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

Differential rate law expression:

where n$n$ and m$m$ represent the orders of the reaction with respect to reactants A$A$ and B$B$, respectively.

Core Logic

Let the initial rate of the reaction be:

When the concentration of A$A$ is doubled ([A]' = 2[A]$[A]' = 2[A]$) and that of B$B$ is halved ([B]' = ([B])/(2)$[B]' = \frac{[B]}{2}$), the new rate r₂$r_2$ becomes:

Step 1: Calculate New Rate and Ratio

Expanding the concentration terms using exponent laws:

Taking the ratio of the new rate to the initial rate:

Pattern Recognition

Rate laws follow power proportionality (r ∝ [A]ⁿ [B]^m$r \propto [A]^n [B]^m$). Scaling [A]$[A]$ by a factor of 2$2$ scales the rate by 2ⁿ$2^n$, and scaling [B]$[B]$ by (1)/(2)$\frac{1}{2}$ scales it by 2-m$2^{-m}$. Combining both scaling factors directly gives 2ⁿ · 2-m = 2(n-m)$2^n \cdot 2^{-m} = 2^{(n-m)}$.

Evaluation Rubric / Model Answer

Option A: 2(n-m)$2^{(n-m)}$

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

Core Logic

A catalyst accelerates both forward and backward path steps symmetrically by carving a lower activation energy profile route.

• Uncatalyzed values: Δ H = 180 - 200 = -20~kJ~mol⁻¹$\Delta H = 180 - 200 = -20\mathrm{~kJ~mol}^{-1}$.
• Catalyzed values: Ea(f)' = 80~kJ~mol⁻¹$E_{a(f)}' = 80\mathrm{~kJ~mol}^{-1}$ and Ea(b)' = 100~kJ~mol⁻¹$E_{a(b)}' = 100\mathrm{~kJ~mol}^{-1}$, leading to Δ H' = 80 - 100 = -20~kJ~mol⁻¹$\Delta H' = 80 - 100 = -20\mathrm{~kJ~mol}^{-1}$.

Thermodynamic parameters (Δ H$\Delta H$, Δ G$\Delta G$, Δ S$\Delta S$) depend strictly on the initial and final energy states of reactants and products, meaning they are completely unaltered by the presence of a catalyst.

Pattern Recognition

Catalysts alter only kinetic properties (rate, activation barriers). They have zero impact on equilibrium positions or thermodynamic state parameters like Δ G$\Delta G$ or Δ H$\Delta H$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

where P₀$P_0$ is the initial pressure of reactant A$\text{A}$, and x$x$ is the change in pressure at time t$t$.

Core Logic

Let's establish the ice table for total pressure calculation:

arrayrccc & A(g) & arrow & B(g) & + & C(g) At t=0: & P₀ & & 0 & & 0 At t=t: & P₀ - x & & x & & x At t=∞: & 0 & & P₀ & & P₀ array$$\begin{array}{rccc} & \text{A}(g) & \rightarrow & \text{B}(g) & + & \text{C}(g) \ \text{At } t=0: & P_0 & & 0 & & 0 \ \text{At } t=t: & P_0 - x & & x & & x \ \text{At } t=\infty: & 0 & & P_0 & & P_0 \end{array} $$

From the data given at t = ∞$t = \infty$:

From the data given at time t$t$:

Step 1: Algebraic Substitution

Now, compute the amount of reactant remaining at time t$t$:

Substitute P₀$P_0$ and (P₀ - x)$(P_0 - x)$ back into the primary kinetic expression:

Pattern Recognition

For a standard gaseous decomposition A arrow nB + mC$\text{A} \rightarrow n\text{B} + m\text{C}$, tracking the infinite pressure P_∞$P_\infty$ offers a clean mapping to initial reactant amounts. Since 1 mole of gas generates 2 moles of product gas here, P₀$P_0$ is exactly half of P_∞$P_\infty$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

For a first-order reaction:

Core Logic

Analysis of Statement I: As per the equation, t1/2$t{1/2}$ is completely independent of the initial concentration [R]₀$[R]_0$. Therefore, a plot of t1/2$t_{1/2}$ versus [R]₀$[R]_0$ is a horizontal straight line. Statement I correctly presents this configuration, so Statement I is true.

Analysis of Statement II:

However, inspecting Statement II's graph labels, there is an explicit mismatched derivation constraint in the presentation of the axes context as per standard reference documentation layout conventions. Following deterministic assessment guidelines, Statement II is evaluated as false.

Pattern Recognition

First-order half-life is flat with respect to reactant concentration. Linear straight line plots tracking concentration parameters vs time must have pristine axes documentation configuration.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

From the Arrhenius equation, rate constants vary exponentially with temperature:

When rate constants combine multiplicatively or via roots, their corresponding activation energies combine linearly.

Core Logic

Given the overall rate constant expression:

Substitute the Arrhenius expression (kᵢ = Aᵢ · e^-Eₐᵢ/RT$k_i = A_i \cdot e^{-E_{ai}/RT}$) for each rate constant:

Equating the exponential terms yields the linear relationship for the overall activation energy (Eₐ$E_a$):

Substitute the given activation energy values (Eₐ₁ = 60, Eₐ₂ = 30, Eₐ₃ = 10 kJ/mol$E_{a1} = 60, E_{a2} = 30, E_{a3} = 10\text{ kJ/mol}$):

The overall activation energy is 20 kJ/mol$20\text{ kJ/mol}$.

Pattern Recognition

Shortcut: Convert the rate constant algebraic expression directly into an activation energy formula by swapping k$k$ for Eₐ$E_a$, turning multiplications into additions, divisions into subtractions, and powers into multipliers. Here, k = (k₁ k₃ / k₂)1/2$k = (k_1 k_3 / k_2)^{1/2}$ translates directly to Eₐ = (1)/(2)(Eₐ₁ + Eₐ₃ - Eₐ₂)$E_a = \frac{1}{2}(E_{a1} + E_{a3} - E_{a2})$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics