For a given reaction Rarrow P,t1 / 2$\mathrm{R}\rightarrow \mathrm{P},\mathrm{t}_{1 / 2}$ is related to [A]₀$[\mathrm{A}]_0$ 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$\log 2 = 0.30$ Which of the following is true? A. The order of the reaction is (1)/(2)$\frac{1}{2}$ . B. If [A]₀$[\mathrm{A}]_0$ is 1M$1\mathrm{M}$ , then t1/2$\mathrm{t}_{1/2}$ is 200√(10)$200\sqrt{10}$ min C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M$0.100 \mathrm{M}$ to 0.500 M$0.500 \mathrm{M}$ . D. t1 / 2$t_{1 / 2}$ is 800 ~min$800 \mathrm{~min}$ for [A]₀ = 1.6 M$[\mathrm{A}]_0 = 1.6 \mathrm{M}$ Choose the correct answer from the options given below:

Related Formula

For a first-order reaction:

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

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:

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

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

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)$\text{Initial} \xrightarrow{t_{50\%}} \frac{1}{2} \xrightarrow{t_{50\%}} \frac{1}{4} \quad \text{(Total } 2 \text{ half-lives)}$ (1)/(4) t50% (1)/(8) (Total 3 half-lives)$\frac{1}{4} \xrightarrow{t_{50\%}} \frac{1}{8} \quad \text{(Total } 3 \text{ half-lives)}$ Therefore, the ratio is simply the ratio of the number of half-lives: 2 : 3$2 : 3$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

Core Logic

Let's track concentration reductions:

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

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

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

Related Formula

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:

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

Related Formula

Core Logic

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

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

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