To calculate the rate constant at 313 K, we can again use the Arrhenius equation in its comparative form:

$\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Given:

• $k_1 = 0.01439 \, \text{min}^{-1}$ (rate constant at 298 K),
• $T_1 = 298 \, \text{K}$,
• $T_2 = 313 \, \text{K}$,
• $E_a = 70000 \, \text{J/mol}$ (activation energy),
• $R = 8.314 \, \text{J/mol·K}$.

Step 1: Calculate the term $\frac{E_a}{R}$

$\frac{E_a}{R} = \frac{70000 \, \text{J/mol}}{8.314 \, \text{J/mol·K}} = 8418.32 \, \text{K}$

Step 2: Calculate $\frac{1}{T_2} - \frac{1}{T_1}$

$\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{313 \, \text{K}} - \frac{1}{298 \, \text{K}} = \frac{298 - 313}{298 \times 313} \, \text{K}^{-1} = -\frac{15}{93274} \, \text{K}^{-1} \approx -0.0001608 \, \text{K}^{-1}$

Step 3: Calculate $\ln\left(\frac{k_2}{k_1}\right)$

$\ln\left(\frac{k_2}{k_1}\right) = -8418.32 \, \text{K} \times -0.0001608 \, \text{K}^{-1} \approx 1.353$

Step 4: Solve for $k_2$

$\frac{k_2}{k_1} = e^{1.353} \approx 3.869$

$k_2 = k_1 \times 3.869 = 0.01439 \, \text{min}^{-1} \times 3.869 \approx 0.0557 \, \text{min}^{-1}$

So, the rate constant of the reaction at 313 K is approximately $0.0557 \, \text{min}^{-1}$.

Would you like further details or have any other questions?

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Here are some related questions:

1. How does the activation energy affect the rate constant at different temperatures?
2. What are some methods to experimentally determine the activation energy of a reaction?
3. How does the Arrhenius equation apply to non-ideal systems?
4. What happens to the rate constant if the temperature increases by a larger amount, say to 350 K?
5. Can the Arrhenius equation be used to predict reaction behavior at very low temperatures?

Tip: Always check the consistency of your units when using the Arrhenius equation, especially when switching between time units like minutes and seconds.