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)$$X \rightarrow Y \quad k_{1} = 10^{6}e^{\frac{-30000}{T}}$$ P arrow Q k₂ = 10⁴e(-24000)/(T)$$P \rightarrow Q \quad k_{2} = 10^{4}e^{\frac{-24000}{T}}$$ Given: ln 10 = 2.303$\ln 10 = 2.303$

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

The dependence of half-life on initial concentration is given by:

Step 1: Finding the reaction order (n)

Using the values provided:

Hence, statement A is correct.

Step 2: Checking half-life at other concentrations

Since n = (1)/(2)$n = \frac{1}{2}$, t1/2 ∝ √([A]₀)$t_{1/2} \propto \sqrt{[A]_0}$.

• For [A]₀ = 1 M$[A]_0 = 1\,\mathrm{M}$:

Hence, statement B is correct.

• For [A]₀ = 1.6 M$[A]_0 = 1.6\,\mathrm{M}$:

Hence, statement D is correct.

Pattern Recognition

Sees: Half-life reducing as initial concentration decreases. Trap: Assuming all reactions are first or zero order without calculations. Shortcut: Reduction of [A]₀$[A]_0$ by 4 causes reduction of t1/2$t_{1/2}$ by 2 arrow$\rightarrow$ indicates a square root dependence (t1/2 ∝ √(A₀)$t_{1/2} \propto \sqrt{A_0}$), which implies n = 0.5$n = 0.5$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

Core Logic

Equating concentrations at time t$t$:

Given [A]₀ = 8[B]₀$[\mathrm{A}]_0 = 8[\mathrm{B}]_0$:

Step 1: Calculation

Substitute kA = (ln 2)/(10)$k_{\mathrm{A}} = \frac{\ln 2}{10}$ and kB = (ln 2)/(40)$k_{\mathrm{B}} = \frac{\ln 2}{40}$:

Pattern Recognition

Equal concentration point in first-order kinetics: Ratio of initial concentrations = 2^t/t1/2, A - t/t1/2, B$= 2^{t/t_{1/2, A} - t/t_{1/2, B}}$. 8 = 2³ (t)/(10) - (t)/(40) = 3 t = 40 min$8 = 2^3 \implies \frac{t}{10} - \frac{t}{40} = 3 \implies t = 40\text{ min}$.

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