To solve this problem, we can make use of the Arrhenius equation, which relates the rate constant of a reaction to its temperature. The Arrhenius equation is given by:

$$k = A \exp\left(-\frac{E_a}{RT}\right)$$

where:

$$k$$ is the rate constant
$$A$$ is the pre-exponential factor (frequency factor)
$$E_a$$ is the activation energy
$$R$$ is the gas constant
$$T$$ is the temperature in Kelvin

For two different temperatures, the Arrhenius equation can be written as:

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

Given:

$$k_1 = 0.004 \, \text{s}^{-1}$$ at $$T_1 = 500 \, \text{K}$$
$$k_2 = 0.014 \, \text{s}^{-1}$$
$$E_a = 18.231 \, \text{kJ/mol}$$ (which is $$18,231 \, \text{J/mol}$$)
$$R = 8.314 \, \text{J/mol K}$$

We need to find the new temperature $$T_2$$. Rearranging the Arrhenius equation to solve for $$T_2$$, we get:

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

Substituting the given values:

$$\ln\left(\frac{0.014}{0.004}\right) = \frac{18,231}{8.314}\left(\frac{1}{500} - \frac{1}{T_2}\right)$$

Calculating the left-hand side:

$$\ln\left(\frac{0.014}{0.004}\right) = \ln(3.5) \approx 1.2528$$

Thus, the equation becomes:

$$1.2528 = \frac{18,231}{8.314}\left(\frac{1}{500} - \frac{1}{T_2}\right)$$

Simplifying further:

$$1.2528 = \frac{18,231}{8.314} \times \left(\frac{1}{500} - \frac{1}{T_2}\right)$$

$$1.2528 = 2,192.6 \times \left(\frac{1}{500} - \frac{1}{T_2}\right)$$

$$\frac{1}{500} - \frac{1}{T_2} = \frac{1.2528}{2192.6} \approx 0.0005714$$

$$\frac{1}{T_2} = \frac{1}{500} - 0.0005714$$

$$\frac{1}{T_2} \approx 0.002 - 0.0005714 = 0.0014286$$

$$T_2 = \frac{1}{0.0014286} \approx 700 \, \text{K}$$

Thus, the correct option is Option C: 700 K.

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