To calculate the rate constant of the reaction at 700 K, we can use the Arrhenius equation, which is given by:

$k = A \cdot e^{-\frac{E_a}{RT}}$

where:

• $k$ is the rate constant,
• $A$ is the pre-exponential factor,
• $E_a$ is the activation energy,
• $R$ is the gas constant, and
• $T$ is the temperature in Kelvin.

However, since we are comparing the rate constants at two different temperatures, we can use the following form of the Arrhenius equation:

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

Where:

• $k_1 = 1.60 \times 10^{-5} \, \text{s}^{-1}$ (rate constant at 600 K),
• $T_1 = 600 \, \text{K}$,
• $T_2 = 700 \, \text{K}$,
• $E_a = 209 \, \text{kJ/mol} = 209 \times 10^3 \, \text{J/mol}$,
• $R = 8.314 \, \text{J/mol·K}$.

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

$\frac{E_a}{R} = \frac{209 \times 10^3 \, \text{J/mol}}{8.314 \, \text{J/mol·K}} = 25141.98 \, \text{K}$

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

$\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{700 \, \text{K}} - \frac{1}{600 \, \text{K}} = \frac{600 - 700}{700 \times 600} \, \text{K}^{-1} = -\frac{100}{420000} \, \text{K}^{-1} = -0.0002381 \, \text{K}^{-1}$

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

$\ln\left(\frac{k_2}{k_1}\right) = -25141.98 \, \text{K} \times -0.0002381 \, \text{K}^{-1} = 5.984$

Step 4: Solve for $k_2$

$\frac{k_2}{k_1} = e^{5.984} \approx 397.73$

$k_2 = k_1 \times 397.73 = 1.60 \times 10^{-5} \, \text{s}^{-1} \times 397.73 \approx 6.36 \times 10^{-3} \, \text{s}^{-1}$

So, the rate constant of the reaction at 700 K is approximately $6.36 \times 10^{-3} \, \text{s}^{-1}$.

Would you like more details on any part of this calculation or have any further questions?

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Here are some related questions:

1. What is the significance of the activation energy in a chemical reaction?
2. How does the rate constant change with temperature for a first-order reaction?
3. What are some practical applications of the Arrhenius equation in chemical engineering?
4. How would the rate constant be affected if the activation energy were lower?
5. Can the Arrhenius equation be applied to reactions other than first-order reactions?

Tip: When working with the Arrhenius equation, always ensure the activation energy is in the same units as the gas constant to avoid unit conversion errors.