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

Q71 jee_main_2026_21_jan_morning Arrhenius Equation

Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by

20 kJ mol⁻¹

. If

k₁

and

k₂

are the rate constants of first and second reaction respectively at 300 K, then

ln k₂k₁

will be ..... (nearest integer)

[R=8.3 J K⁻¹ mol⁻¹]

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

ln k = ln A - (Eₐ)/(RT)

Core Logic

For Reaction 1: $ln k₁ = ln A - (E₁)/(RT)$ $\ln k_1 = \ln A - \frac{E_1}{RT}$ For Reaction 2: $ln k₂ = ln A - (E₂)/(RT)$ $\ln k_2 = \ln A - \frac{E_2}{RT}$ (Pre-exponential factor $A$ $A$ is the same).

Subtracting the first from the second:

ln k₂ - ln k₁ = -(E₂)/(RT) - (-(E₁)/(RT))

ln ((k₂)/(k₁)) = (E₁ - E₂)/(RT)

Given that $E₁$ $E_1$ exceeds $E₂$ $E_2$ by $20 kJ mol⁻¹$ $20\text{ kJ mol}^{-1}$, $E₁ - E₂ = 20000 J mol⁻¹$ $E_1 - E_2 = 20000\text{ J mol}^{-1}$. $T = 300 K$ $T = 300\text{ K}$, $R = 8.3 J K⁻¹ mol⁻¹$ .

ln ((k₂)/(k₁)) = (20000)/(8.3 \times 300) = (200)/(8.3 \times 3) = (200)/(24.9)

ln ((k₂)/(k₁)) = 8.032

Rounding off to nearest integer gives 8.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q58 jee_main_2026_21_jan_evening First Order Reactions and Rate Constant

Decomposition of A is a first order reaction at T(K) and is given by

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

. In a closed 1 L vessel, 1 bar

A(g)

is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant (in

min⁻¹

) of the reaction?

( 2 = 0.3)

A. $(1) 6.9 \times 10⁻¹$ $(1) \ 6.9 \times 10^{-1}$
B. $(2) 6.9 \times 10⁻³$ $(2) \ 6.9 \times 10^{-3}$
C. $(3) 6.9 \times 10⁻²$ $(3) \ 6.9 \times 10^{-2}$
D. $(4) 6.9 \times 10⁻⁴$ $(4) \ 6.9 \times 10^{-4}$

Solution

Related Formula

$k = (2.303)/(t) ( P₀2P₀ - Ptotal)$ or equivalent first-order expression.

Core Logic

For $A(g) arrow B(g) + C(g)$ :

Initial pressure: $P₀ = 1 bar$ $P_0 = 1 \text{ bar}$
At time $t = 100 min$ $t = 100 \text{ min}$, pressure of A remaining $= 1 - P$ $= 1 - P$, pressures of B and C $= P$ $= P$.
Total pressure $Ptotal = 1 - P + P + P = 1 + P = 1.5 bar P = 0.5 bar$ .

Remaining pressure of A $= 1 - 0.5 = 0.5 bar$ $= 1 - 0.5 = 0.5 \text{ bar}$.

Step 1: Calculating Rate Constant

k = (1)/(100) ln((1)/(0.5)) = (0.693)/(100) = 6.9 \times 10⁻³ min⁻¹

Pattern Recognition

Sees: gaseous phase first-order kinetics with total pressure data. Trap: Confusing partial pressure of reactant with total pressure in rate expressions.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q62 jee_main_2026_22_january_morning First Order Reaction Rates

A

arrow

products (First order reaction). Three sets of experiment were performed for a reaction under similar experimental conditions. Run 1

\Rightarrow

100 mL of 10 M solution of reactant A Run 2

\Rightarrow

200 mL of 10 M solution of reactant A Run 3

\Rightarrow

100 mL of 10 M solution of reactant A + 100 mL of

H₂O

added. The correct variation of rate of reaction is

A. $Run 1 = Run 2 = Run 3$ $\text{Run 1 = Run 2 = Run 3}$
B. $Run 3 < Run 1 = Run 2$ $\text{Run 3 < Run 1 = Run 2}$
C. $Run 3 < Run 1 < Run 2$ $\text{Run 3 < Run 1 < Run 2}$
D. $Run 1 < Run 2 < Run 3$ $\text{Run 1 < Run 2 < Run 3}$

Solution

Related Formula

Rate = k[A]

Where $[A]$ $[A]$ is the molar concentration of reactant A.

Core Logic

For a first order reaction, the rate is directly proportional to the concentration of the reactant, not the total volume or the total number of moles.

In Run 1: $[A] = 10 M$ $[A] = 10\text{ M}$. $Rate₁ = k(10)$ $\text{Rate}_1 = k(10)$. In Run 2: $[A] = 10 M$ $[A] = 10\text{ M}$. $Rate₂ = k(10)$ $\text{Rate}_2 = k(10)$. (Volume increased, but molarity is identical). In Run 3: 100 mL of 10 M solution is diluted with 100 mL water. New volume is 200 mL. $M₁ V₁ = M₂ V₂ 10 \times 100 = M₂ \times 200 M₂ = 5 M$ . $[A] = 5 M$ $[A] = 5\text{ M}$. $Rate₃ = k(5)$ $\text{Rate}_3 = k(5)$.

Step 1: Final Conclusion

Therefore, $Rate₃ < Rate₁ = Rate₂$ .

Pattern Recognition

Rate laws depend exclusively on molar concentration (M). Diluting the solution decreases rate, while just taking a larger volume of the same stock solution keeps the rate identical.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q73 jee_main_2026_22_january_morning Arrhenius Equation

The temperature at which the rate constants of the given below two gaseous reactions become equal is ____ K. (Nearest integer).

X arrow Y k₁ = 10⁶e(-30000)/(T)

P arrow Q k₂ = 10⁴e(-24000)/(T)

Given:

ln 10 = 2.303

Numerical Answer. Answer: 1303 to 1303

Solution

Related Formula

k = A e-Eₐ/RT

Core Logic

Equate the two rate constants: $k₁ = k₂$ $k_{1} = k_{2}$

10⁶e(-30000)/(T) = 10⁴e(-24000)/(T)

Divide both sides by $10⁴$ $10^4$:

10²e(-30000)/(T) = e(-24000)/(T)

Divide both sides by $e(-30000)/(T)$ $e^{\frac{-30000}{T}}$:

100 = e(-24000)/(T)e(-30000)/(T)

100 = e(6000)/(T)

Step 1: Solve for T

Take the natural logarithm ( $ln$ $\ln$) on both sides:

ln(100) = (6000)/(T)

2 ln(10) = (6000)/(T)

Substitute $ln 10 = 2.303$ $\ln 10 = 2.303$:

2 \times 2.303 = (6000)/(T)

4.606 = (6000)/(T)

T = (6000)/(4.606) = 1302.64 K

Step 2: Rounding

Nearest integer is 1303.

Pattern Recognition

Simple exponential equating. Group powers of 10 on one side and exponentials on the other, then apply natural log.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q57 jee_main_2026_22_january_evening Arrhenius Equation and Temperature Dependence

Correct statements regarding Arrhenius equation among the following are: (A) Factor

e-Eₐ/RT

corresponds to fraction of molecules having kinetic energy less than

Eₐ

. (B) At a given temperature, lower the

Eₐ

, faster is the reaction. (C) Increase in temperature by about

10^

doubles the rate of reaction. (D) Plot of

k

vs

(1)/(T)

gives a straight line with

slope = -(Eₐ)/(R)

. Choose the correct answer from the options given below:

A. B and D only
B. A and B only
C. A and C only
D. B and C only

Solution

Related Formula

k = A e-Eₐ/RT

ln k = ln A - (Eₐ)/(RT) k = A - (Eₐ)/(2.303 R T)

Core Logic

Statement (A): BANNED - $e-Eₐ/RT$ $e^{-E_a/RT}$ represents fraction of molecules with energy $\geq Eₐ$ $\ge E_a$ (not less).

Statement (B): CORRECT - Lower activation energy $Eₐ$ $E_a$ increases the rate constant $k$ $k$, speeding up the reaction.

Statement (C): CORRECT - For most reactions, a $10^$ $10^\circ\text{C}$ rise in temperature doubles the rate coefficient.

Statement (D): INCORRECT - Plot of $k$ $\log k$ vs $1/T$ $1/T$ has slope equal to $-(Eₐ)/(2.303 R)$ $-\frac{E_a}{2.303 R}$ (the factor $2.303$ $2.303$ is missing).

Pattern Recognition

Sees: Arrhenius statements. Shortcut: Watch for missing $2.303$ $2.303$ in $\log$ slope equation and 'less than' vs 'greater than' in exponential fraction definition.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More Chemical Kinetics Previous-Year Questions

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Rate Constant

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for T

Step 2: Rounding

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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