Introduction:
This question relates to the activation energy and standard enthalpy of the activated complex in a first-order isomerization reaction of an organic compound. The activation energy is given as 108.4 kJ/mol, and we need to determine the standard enthalpy of the activated complex. The correct answer is expected to be between 102.0 and 110.0 kJ/mol.

Understanding First-Order Isomerization:
First-order isomerization is a type of reaction where a molecule undergoes a structural rearrangement to form an isomer. The rate of this reaction is proportional to the concentration of the starting material, and the reaction follows first-order kinetics.

Determining the Standard Enthalpy of Activated Complex:
The standard enthalpy of activation (ΔH‡) can be determined using the Arrhenius equation:

k = Ae^(-ΔH‡/RT)

Where:
- k is the rate constant
- A is the pre-exponential factor
- ΔH‡ is the activation energy
- R is the gas constant
- T is the temperature in Kelvin

In this case, the temperature is given as 1300°C, which needs to be converted to Kelvin:

T = 1300 + 273.15 = 1573.15 K

Calculating the Rate Constant:
To determine the standard enthalpy of the activated complex, we need to calculate the rate constant (k) at the given temperature. Rearranging the Arrhenius equation:

k = Ae^(-ΔH‡/RT)

We can express the equation as:

ln(k) = -ΔH‡/RT + ln(A)

The rate constant can be determined experimentally, or if not given, we can assume a value. Let's assume a hypothetical value of k = 1 s^-1 for simplicity.

ln(1) = -ΔH‡/(1573.15 K * R) + ln(A)

0 = -ΔH‡/(1573.15 K * R) + ln(A)

Determining the Activation Energy:
We can rearrange the equation to solve for ΔH‡:

ΔH‡ = -ln(A) * (1573.15 K * R)

To calculate ΔH‡, we need the value of A. Unfortunately, it is not provided in the question, so we cannot determine the exact value of ΔH‡. However, we can estimate a range based on typical values of A in chemical reactions.

Estimating the Range:
The pre-exponential factor (A) typically falls within a range of 10^10 to 10^15 s^-1. Let's assume a value of A = 10^12 s^-1.

Using this value, we can calculate the activation energy:

ΔH‡ = -ln(10^12) * (1573.15 K * R)

Using the gas constant R = 8.314 J/(mol·K), we can calculate ΔH‡:

ΔH‡ = -27.631 * (1573.15 K * 8.314 J/(mol·K))

ΔH‡ ≈ -366.4 kJ/mol

Since the activation energy