Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by 20 kJ mol⁻¹$20\text{ kJ mol}^{-1}$. If k₁$k_{1}$ and k₂$k_{2}$ are the rate constants of first and second reaction respectively at 300 K, then ln k₂k₁$\ln \frac{k_{2}}{k_{1}}$ will be ..... (nearest integer) [R=8.3 J K⁻¹ mol⁻¹]$[R=8.3\text{ J K}^{-1}\text{ mol}^{-1}]$

Related Formula

k = (2.303)/(t) ( P₀2P₀ - Ptotal)$$k = \frac{2.303}{t} \log\left(\frac{P_0}{2P_0 - P_{\text{total}}}\right)$$ or equivalent first-order expression.

Core Logic

For A(g) arrow B(g) + C(g)$\text{A}(\text{g}) \rightarrow \text{B}(\text{g}) + \text{C}(\text{g})$:

• Initial pressure: P₀ = 1 bar$P_0 = 1 \text{ bar}$
• At time t = 100 min$t = 100 \text{ min}$, pressure of A remaining = 1 - P$= 1 - P$, pressures of B and C = P$= P$.
• Total pressure Ptotal = 1 - P + P + P = 1 + P = 1.5 bar P = 0.5 bar$P_{\text{total}} = 1 - P + P + P = 1 + P = 1.5 \text{ bar} \implies P = 0.5 \text{ bar}$.

Remaining pressure of A = 1 - 0.5 = 0.5 bar$= 1 - 0.5 = 0.5 \text{ bar}$.

Step 1: Calculating Rate Constant

Pattern Recognition

Sees: gaseous phase first-order kinetics with total pressure data. Trap: Confusing partial pressure of reactant with total pressure in rate expressions.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

Where [A]$[A]$ is the molar concentration of reactant A.

Core Logic

For a first order reaction, the rate is directly proportional to the concentration of the reactant, not the total volume or the total number of moles.

In Run 1: [A] = 10 M$[A] = 10\text{ M}$. Rate₁ = k(10)$\text{Rate}_1 = k(10)$. In Run 2: [A] = 10 M$[A] = 10\text{ M}$. Rate₂ = k(10)$\text{Rate}_2 = k(10)$. (Volume increased, but molarity is identical). In Run 3: 100 mL of 10 M solution is diluted with 100 mL water. New volume is 200 mL. M₁ V₁ = M₂ V₂ 10 × 100 = M₂ × 200 M₂ = 5 M$M_1 V_1 = M_2 V_2 \implies 10 \times 100 = M_2 \times 200 \implies M_2 = 5\text{ M}$. [A] = 5 M$[A] = 5\text{ M}$. Rate₃ = k(5)$\text{Rate}_3 = k(5)$.

Step 1: Final Conclusion

Therefore, Rate₃ < Rate₁ = Rate₂$\text{Rate}_3 < \text{Rate}_1 = \text{Rate}_2$.

Pattern Recognition

Rate laws depend exclusively on molar concentration (M). Diluting the solution decreases rate, while just taking a larger volume of the same stock solution keeps the rate identical.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

Core Logic

Equate the two rate constants: k₁ = k₂$k_{1} = k_{2}$

Divide both sides by 10⁴$10^4$:

Divide both sides by e(-30000)/(T)$e^{\frac{-30000}{T}}$:

Step 1: Solve for T

Take the natural logarithm (ln$\ln$) on both sides:

Substitute ln 10 = 2.303$\ln 10 = 2.303$:

Step 2: Rounding

Nearest integer is 1303.

Pattern Recognition

Simple exponential equating. Group powers of 10 on one side and exponentials on the other, then apply natural log.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

Core Logic

Statement (A): BANNED - e-Eₐ/RT$e^{-E_a/RT}$ represents fraction of molecules with energy ≥ Eₐ$\ge E_a$ (not less).

Statement (B): CORRECT - Lower activation energy Eₐ$E_a$ increases the rate constant k$k$, speeding up the reaction.

Statement (C): CORRECT - For most reactions, a 10^$10^\circ\text{C}$ rise in temperature doubles the rate coefficient.

Statement (D): INCORRECT - Plot of k$\log k$ vs 1/T$1/T$ has slope equal to -(Eₐ)/(2.303 R)$-\frac{E_a}{2.303 R}$ (the factor 2.303$2.303$ is missing).

Pattern Recognition

Sees: Arrhenius statements. Shortcut: Watch for missing 2.303$2.303$ in $\log$ slope equation and 'less than' vs 'greater than' in exponential fraction definition.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics