A rightarrow B (first reaction)
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(C) For the first reaction $A \xrightarrow{k_1} B$:
Using the Arrhenius equation: $\ln\left(\frac{k_{1, 500}}{k_{1, 300}}\right) = \frac{E_{a1}}{R} \

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(C) For the first reaction $A \xrightarrow{k_1} B$:
Using the Arrhenius equation: $\ln\left(\frac{k_{1, 500}}{k_{1, 300}}ight) = \frac{E_{a1}}{R} \left(\frac{1}{300} - \frac{1}{500}ight)$.
Given $k_{1, 500} = 2 k_{1, 300}$, so $\ln(2) = \frac{E_{a1}}{R} \left(\frac{2}{1500}ight) \implies E_{a1} = \frac{1500 R \ln 2}{2} = 750 R \ln 2$.
For the second reaction $C \xrightarrow{k_2} D$, $E_{a2} = \frac{E_{a1}}{2} = 375 R \ln 2$.
At $500 \ K$, for the first reaction, $t_{1/2} = 2 \ hours$, so $k_{1, 500} = \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{2} \approx 0.3466 \ hour^{-1}$.
Given $k_{2, 500} = 2 k_{1, 500} = 2 	imes \frac{\ln 2}{2} = \ln 2 \approx 0.6931 \ hour^{-1}$.
Using the Arrhenius equation for the second reaction:
$\ln\left(\frac{k_{2, 500}}{k_{2, 300}}ight) = \frac{E_{a2}}{R} \left(\frac{1}{300} - \frac{1}{500}ight) = \frac{375 R \ln 2}{R} \left(\frac{2}{1500}ight) = 375 \ln 2 	imes \frac{1}{750} = \frac{\ln 2}{2} = \ln(\sqrt{2})$.
Therefore, $\frac{k_{2, 500}}{k_{2, 300}} = \sqrt{2} \implies k_{2, 300} = \frac{k_{2, 500}}{\sqrt{2}} = \frac{\ln 2}{\sqrt{2}} \approx \frac{0.6931}{1.414} \approx 0.4902 \ hour^{-1}$.
$k_{2, 300} = 4.9 	imes 10^{-1} \ hour^{-1}$.

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