Final answer:
The reaction in question proceeded with a decrease in entropy under standard conditions.
Step-by-step explanation:
In this reaction, the change in entropy (ΔS) can be determined using the enthalpy change (ΔH) and the temperature. Since the enthalpy change (ΔH) is negative (-750 kcal/mol) and the temperature is positive, the entropy change (ΔS) will be negative. This means that the reaction proceeded with a decrease in entropy under standard conditions.