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

Q66 jee_main_2026_24_january_morning Arrhenius Equation and Catalyst Effect

At

27°C

in presence of a catalyst, activation energy of a reaction is lowered by

10 kJ mol⁻¹

. The logarithm ratio of

k(catalysed)k(uncatalysed)

is (Consider that the frequency factor for both the reactions is same)

A. $17.41$ $17.41$
B. $1.741$ $1.741$
C. $3.482$ $3.482$
D. $0.1741$ $0.1741$

Solution

Related Formula

k = A e-(Eₐ)/(RT)

Core Logic

Let the activation energy of the uncatalyzed reaction be $Eₐ$ $E_a$. Activation energy of the catalyzed reaction $= Eₐ - 10 kJ mol⁻¹$ $= E_a - 10 \text{ kJ mol}^{-1}$. $kuncatalyzed = A e-(Eₐ)/(RT)$ $kcatalyzed = A e-((Eₐ - 10000))/(RT)$

Dividing the two rate constants: $kcatalyzedkuncatalyzed = e(Δ Eₐ)/(RT)$ where $Δ Eₐ = 10 kJ mol⁻¹ = 10000 J mol⁻¹$ .

Step 1: Calculate Logarithm Ratio

Taking natural logarithm on both sides:

ln ( kcatalyzedkuncatalyzed) = (Δ Eₐ)/(RT)

Converting to base 10 logarithm:

₁₀ ( kcatalyzedkuncatalyzed) = (Δ Eₐ)/(2.303 RT)

Substitute the values ( $R = 8.314 J K⁻¹ mol⁻¹$ , $T = 300 K$ $T = 300 \text{ K}$):

₁₀ ( kcatalyzedkuncatalyzed) = (10000)/(2.303 \times 8.314 \times 300)

₁₀ ( kcatalyzedkuncatalyzed) = (10000)/(5744.14) \approx 1.741

Pattern Recognition

Remember that lowering activation energy by $Δ Eₐ$ $\Delta E_a$ increases the rate by a factor of $eΔ Eₐ / RT$ $e^{\Delta E_a / RT}$. Taking log base 10 simply scales this exponent by $1/2.303$ $1/2.303$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q72 jee_main_2026_24_january_evening First Order Reactions

The half-life of

⁶⁵

Zn is 245 days. After x days, 75% of original activity remained. The value of x in days is ____. (Nearest integer) (Given : log 3 = 0.4771 and log 2 = 0.3010)

Numerical Answer. Answer: 102 to 102

Solution

Related Formula

t = (2.303)/(K) ((a₀)/(aₜ))

where $K = ln 2t1/2$ $K = \frac{\ln 2}{t_{1/2}}$

Core Logic

Given $t1/2 = 245 days$ $t_{1/2} = 245 \text{ days}$, so decay constant $K = (ln 2)/(245)$ $K = \frac{\ln 2}{245}$. If 75% activity remained, the fraction remaining is $aₜ / a₀ = 3/4$ $a_t / a_0 = 3/4$ (or $75/100$ $75/100$). So, $(a₀)/(aₜ) = (4)/(3)$ $\frac{a_0}{a_t} = \frac{4}{3}$.

Step 1: Substitute Values

t = (1)/(K) ln((4)/(3))

t = (245)/(ln 2) ln((4)/(3))

Converting to base 10 logs:

t = 245 \times ( (4/3))/( 2)

t = 245 [ (2 2 - 3)/( 2) ]

Step 2: Arithmetic Evaluation

t = 245 [ (2(0.3010) - 0.4771)/(0.3010) ]

t = 245 [ (0.6020 - 0.4771)/(0.3010) ]

t = 245 [ (0.1249)/(0.3010) ] = 245 \times 0.4149

t 101.66 days

Rounded to nearest integer = 102.

Pattern Recognition

All radioactive decays follow first-order kinetics. Don't waste time deriving formulas—plug $aₜ = 0.75 a₀$ $a_t = 0.75 a_0$ straight into the standard integrated rate law.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics Class 12 Physics: Nuclei

Q57 jee_main_2026_28_january_morning First Order Reaction Decomposition

An organic compound undergoes first order decomposition. The time taken for decomposition to

((1)/(8))th

and

((1)/(10))th

of its initial concentration are

t1/8

and

t1/10

respectively. What is the value of

t1/8t1/10\times 10

? (

2 = 0.3

)

A. $9$ $9$
B. $0.9$ $0.9$
C. $3$ $3$
D. $30$ $30$

Solution

Related Formula

t = (1)/(k)ln((A₀)/(Aₜ))

Step 1: Formulate Time Ratios

For $Aₜ = A₀/8$ $A_t = A_0/8$:

t1/8 = (1)/(k)ln((A₀)/(A₀/8)) = (1)/(k)ln(8)

For $Aₜ = A₀/10$ $A_t = A_0/10$:

t1/10 = (1)/(k)ln((A₀)/(A₀/10)) = (1)/(k)ln(10)

Step 2: Find the Ratio

t1/8t1/10 = (ln 8)/(ln 10) = ( 8)/( 10)

\n

= (3 2)/(1) = 3(0.3) = 0.9

\nTherefore:\n

t1/8t1/10 \times 10 = 0.9 \times 10 = 9

Pattern Recognition

For first order kinetics, ratio of completion times is independent of $k$ $k$ and strictly equals the ratio of logs of initial-to-final concentration fractions.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q72 jee_main_2026_28_january_evening Arrhenius Equation And Activation Energy

A arrow B

(first reaction)

C arrow D

(second reaction) Consider the above two first-order reactions. The rate constant for first reaction at

500 K

is double of the same at

300 K

. At

500 K

,

50%

of the reaction becomes complete in

2 hour

. The activation energy of the second reaction is half of that of first reaction. If the rate constant at

500 K

of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at

300 K

is ____

\times 10⁻¹ hour⁻¹

(nearest integer).

Numerical Answer. Answer: 5 to 5

Solution

Related Formula

ln( KT₂KT₁) = (Eₐ)/(R) [ (1)/(T₁) - (1)/(T₂) ]

K = ln 2t1/2 (First Order)

Core Logic

For Reaction 1 ( $A K₁ B$ $A \xrightarrow{K_1} B$): $(K₁)500K = 2(K₁)300K$ . $ln(2) = Eₐ₁R [ (1)/(300) - (1)/(500) ] = Eₐ₁R [ (200)/(150000) ] = Eₐ₁R [ (2)/(1500) ]$ $Eₐ₁ = (ln 2 \times R \times 1500)/(2)$

Also for Reaction 1 at 500K, half-life is 2 hours, so: $(K₁)500K = (ln 2)/(2) hr⁻¹$

For Reaction 2 ( $C K₂ D$ $C \xrightarrow{K_2} D$): Given $Eₐ₂ = Eₐ₁2 = (ln 2 \times R \times 1500)/(4)$ Given $(K₂)500K = 2 \times (K₁)500K = 2 \times (ln 2)/(2) = ln 2$

Applying Arrhenius equation for Reaction 2 from 300K to 500K: $ln [ (K₂)500K(K₂)300K ] = Eₐ₂R [ (1)/(300) - (1)/(500) ]$ $ln [ ln 2(K₂)300K ] = (((ln 2 \times R \times 1500)/(4)))/(R) \times (2)/(1500) = (ln 2)/(2) = ln(\sqrt(2))$

Step 1: Solve for K2 at 300K

Equating the arguments of the natural logs: $ln 2(K₂)300K = \sqrt(2)$ $(K₂)300K = ln 2\sqrt(2) = (0.693)/(1.414) = 0.49 hr⁻¹$ $(K₂)300K = 4.9 \times 10⁻¹ hr⁻¹ \approx 5 \times 10⁻¹ hr⁻¹$ (rounded to nearest integer).

Pattern Recognition

Sequential tracking of unknowns through Arrhenius. $Eₐ$ $E_a$ ratios dictate log ratios between temperatures directly.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q39 jee_main_2025_02_april_evening Reaction Mechanism and Energy Profiles

Reactant A converts to product D through the given mechanism (with the net evolution of heat):

A arrow B

slow;

Δ H = +ve

B arrow C

fast;

Δ H = -ve

C arrow D

fast;

Δ H = -ve

Which of the following represents the above reaction mechanism?

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

Related Formula

k = A e^-Eₐ/RT

Rate \propto 1Eₐ

Core Logic

Let us break down each step of the mechanism:

Step 1: $A arrow B$ $\mathrm{A} \rightarrow \mathrm{B}$ is slow.
Being the rate-determining step, it must have the highest activation energy barrier ( $Eₐ₁$ $E_{\mathrm{a1}}$).
Since $Δ H = +ve$ $\Delta H = +\mathrm{ve}$ (endothermic), the energy level of intermediate state B must be higher than the reactant state A.
Step 2: $B arrow C$ $\mathrm{B} \rightarrow \mathrm{C}$ is fast.
It has a much lower activation energy barrier ( $Eₐ₂$ $E_{\mathrm{a2}}$).
Since $Δ H = -ve$ $\Delta H = -\mathrm{ve}$ (exothermic), the energy level of intermediate C is lower than state B.
Step 3: $C arrow D$ $\mathrm{C} \rightarrow \mathrm{D}$ is fast.
It has a low activation energy barrier ( $Eₐ₃$ $E_{\mathrm{a3}}$).
Since $Δ H = -ve$ $\Delta H = -\mathrm{ve}$ (exothermic), the energy level of final state D is lower than state C.
Net Reaction: Exothermic with "net evolution of heat".
The potential energy of the final product D is lower than the initial potential energy of reactant A.

Step 1: Check Potential Energy Profile

Evaluating the transition states and relative energy levels in Graph (1):

The first peak (transition state 1) is clearly the highest ( $Eₐ₁ > Eₐ₂, Eₐ₃$ ) $\implies$ Step 1 is the slowest.
The intermediate B is higher in energy than A.
Intermediates C and product D are progressively lower in energy.
Product D has lower energy than reactant A (net exothermic).

Annotated reaction mechanism coordinate graph showing relative activation energies

This perfectly corresponds to Graph (1).

Pattern Recognition

Kinetics shortcut: Slow step = tallest peak. Exothermic step = drop in energy levels of products/intermediates. Endothermic step = climb in energy levels. Use these rules to visually scan energy profiles in under 5 seconds.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More Chemical Kinetics Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Calculate Logarithm Ratio

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Substitute Values

Step 2: Arithmetic Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Formulate Time Ratios

Step 2: Find the Ratio

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for K2 at 300K

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Check Potential Energy Profile

Pattern Recognition

Chapter Mix

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