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

Core Logic

Let the activation energy of the uncatalyzed reaction be Eₐ$E_a$. Activation energy of the catalyzed reaction = Eₐ - 10 kJ mol⁻¹$= E_a - 10 \text{ kJ mol}^{-1}$. kuncatalyzed = A e-(Eₐ)/(RT)$k_{\text{uncatalyzed}} = A e^{-\frac{E_a}{RT}}$ kcatalyzed = A e-((Eₐ - 10000))/(RT)$k_{\text{catalyzed}} = A e^{-\frac{(E_a - 10000)}{RT}}$

Dividing the two rate constants: kcatalyzedkuncatalyzed = e(Δ Eₐ)/(RT)$$\frac{k_{\text{catalyzed}}}{k_{\text{uncatalyzed}}} = e^{\frac{\Delta E_a}{RT}}$$ where Δ Eₐ = 10 kJ mol⁻¹ = 10000 J mol⁻¹$\Delta E_a = 10 \text{ kJ mol}^{-1} = 10000 \text{ J mol}^{-1}$.

Step 1: Calculate Logarithm Ratio

Taking natural logarithm on both sides:

Converting to base 10 logarithm:

Substitute the values (R = 8.314 J K⁻¹ mol⁻¹$R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$, T = 300 K$T = 300 \text{ K}$):

Pattern Recognition

Remember that lowering activation energy by Δ Eₐ$\Delta E_a$ increases the rate by a factor of eΔ Eₐ / RT$e^{\Delta E_a / RT}$. Taking log base 10 simply scales this exponent by 1/2.303$1/2.303$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

where K = ln 2t1/2$K = \frac{\ln 2}{t_{1/2}}$

Core Logic

Given t1/2 = 245 days$t_{1/2} = 245 \text{ days}$, so decay constant K = (ln 2)/(245)$K = \frac{\ln 2}{245}$. If 75% activity remained, the fraction remaining is aₜ / a₀ = 3/4$a_t / a_0 = 3/4$ (or 75/100$75/100$). So, (a₀)/(aₜ) = (4)/(3)$\frac{a_0}{a_t} = \frac{4}{3}$.

Step 1: Substitute Values

Converting to base 10 logs:

Step 2: Arithmetic Evaluation

Rounded to nearest integer = 102.

Pattern Recognition

All radioactive decays follow first-order kinetics. Don't waste time deriving formulas—plug aₜ = 0.75 a₀$a_t = 0.75 a_0$ straight into the standard integrated rate law.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics Class 12 Physics: Nuclei

Related Formula

Step 1: Formulate Time Ratios

For Aₜ = A₀/8$A_t = A_0/8$:

For Aₜ = A₀/10$A_t = A_0/10$:

Step 2: Find the Ratio

Therefore:

Pattern Recognition

For first order kinetics, ratio of completion times is independent of k$k$ and strictly equals the ratio of logs of initial-to-final concentration fractions.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

Core Logic

For Reaction 1 (A K₁ B$A \xrightarrow{K_1} B$): (K₁)500K = 2(K₁)300K$(K_1)_{500\text{K}} = 2(K_1)_{300\text{K}}$. ln(2) = Eₐ₁R [ (1)/(300) - (1)/(500) ] = Eₐ₁R [ (200)/(150000) ] = Eₐ₁R [ (2)/(1500) ]$\ln(2) = \frac{E_{a1}}{R} \left[ \frac{1}{300} - \frac{1}{500} \right] = \frac{E_{a1}}{R} \left[ \frac{200}{150000} \right] = \frac{E_{a1}}{R} \left[ \frac{2}{1500} \right]$ Eₐ₁ = (ln 2 × R × 1500)/(2)$E_{a1} = \frac{\ln 2 \times R \times 1500}{2}$

Also for Reaction 1 at 500K, half-life is 2 hours, so: (K₁)500K = (ln 2)/(2) hr⁻¹$(K_1)_{500\text{K}} = \frac{\ln 2}{2}\text{ hr}^{-1}$

For Reaction 2 (C K₂ D$C \xrightarrow{K_2} D$): Given Eₐ₂ = Eₐ₁2 = (ln 2 × R × 1500)/(4)$E_{a2} = \frac{E_{a1}}{2} = \frac{\ln 2 \times R \times 1500}{4}$ Given (K₂)500K = 2 × (K₁)500K = 2 × (ln 2)/(2) = ln 2$(K_2)_{500\text{K}} = 2 \times (K_1)_{500\text{K}} = 2 \times \frac{\ln 2}{2} = \ln 2$

Applying Arrhenius equation for Reaction 2 from 300K to 500K: ln [ (K₂)500K(K₂)300K ] = Eₐ₂R [ (1)/(300) - (1)/(500) ]$\ln \left[ \frac{(K_2)_{500\text{K}}}{(K_2)_{300\text{K}}} \right] = \frac{E_{a2}}{R} \left[ \frac{1}{300} - \frac{1}{500} \right]$ ln [ ln 2(K₂)300K ] = (((ln 2 × R × 1500)/(4)))/(R) × (2)/(1500) = (ln 2)/(2) = ln(√(2))$\ln \left[ \frac{\ln 2}{(K_2)_{300\text{K}}} \right] = \frac{\left(\frac{\ln 2 \times R \times 1500}{4}\right)}{R} \times \frac{2}{1500} = \frac{\ln 2}{2} = \ln(\sqrt{2})$

Step 1: Solve for K2 at 300K

Equating the arguments of the natural logs: ln 2(K₂)300K = √(2)$\frac{\ln 2}{(K_2)_{300\text{K}}} = \sqrt{2}$ (K₂)300K = ln 2√(2) = (0.693)/(1.414) = 0.49 hr⁻¹$(K_2)_{300\text{K}} = \frac{\ln 2}{\sqrt{2}} = \frac{0.693}{1.414} = 0.49\text{ hr}^{-1}$ (K₂)300K = 4.9 × 10⁻¹ hr⁻¹ ≈ 5 × 10⁻¹ hr⁻¹$(K_2)_{300\text{K}} = 4.9 \times 10^{-1}\text{ hr}^{-1} \approx 5 \times 10^{-1}\text{ hr}^{-1}$ (rounded to nearest integer).

Pattern Recognition

Sequential tracking of unknowns through Arrhenius. Eₐ$E_a$ ratios dictate log ratios between temperatures directly.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

Core Logic

Let us break down each step of the mechanism:

• Step 1: A arrow B$\mathrm{A} \rightarrow \mathrm{B}$ is slow.
• Being the rate-determining step, it must have the highest activation energy barrier (Eₐ₁$E_{\mathrm{a1}}$).
• Since Δ H = +ve$\Delta H = +\mathrm{ve}$ (endothermic), the energy level of intermediate state B must be higher than the reactant state A.
• Step 2: B arrow C$\mathrm{B} \rightarrow \mathrm{C}$ is fast.
• It has a much lower activation energy barrier (Eₐ₂$E_{\mathrm{a2}}$).
• Since Δ H = -ve$\Delta H = -\mathrm{ve}$ (exothermic), the energy level of intermediate C is lower than state B.
• Step 3: C arrow D$\mathrm{C} \rightarrow \mathrm{D}$ is fast.
• It has a low activation energy barrier (Eₐ₃$E_{\mathrm{a3}}$).
• Since Δ H = -ve$\Delta H = -\mathrm{ve}$ (exothermic), the energy level of final state D is lower than state C.
• Net Reaction: Exothermic with "net evolution of heat".
• The potential energy of the final product D is lower than the initial potential energy of reactant A.

Step 1: Check Potential Energy Profile

Evaluating the transition states and relative energy levels in Graph (1):

• The first peak (transition state 1) is clearly the highest (Eₐ₁ > Eₐ₂, Eₐ₃$E_{\mathrm{a1}} > E_{\mathrm{a2}}, E_{\mathrm{a3}}$) $\implies$ Step 1 is the slowest.
• The intermediate B is higher in energy than A.
• Intermediates C and product D are progressively lower in energy.
• Product D has lower energy than reactant A (net exothermic).



This perfectly corresponds to Graph (1).

Pattern Recognition

Kinetics shortcut: Slow step = tallest peak. Exothermic step = drop in energy levels of products/intermediates. Endothermic step = climb in energy levels. Use these rules to visually scan energy profiles in under 5 seconds.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics