5) = 20$ $\frac{50}{2.5} = 20$. Since $2⁴ = 16$ $2^4 = 16$ (takes 40 s) and $2⁵ = 32$ $2^5 = 32$ (takes 50 s), a drop of 20 times must take slightly over 40 seconds. This confirms our calculation of 43 seconds.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q jee_main_2025_02_april_morning Zero Order Reaction Dynamics

For the reaction

Aarrow

products.

Half-life concentration dependence graph for Q50 — The graph plots half-life period t_1/2 along the vertical axis against starting substance concentration [A]_0 on the horizontal axis, displaying a straight line passing through the origin with an explicit slope value of 76.92.

The concentration of A at 10 minutes is

\times 10⁻³ ~mol ~L⁻¹

(nearest integer). The reaction was started with

2.5mol ⁻¹

of A.

Numerical Answer. Answer: 2435 to 2435

Solution

Related Formula

Half-life equation for a Zero-Order reaction layout:

t1/2 = [A]₀2K

Integrated rate law configuration expression:

[A]ₜ = -Kt + [A]₀

Core Logic

Let's extract kinetic constants step-by-step:

The given plot displays a perfect linear variation passing through the origin: $t1/2 \propto [A]₀$ $t_{1/2} \propto [\mathrm{A}]_0$. This confirms the transformation process follows Zero-Order kinetics.
The visual slope equation evaluates as:

Slope = (1)/(2K) = 76.92 K = (1)/(2 \times 76.92) = 0.0065 ~mol\cdot L⁻¹\cdot min⁻¹

Step 1: Compute Concentration at t = 10 min

Apply the values to the integrated rate law formula track ( $t = 10~min$ $t = 10\mathrm{~min}$, $[A]₀ = 2.5 ~mol\cdot L⁻¹$ ):

[A]₁₀ = -((1)/(2 \times 76.92)) \times 10 + 2.5

[A]₁₀ = -0.0650 + 2.5 = 2.435 ~mol\cdot L⁻¹ = 2435 \times 10⁻³ ~mol\cdot L⁻¹

Hence, the target value for the blank field is 2435.

Pattern Recognition

Always identify the reaction order first by inspecting the axes layout: a linear plot of $t1/2$ $t_{1/2}$ versus $[A]₀$ $[A]_0$ uniquely identifies zero-order behavior, whereas a horizontal flat line implies a first-order path.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q45 jee_main_2025_03_april_evening Temperature Dependence of Reaction Rate

Consider the following statements related to temperature dependence of rate constants. Identify the correct statements, A. The Arrhenius equation holds true only for an elementary homogenous reaction. B. The unit of A is same as that of k in Arrhenius equation. C. At a given temperature, a low activation energy means a fast reaction. D. A and Ea as used in Arrhenius equation depend on temperature. E. When Ea >> RT, A and Ea become interdependent. Choose the correct answer from the options given below:

A. A, C and D Only
B. B, D and E Only
C. B and C Only
D. A and B Only

Solution

Related Formula

Arrhenius equation is given by:

k = A e-(Eₐ)/(R T)

where:

$k$ $k$ is the rate constant
$A$ $A$ is the pre-exponential factor (frequency factor)
$Eₐ$ $E_a$ is the activation energy

Core Logic

Evaluate each statement:

A: Arrhenius equation is an empirical relation that works well for both elementary and complex homogeneous reactions $arrow$ $\rightarrow$ Incorrect.
B: Since the exponential term $e-Eₐ/RT$ $e^{-E_a/RT}$ is dimensionless, the pre-exponential factor $A$ $A$ has the exact same unit as the rate constant $k$ $k$ $arrow$ $\rightarrow$ Correct.
C: For low $Eₐ$ $E_a$, the term $e-Eₐ/RT$ $e^{-E_a/RT}$ is large, giving a high rate constant $k$ $k$ and a fast reaction $arrow$ $\rightarrow$ Correct.
D: $A$ $A$ and $Eₐ$ $E_a$ are assumed to be independent of temperature over a narrow range $arrow$ $\rightarrow$ Incorrect.
E: $A$ $A$ and $Eₐ$ $E_a$ remain independent parameters of the system, not interdependent $arrow$ $\rightarrow$ Incorrect.

Step 1: Select correct statements

Statements B and C are correct, matching Option (3).

Pattern Recognition

The exponential factor $e-Eₐ/RT$ $e^{-E_a/RT}$ represents the fraction of collisions with energy greater than the activation barrier. As $Eₐ$ $E_a$ decreases, this fraction grows exponentially, explaining why low-activation pathways (like catalyzed reactions) run much faster.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q jee_main_2025_07_april_morning First Order Reactions

Reaction

A(g) arrow 2B(g) + C(g)

is a first order reaction. It was started with pure A.

$t / min$ $t / \text{min}$	Pressure of system at time $t / mm Hg$ $t / \text{mm Hg}$
10	160
$\infty$ $\infty$	240

Which of the following options is incorrect?

A. $Initial pressure of A is 80 mm Hg$
B. $The reaction never goes to completion$
C. $Rate constant of the reaction is 1.693 min⁻¹$
D. $Partial pressure of A after 10 minute is 40 mm Hg$

Solution

Related Formula

k = (2.303)/(t) ((P₀)/(PA))

Core Logic

For the reaction: $A(g) arrow 2B(g) + C(g)$

At $t=0$ $t=0$, pressure of $A = P₀$ $A = P_0$, while $B = 0$ $B = 0$ and $C = 0$ $C = 0$.
At $t=\infty$ $t=\infty$, $A$ $A$ is completely consumed, leaving $2P₀$ $2P_0$ of $B$ $B$ and $P₀$ $P_0$ of $C$ $C$.

P\infty = 3P₀ = 240 mm Hg P₀ = 80 mm Hg

This confirms option (A) is correct.

At any time $t$ $t$, pressure of $A = P₀ - x$ $A = P_0 - x$, $B = 2x$ $B = 2x$, $C = x$ $C = x$.

Pₜ = P₀ + 2x = 80 + 2x

At $t=10 min$ $t=10\text{ min}$, $P₁₀ = 160 mm Hg$ $P_{10} = 160\text{ mm Hg}$:

80 + 2x = 160 x = 40 mm Hg

Thus, partial pressure of $A$ $A$ after $10 min$ $10\text{ min}$ is:

PA = P₀ - x = 80 - 40 = 40 mm Hg

This confirms option (D) is correct.

Now, calculate the rate constant $k$ $k$:

k = (1)/(10) ln((80)/(40)) = (ln 2)/(10) = 0.0693 min⁻¹

Therefore, option (C) which states $k = 1.693 min⁻¹$ $k = 1.693\text{ min}^{-1}$ is incorrect.

Pattern Recognition

At $t=\infty$ $t=\infty$, the total pressure is $3$ $3$ times the initial pressure of $A$ $A$. So, $P₀ = P_\infty / 3 = 80 mm Hg$ $P_0 = P_\infty / 3 = 80\text{ mm Hg}$. Half-life $t1/2 = 10 min$ $t_{1/2} = 10\text{ min}$ since $PA$ $P_A$ drops from $80$ $80$ to $40$ $40$ in $10 min$ $10\text{ min}$. Thus, $k = 0.693 / 10 = 0.0693 min⁻¹$ .

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More Chemical Kinetics Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Calculate the Rate Constant (k)

Step 2: Calculate the Time for Decay to 2.5 g/L

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Concentration at t = 10 min

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Select correct statements

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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