To solve this problem, we will use the Arrhenius equation, which relates the rate constant ($k$) of a chemical reaction to the temperature ($T$) and the activation energy ($E_a$). The equation in its logarithmic form is:

$\log k = \log A - \frac{E_a}{2.303RT}$

where:

• $k$ is the rate constant,
• $A$ is the pre-exponential factor (frequency factor),
• $E_a$ is the activation energy,
• $R$ is the universal gas constant ($$8.314 \, \text{J mol}^{-1} \text{K}^{-1}$$),
• $T$ is the temperature in Kelvin,
• $2.303$ is the factor to convert from natural log to common log.

According to the problem, the rate of the reaction quadruples ($k_2 = 4k_1$) when the temperature increases from $27^{\circ}C$ (which equals $$27 + 273.15 = 300.15 \, K$$) to $57^{\circ}C$ (which equals $$57 + 273.15 = 330.15 \, K$$). Using the logarithmic form of the Arrhenius equation, for two different temperatures we have:

$\log k_1 = \log A - \frac{E_a}{2.303R \times 300.15}$

$\log k_2 = \log A - \frac{E_a}{2.303R \times 330.15}$

With $k_2 = 4k_1$, substituting in the values and taking their difference:

$\log 4k_1 - \log k_1 = \left(\log A - \frac{E_a}{2.303 \times 8.314 \times 330.15}\right) - \left(\log A - \frac{E_a}{2.303 \times 8.314 \times 300.15}\right)$

$\log 4 = \frac{E_a}{2.303 \times 8.314} \left(\frac{1}{300.15} - \frac{1}{330.15}\right)$

Given that $\log 4 = 0.6021$, we now solve for $E_a$:

$0.6021 = \frac{E_a}{2.303 \times 8.314} \left(\frac{1}{300.15} - \frac{1}{330.15}\right)$

First, calculate $$\frac{1}{300.15} - \frac{1}{330.15}$$:

$\frac{1}{300.15} \approx 0.003332 \quad \text{and} \quad \frac{1}{330.15} \approx 0.003029$

$\frac{1}{300.15} - \frac{1}{330.15} = 0.003332 - 0.003029 = 0.000303$

So,

$0.6021 = \frac{E_a}{19.141872} \times 0.000303$

$\frac{E_a}{19.141872} = \frac{0.6021}{0.000303} = 1987.05$

Solve for $E_a$:

$E_a = 1987.05 \times 19.141872 = 38067 \, \text{J/mol} \approx 38.067 \, \text{kJ/mol}$

Therefore, the activation energy $E_a$ is $$38.07 \, \text{kJ/mol}$$, which best matches Option A:

$38.04 \, \text{kJ/mol}$