A arrow B$A \rightarrow B$ (first reaction) C arrow D$C \rightarrow D$ (second reaction) Consider the above two first-order reactions. The rate constant for first reaction at 500 K$500\text{ K}$ is double of the same at 300 K$300\text{ K}$. At 500 K$500\text{ K}$, 50%$50\%$ of the reaction becomes complete in 2 hour$2\text{ hour}$. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K$500\text{ K}$ of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K$300\text{ K}$ is ____ × 10⁻¹ hour⁻¹$\times 10^{-1}\text{ hour}^{-1}$ (nearest integer).

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