To solve this problem, we need to use the Arrhenius equation in its logarithmic form to compare the rate constants at two different temperatures. The equation is given as:

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

Where:

• $k_1$ is the rate constant at temperature $T_1 = 600.0 \, \text{K}$,
• $k_2$ is the rate constant at temperature $T_2 = 770.0 \, \text{K}$,
• $E_a$ is the activation energy, $262 \, \text{kJ/mol} = 262 \times 10^3 \, \text{J/mol}$,
• $R$ is the gas constant, $8.314 \, \text{J/mol·K}$.

Given:

• $k_1 = 6.1 \times 10^{-8} \, \text{s}^{-1}$,
• $T_1 = 600.0 \, \text{K}$,
• $T_2 = 770.0 \, \text{K}$,
• $E_a = 262 \times 10^3 \, \text{J/mol}$.

Step-by-step solution:

1. Calculate the difference in reciprocal temperatures: $\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{600.0} - \frac{1}{770.0} = 0.0016667 - 0.0012987 = 0.000368 \, \text{K}^{-1}$

2. Multiply by the activation energy and divide by the gas constant $R$: $\frac{E_a}{R} = \frac{262 \times 10^3}{8.314} = 31,515.77 \, \text{K}$ $\frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) = 31,515.77 \times 0.000368 = 11.59$

3. Now, calculate the logarithm of the ratio of rate constants: $\ln\left(\frac{k_2}{k_1}\right) = 11.59$ $\frac{k_2}{k_1} = e^{11.59} = 107,742$

4. Solve for $k_2$: $k_2 = k_1 \times 107,742 = (6.1 \times 10^{-8}) \times 107,742 = 6.57 \times 10^{-3} \, \text{s}^{-1}$

Thus, the rate constant at $770.0 \, \text{K}$ is approximately: $k_2 = 6.57 \times 10^{-3} \, \text{s}^{-1}$

Would you like more details on any step, or have any questions?

Here are 5 related questions to expand your knowledge:

1. How does the Arrhenius equation explain the effect of temperature on reaction rates?
2. What is the significance of activation energy in a chemical reaction?
3. Can you derive the Arrhenius equation from first principles of chemical kinetics?
4. How would the rate constant change if the temperature were increased to 1000 K?
5. What happens to the rate of a reaction if the activation energy is very low?

Tip: In problems involving energy, always check unit conversions carefully—especially for activation energy.