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

Step 1: Calculate Eₐ₁$E_{a1}$ from the Arrhenius equation slope:

Step 2: Calculate Eₐ₂$E_{a2}$:

Pattern Recognition

Sees: Arrhenius log equation comparing slopes. Shortcut: Eₐ₁ = 1.5 × 10⁴ × 2.303 × 8.314 = 287.2 kJ$E_{a1} = 1.5 \times 10^4 \times 2.303 \times 8.314 = 287.2\text{ kJ}$; divide by 5 gives 57.44 ≈ 57 kJ/mol$57.44 \approx 57\text{ kJ/mol}$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

Core Logic

The graph of concentration vs time is a straight line with a negative slope, which is the hallmark of a zero-order reaction. The slope of the line equals -k$-k$.

Step 1: Finding Rate Constant (k)

From the given graph coordinates: At t = 0$t = 0$, [AB]₀ = 0.60$[AB]_0 = 0.60$ M At t = 100$t = 100$ s, [AB]₁₀₀ = 0.55$[AB]_{100} = 0.55$ M

Step 2: Calculating Half-Life

Converting to minutes:

Pattern Recognition

Linear [A]$[A]$ vs t$t$ plot arrow$\rightarrow$ Zero order. Half-life for zero order is proportional to initial concentration. t1/2 = [A]₀ / 2k$t_{1/2} = [A]_0 / 2k$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

Core Logic

For the reaction A arrow nB$A \rightarrow nB$, the stoichiometry determines the relationship between the change in concentration of A$A$ and B$B$.

From the graph, at t = 0 min$t = 0 \text{ min}$: [A] = 0.05 M$[A] = 0.05 \text{ M}$ [B] = 0 M$[B] = 0 \text{ M}$

At t = 10 min$t = 10 \text{ min}$: [A] = 0.04 M$[A] = 0.04 \text{ M}$ Change in [A] = 0.05 - 0.04 = 0.01 M$[A] = 0.05 - 0.04 = 0.01 \text{ M}$

Since 1$1$ mole of A$A$ dissociates to give n$n$ moles of B$B$, the concentration of B$B$ formed will be n × change in [A]$n \times \text{change in } [A]$. Δ[B] = 0.01 × n$\Delta[B] = 0.01 \times n$

Step 1: Calculating n

From the graph, at t = 10 min$t = 10 \text{ min}$, the concentration of B$B$ is 0.03 M$0.03 \text{ M}$.

Equating this to our stoichiometric expression:

n = 3$n = 3$

Pattern Recognition

Visually compare the vertical drops/rises at the first time interval. A drops by 1 unit, B rises by 3 units. The ratio directly gives the stoichiometric coefficient n = 3$n = 3$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

Core Logic

At t = 10$t = 10$ minutes (10:10 am$10:10 \text{ am}$): For reaction II: 5Br⁻ + BrO₃⁻ + 6H⁺ arrow 3Br₂ + 3H₂O$5Br^{-} + BrO_{3}^{-} + 6H^{+} \rightarrow 3Br_{2} + 3H_{2}O$ The overall rate of reaction II is expressed by dividing the rate of disappearance of Br^-$Br^-$ by its stoichiometric coefficient:

Given - Δ[Br⁻]Δ t = 2 × 10⁻⁴ mol L⁻¹ min⁻¹$-\frac{\Delta[Br^{-}]}{\Delta t} = 2 \times 10^{-4} \text{ mol L}^{-1} \text{ min}^{-1}$, we have:

Step 1: Equating Rates

The problem states that the rates of both reactions are identical at this time. Therefore, RateI = RateII = 4 × 10⁻⁵ mol L⁻¹ min⁻¹$\text{Rate}_{I} = \text{Rate}_{II} = 4 \times 10^{-5} \text{ mol L}^{-1} \text{ min}^{-1}$.

For the first-order reaction I (A arrow products$A \rightarrow \text{products}$):

Step 2: Calculating Rate Constant

Substitute the known values at t = 10 min$t = 10 \text{ min}$:

Pattern Recognition

Remember to divide the given rate of disappearance of a species by its stoichiometric coefficient to find the true, normalized "Rate of Reaction" before equating it to another reaction's rate.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Core Logic

Given the overall reaction A arrow D$\text{A} \rightarrow \text{D}$ is endothermic:

Analyzing the mechanism steps: Step (i): A arrow B$\text{A} \rightarrow \text{B}$, Δᵣ H = +ve EB > EA$\Delta_r H = +\text{ve} \implies E_B > E_A$ Step (ii): B arrow C$\text{B} \rightarrow \text{C}$, Δᵣ H = -ve EC < EB$\Delta_r H = -\text{ve} \implies E_C < E_B$ Step (iii): C arrow D$\text{C} \rightarrow \text{D}$, Δᵣ H = -ve ED < EC$\Delta_r H = -\text{ve} \implies E_D < E_C$

Putting it all together, we must have a profile where the final energy ED$E_D$ is higher than the initial energy EA$E_A$, with intermediate dips corresponding to B$B$ and C$C$ as guided by the individual step enthalpies.

Step 1: Final Conclusion

Graph 3 matches all criteria: Overall endothermic (D > A$D > A$), first step endothermic (B > A$B > A$), followed by two exothermic steps (C < B$C < B$ and D < C$D < C$).

Pattern Recognition

Endothermic overall means the tail end (products) sits higher than the origin (reactants). Exothermic intermediate steps mean the curve drops from the previous intermediate well.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics Class 11 Chemistry: Thermodynamics