In a reaction A + B arrow C$\mathrm{A} + \mathrm{B} \rightarrow \mathrm{C}$, initial concentrations of A$\mathrm{A}$ and B$\mathrm{B}$ are related as [A]₀ = 8[B]₀$[\mathrm{A}]_0 = 8[\mathrm{B}]_0$. The half lives of A$\mathrm{A}$ and B$\mathrm{B}$ are 10~min$10\mathrm{~min}$ and 40~min$40\mathrm{~min}$ respectively. If they start to disappear at the same time, both following first order kinetics, after how much time will the concentration of both the reactants be same?

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

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