Question

Consider the following two first-order reactions: A \(\to\) B (first reaction) C \(\to\) D (second reaction) The rate constant for first reaction at 500 K is double of the same at 300 K. At 500 K, 50% of the reaction becomes complete in 2 hours. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 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 is ______ \(\times 10^{-3}\,\text{hour}^{-1}\) (nearest integer).

Hint:

For Arrhenius problems: - Use half-life to find rate constant. - Apply ratio of rate constants at two temperatures. - Activation energy scaling helps compare different reactions.

Answer 1

Correct Answer: 22

Step 1: Rate constant of first reaction at 500 K. For first-order kinetics: \[ t_{1/2} = \frac{0.693}{k} \] Given \(t_{1/2} = 2 \,\text{h}\): \[ k_{500}^{(1)} = \frac{0.693}{2} = 0.3465 \,\text{h}^{-1} \]
Step 2: Rate constant at 300 K for first reaction. Given: \(k_{500}^{(1)} = 2k_{300}^{(1)}\). \[ k_{300}^{(1)} = \frac{0.3465}{2} = 0.17325 \,\text{h}^{-1} \]
Step 3: Relation of activation energies. Activation energy of second reaction = \(\frac{1}{2}\) of first reaction.
Step 4: Rate constant of second reaction at 500 K. Given: \(k_{500}^{(2)} = 2k_{500}^{(1)} = 2 \times 0.3465 = 0.693 \,\text{h}^{-1}\).
Step 5: Use Arrhenius relation. \[ \ln\left(\frac{k_{500}}{k_{300}}\right) = \frac{E_a}{R}\left(\frac{1}{300} - \frac{1}{500}\right) \] For first reaction: \[ \ln\left(\frac{0.3465}{0.17325}\right) = \ln(2) = 0.693 \] \[ \Rightarrow \frac{E_a^{(1)}}{R}\left(\frac{1}{300} - \frac{1}{500}\right) = 0.693 \] \[ \frac{E_a^{(1)}}{R} \cdot \frac{200}{150000} = 0.693 \quad \Rightarrow \quad \frac{E_a^{(1)}}{R} = 5197 \]
Step 6: Activation energy of second reaction. \[ \frac{E_a^{(2)}}{R} = \frac{5197}{2} = 2598.5 \]
Step 7: Calculate \(k_{300}^{(2)\).} \[ \ln\left(\frac{k_{500}^{(2)}}{k_{300}^{(2)}}\right) = \frac{E_a^{(2)}}{R}\left(\frac{1}{300} - \frac{1}{500}\right) \] \[ \ln\left(\frac{0.693}{k_{300}^{(2)}}\right) = 2598.5 \times \frac{200}{150000} = 3.465 \] \[ \frac{0.693}{k_{300}^{(2)}} = e^{3.465} \approx 32 \] \[ k_{300}^{(2)} = \frac{0.693}{32} \approx 0.0217 \,\text{h}^{-1} \] Final Answer: \[ k_{300}^{(2)} \approx 22 \times 10^{-3} \,\text{h}^{-1} \]