First-Order Reaction Analysis: Rate Constant Calculation

This problem involves calculating the rate constant for a first-order reaction at a specific temperature using information about its temperature dependence and half-life.

Calculating Reaction 1 Rate Constant

1. At 500 K: The half-life ($t_{1/2}$) for a first-order reaction is given by $t_{1/2} = \frac{\ln(2)}{k}$. Given $t_{1/2} = 2$ hours at $500\text{ K}$ for the first reaction:

   $k_1(500\text{ K}) = \frac{\ln(2)}{2\text{ hour}} \approx \frac{0.693}{2} = 0.3465\text{ hour}^{-1}$

2. At 300 K: The problem states $k_1(500\text{ K}) = 2 \times k_1(300\text{ K})$. Therefore:

   $k_1(300\text{ K}) = \frac{k_1(500\text{ K})}{2} = \frac{0.3465\text{ hour}^{-1}}{2} = 0.17325\text{ hour}^{-1}$

Determining Reaction 2 Rate Constant at 500 K

The problem states that at $500\text{ K}$, the rate constant for the second reaction ($k_2$) is double that of the first reaction ($k_1$):

$k_2(500\text{ K}) = 2 \times k_1(500\text{ K}) = 2 \times 0.3465\text{ hour}^{-1} = 0.693\text{ hour}^{-1}$

Using Arrhenius Equation for Temperature Dependence

The Arrhenius equation relates rate constants at different temperatures: $\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$.

For reaction 1:

$\ln\left(\frac{k_1(500\text{ K})}{k_1(300\text{ K})}\right) = \frac{E_{a1}}{R}\left(\frac{1}{300\text{ K}} - \frac{1}{500\text{ K}}\right)$

Since $\frac{k_1(500\text{ K})}{k_1(300\text{ K})} = 2$:

$\ln(2) = \frac{E_{a1}}{R}\left(\frac{500 - 300}{300 \times 500}\right) = \frac{E_{a1}}{R}\left(\frac{200}{150000}\right) = \frac{E_{a1}}{R}\left(\frac{1}{750}\right)$

So, $\frac{E_{a1}}{R} = 750 \times \ln(2)$.

For reaction 2, we are given $E_{a2} = \frac{E_{a1}}{2}$. Thus, $\frac{E_{a2}}{R} = \frac{E_{a1}}{2R} = \frac{750 \times \ln(2)}{2} = 375 \times \ln(2)$.

Now, apply the Arrhenius equation for reaction 2:

$\ln\left(\frac{k_2(500\text{ K})}{k_2(300\text{ K})}\right) = \frac{E_{a2}}{R}\left(\frac{1}{300\text{ K}} - \frac{1}{500\text{ K}}\right) = \frac{E_{a2}}{R}\left(\frac{1}{750}\right)$

Substitute the value of $\frac{E_{a2}}{R}$:

$\ln\left(\frac{k_2(500\text{ K})}{k_2(300\text{ K})}\right) = (375 \times \ln(2)) \times \left(\frac{1}{750}\right) = \frac{375}{750} \times \ln(2) = \frac{1}{2} \times \ln(2) = \ln(2^{1/2}) = \ln(\sqrt{2})$

Therefore, $\frac{k_2(500\text{ K})}{k_2(300\text{ K})} = \sqrt{2}$.

Final Calculation for $k_2(300\text{ K})$

Rearrange the equation to find $k_2(300\text{ K})$:

$k_2(300\text{ K}) = \frac{k_2(500\text{ K})}{\sqrt{2}} = \frac{0.693\text{ hour}^{-1}}{\sqrt{2}} \approx \frac{0.693}{1.414} \approx 0.4901\text{ hour}^{-1}$

Convert this to the required format:

$0.4901\text{ hour}^{-1} = 4.901 \times 10^{-1}\text{ hour}^{-1}$

Rounding to the nearest integer, the value is $5 \times 10^{-1}\text{ hour}^{-1}$.