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

Core Logic

Since r = k[A]$r = k[A]$, the reaction follows first-order kinetics. The half-life (50%$50\%$ decomposition) is t1/2 = 120 minutes$t_{1/2} = 120\text{ minutes}$.

Step 1: Calculating for 90% decomposition

For 90%$90\%$ completion of the reaction, [A]₀ = 100$[A]_0 = 100$ and [A]ₜ = 100 - 90 = 10$[A]_t = 100 - 90 = 10$.

Rounding off to the nearest integer, we get 399 minutes$399\text{ minutes}$.

Pattern Recognition

For a first order reaction, t90% ≈ 3.32 × t50%$t_{90\%} \approx 3.32 \times t_{50\%}$. 120 × 3.32 = 398.4$120 \times 3.32 = 398.4$, so roughly 399$399$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Core Logic

Consider a general gas phase reaction: A arrow B + C$A \rightarrow B + C$

Initial (t=0$t=0$): Pᵢ 0 0$\quad P_i \quad\quad 0 \quad\quad 0$ At time t$t$: Pᵢ - x x x$\quad P_i - x \quad\quad x \quad\quad x$

Total pressure at time t$t$:

x = Pₜ - Pᵢ$x = P_t - P_i$

Partial pressure of A at time t$t$ (PA$P_A$): PA = Pᵢ - x$P_A = P_i - x$

For a first-order reaction:

Here, P₀ = Pᵢ$P_0 = P_i$ and the pressure of the reactant at time t$t$ is PA$P_A$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics