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?

Exponential growth equation model: N = N0 eKt Normalized configuration formula: Radioactive decay follows a decreasing exponential path (N = N0 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). This results in an exponential curve that starts at (N/N0) = 1 when t = 0 and curves sharply upward over time. Plotting (N/N0) against time shows an upward-clinging exponential profile starting from 1, which perfectly matches the curve in option (4). The expression eKt dictates an exponential increase. Ensure the curve starts from a non-zero value (1) at t=0, as (N0/N0) = 1, rather than starting from the origin (0). Class 12 Chemistry: Chemical 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?

Exponential growth equation model: N = N0 eKt Normalized configuration formula: Radioactive decay follows a decreasing exponential path (N = N0 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). This results in an exponential curve that starts at (N/N0) = 1 when t = 0 and curves sharply upward over time. Plotting (N/N0) against time shows an upward-clinging

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?

Exponential growth equation model: N = N0 eKt Normalized configuration formula: Radioactive decay follows a decreasing exponential path (N = N0 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). This results in an exponential curve that starts at (N/N0) = 1 when t = 0 and curves sharply upward over time. Plotting (N/N0) against time shows an upward-clinging exponential profile starting from 1, which perfectly matches the curve in option (4). The expression eKt dictates an exponential increase. Ensure the curve starts from a non-zero value (1) at t=0, as (N0/N0) = 1, rather than starting from the origin (0). Class 12 Chemistry: Chemical 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?

Exponential growth equation model: N = N0 eKt Normalized configuration formula: Radioactive decay follows a decreasing exponential path (N = N0 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). This results in an exponential curve that starts at (N/N0) = 1 when t = 0 and curves sharply upward over time. Plotting (N/N0) against time shows an upward-clinging

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the…

Year	2026	2025	2024	Total
Questions	14	20	8	42

Q26 jee_main_2025_08_april_evening First Order Reactions

In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are

t₁

and

t₂

(s), respectively. The ratio

t₁ / t₂

will: Choose the correct answer from the options given below:

A. $(4)/(3)$ $\frac{4}{3}$
B. $(3)/(2)$ $\frac{3}{2}$
C. $(3)/(4)$ $\frac{3}{4}$
D. $(2)/(3)$ $\frac{2}{3}$

Solution

Related Formula

For a first-order reaction:

t = (2.303)/(k) ((C₀)/(Cₜ))

Alternatively, using half-life $t50%$ $t_{50\%}$:

Cₜ = (C₀)/(2ⁿ)

where $n$ $n$ is the number of half-lives ( $n = tt50%$ $n = \frac{t}{t_{50\%}}$).

Core Logic

Step 1: Calculate $t₁$ $t_1$ for decomposition to $(1)/(4)$ $\frac{1}{4}$ of initial concentration:

Cₜ = (C₀)/(4) = (C₀)/(2²) n = 2

t₁ = 2 \cdot t50%

Step 2: Calculate $t₂$ $t_2$ for decomposition to $(1)/(8)$ $\frac{1}{8}$ of initial concentration:

Cₜ = (C₀)/(8) = (C₀)/(2³) n = 3

t₂ = 3 \cdot t50%

Step 3: Find the ratio $(t₁)/(t₂)$ $\frac{t_1}{t_2}$:

(t₁)/(t₂) = 2 \cdot t50%3 \cdot t50% = (2)/(3)

Pattern Recognition

For first-order kinetics, every step of concentration halving takes exactly one half-life ( $t50%$ $t_{50\%}$). $Initial t50% (1)/(2) t50% (1)/(4) (Total 2 half-lives)$ $(1)/(4) t50% (1)/(8) (Total 3 half-lives)$ Therefore, the ratio is simply the ratio of the number of half-lives: $2 : 3$ $2 : 3$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q42 jee_main_2025_29_jan_evening First Order Reaction Kinetics

Drug X becomes ineffective after

50%

decomposition. The original concentration of drug in a bottle was

16 mg/mL

which becomes

4 mg/mL

in 12 months. The expiry time of the drug in months is (Assume that the decomposition of the drug follows first order kinetics). (1) 12 (2) 2 (3) 3 (4) 6

A. 12
B. 2
C. 3
D. 4

Solution

Related Formula

Nₜ = N₀ left((1)/(2))ⁿ

Core Logic

Let's track concentration reductions:

16 mg/mL xrightarrowt1/2 8 mg/mL xrightarrowt1/2 4 mg/mL

This total progression constitutes exactly $2$ $2$ half-lives ( $n = 2$ $n = 2$).

2 cdot t1/2 = 12 months implies t1/2 = 6 months

Since the drug becomes ineffective right after $50%$ $50\%$ decomposition, its functional expiry limit is exactly $1$ $1$ half-life period.

Expiry time = t1/2 = 6 months

Pattern Recognition

For multi-step concentration halving, bypass complex integrated logarithmic rate expressions by directly applying integer half-life steps ( $16 arrow 8 arrow 4$ $16 \rightarrow 8 \rightarrow 4$).

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q32 jee_main_2025_28_jan_morning Order of Reaction and Half-life

For a given reaction

Rarrow P,t1 / 2

is related to

[A]₀

as given in table :

$[A]₀ / mol L⁻¹$ $[A]_0 / \mathrm{mol\,L^{-1}}$	$t1/2 / min$ $t_{1/2} / \mathrm{min}$
0.100	200
0.025	100

Given:

2 = 0.30

Which of the following is true? A. The order of the reaction is

(1)/(2)

. B. If

[A]₀

is

1M

, then

t1/2

is

200\sqrt(10)

min C. The order of the reaction changes to 1 if the concentration of reactant changes from

0.100 M

to

0.500 M

. D.

t1 / 2

is

800 ~min

for

[A]₀ = 1.6 M

Choose the correct answer from the options given below:

A. $A and C only$ $\text{A and C only}$
B. $A and B only$ $\text{A and B only}$
C. $A, B and D only$ $\text{A, B and D only}$
D. $C and D only$ $\text{C and D only}$

Solution

Related Formula

The dependence of half-life on initial concentration is given by:

t1/2 \propto 1[A]₀ⁿ⁻¹

Step 1: Finding the reaction order (n)

Using the values provided:

(t1/2)₁(t1/2)₂ = ( [A]0,2[A]0,1 )ⁿ⁻¹

(200)/(100) = ( (0.025)/(0.100) )ⁿ⁻¹ \Rightarrow 2 = ( (1)/(4) )ⁿ⁻¹

2 = 2-2(n-1) \Rightarrow 1 = -2n + 2 \Rightarrow n = (1)/(2)

Hence, statement A is correct.

Step 2: Checking half-life at other concentrations

Since $n = (1)/(2)$ $n = \frac{1}{2}$, $t1/2 \propto \sqrt([A]₀)$ $t_{1/2} \propto \sqrt{[A]_0}$.

For $[A]₀ = 1 M$ $[A]_0 = 1\,\mathrm{M}$:

200t1/2 = \sqrt((0.1)/(1)) \Rightarrow t1/2 = 200\sqrt(10) min

Hence, statement B is correct.

For $[A]₀ = 1.6 M$ $[A]_0 = 1.6\,\mathrm{M}$:

200t1/2 = \sqrt((0.1)/(1.6)) = \sqrt((1)/(16)) = (1)/(4) \Rightarrow t1/2 = 800 min

Hence, statement D is correct.

Pattern Recognition

Sees: Half-life reducing as initial concentration decreases. Trap: Assuming all reactions are first or zero order without calculations. Shortcut: Reduction of $[A]₀$ $[A]_0$ by 4 causes reduction of $t1/2$ $t_{1/2}$ by 2 $arrow$ $\rightarrow$ indicates a square root dependence ( $t1/2 \propto \sqrt(A₀)$ $t_{1/2} \propto \sqrt{A_0}$), which implies $n = 0.5$ $n = 0.5$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q jee_main_2025_04_april_evening Arrhenius Equation and Activation Energy

Consider the following plots of log of rate constant k (log k) vs

1T

for three different reactions. The correct order of activation energies of these reactions is

Arrhenius plots of log k vs 1 over T for Q40 - JEE Main 2025 Evening — The graph depicts three linear curves with distinct negative slopes showing the temperature dependence of rate constants.

A. $Ea₂ > Ea₁ > Ea₃$
B. $Ea₁ > Ea₃ > Ea₂$
C. $Ea₁ > Ea₂ > Ea₃$
D. $Ea₃ > Ea₂ > Ea₁$

Solution

Related Formula

k = A - (Eₐ)/(2.303 R T)

Slope of the line = -(Eₐ)/(2.303 R) |Slope| \propto Eₐ

Core Logic

From the given graph, we look at the steepness (magnitude of the negative slope) of lines 1, 2, and 3:

Line 2 is the steepest, meaning it has the largest slope magnitude.
Line 1 has an intermediate slope.
Line 3 is the flattest, indicating the smallest slope magnitude.

Since the activation energy $Eₐ$ $E_a$ is directly proportional to the magnitude of this slope:

|Slope₂| > |Slope₁| > |Slope₃| Eₐ₂ > Eₐ₁ > Eₐ₃

Pattern Recognition

In Arrhenius coordinates, steepness equals barriers. A steeper line means the reaction rate is highly sensitive to temperature because it has a higher activation energy ( $Eₐ$ $E_a$).

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q45 jee_main_2025_04_april_evening Integrated Rate Equations

Half-life of zero order reaction

A arrow

product is 1 hour, when initial concentration of reaction is

2.0 ~mol L⁻¹

. The time required to decrease concentration of A from 0.50 to

0.25 ~mol L⁻¹

is:

A. 0.5 hour
B. 4 hour
C. 15 min
D. 60 min

Solution

Related Formula

t1/2 = ([A]₀)/(2k) (for Zero-Order रिएक्शन)

t = ([A]₀ - [A]ₜ)/(k)

Core Logic

Find the rate constant $k$ $k$ using the given half-life parameters:

1 hour = 60 min = (2.0)/(2k) k = (2.0)/(2 \times 60) = (1)/(60) ~M \cdot min⁻¹

Calculate the time $t$ $t$ to drop from $0.50 ~mol\cdot L⁻¹$ $0.50 \mathrm{~mol\cdot L^{-1}}$ to $0.25 ~mol\cdot L⁻¹$ $0.25 \mathrm{~mol\cdot L^{-1}}$:

t = (0.50 - 0.25)/(k) = (0.25)/(((1)/(60))) = 0.25 \times 60 = 15 minutes

Pattern Recognition

For zero-order systems, the rate of reaction is entirely independent of concentration. This means the time required to consume a specific quantity of reactant scales linearly with the concentration change ( $t = (Δ C)/(k)$ $t = \frac{\Delta C}{k}$).

Chapter Mix

Class 12 Chemistry: Chemical 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?

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding the Matching Curve

Pattern Recognition

Chapter Mix

Reference Study Guides

More Chemical Kinetics Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Finding the reaction order (n)

Step 2: Checking half-life at other concentrations

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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