Consider the following data: $\Delta H = -185 \, \text{kJ}$, $\Delta S = -483.7 \, \text{kJ}$.

Calculate for the following reaction:

a. -299 kJ
b. 114 kJ
c. 299 kJ
d. -114 kJ

A) (a) -114 kJ, (b) 299 kJ, (c) -299 kJ, (d) 114 kJ
B) (a) 114 kJ, (b) -299 kJ, (c) 299 kJ, (d) -114 kJ
C) (a) -299 kJ, (b) 114 kJ, (c) -114 kJ, (d) 299 kJ
D) (a) 299 kJ, (b) -114 kJ, (c) 114 kJ, (d) -299 kJ

To calculate the standard free energy change (ΔG°), the formula ΔG° = ΔH° - TΔS° is used with the provided enthalpy and entropy values at a presumed temperature of 298 K. The calculated ΔG° is -40.817 kJ, which does not correspond with any of the given answer choices.

Explanation:

To solve for the reaction's standard free energy change (ΔG°), we need to use the provided standard enthalpy change (ΔH° = -185 kJ) and standard entropy change (ΔS° = -483.7 J/K, converted from kJ). The equation for the standard free energy change is ΔG° = ΔH° - TΔS°, where T is the temperature in Kelvin. Assuming the temperature is 298 K (standard temperature), we can perform the calculation:

ΔG° = (-185 kJ) - (298 K * -483.7 J/K * 1 kJ/1000 J)

ΔG° = -185 kJ + (298 K * 0.4837 kJ/K)

ΔG° = -185 kJ + 144.183 kJ

ΔG° = -40.817 kJ (rounded to three significant figures)

However, none of the options provided (a. -299 kJ, b. 114 kJ, c. 299 kJ, d. -114 kJ) match this result, suggesting that either the given options are incorrect, there has been a misunderstanding of the question, or further information is required to match the question details.