To solve this problem, we need to calculate the Gibbs free energy change (ΔG) for the reaction at 298.15 K and then determine if the reaction is spontaneous.

Given Information:

• Temperature (T) = 298.15 K
• Enthalpy change (ΔH) = -1141.1 kJ
• Entropy change (ΔS) = -146.5 J/K

Since ΔH is given in kJ and ΔS in J/K, let's convert ΔS to kJ/K for consistency: $\Delta S = -146.5 \, \text{J/K} = -0.1465 \, \text{kJ/K}$

Step 1: Use the Gibbs Free Energy Equation

The Gibbs free energy change, ΔG, can be calculated using the formula: $\Delta G = \Delta H - T \Delta S$

Step 2: Substitute Values

$\Delta G = -1141.1 \, \text{kJ} - (298.15 \, \text{K}) \times (-0.1465 \, \text{kJ/K})$ $\Delta G = -1141.1 \, \text{kJ} + 43.674475 \, \text{kJ}$ $\Delta G = -1097.425525 \, \text{kJ}$

Step 3: Interpretation

Since ΔG is negative (-1097.43 kJ), the reaction is spontaneous at 298.15 K.

Final Answer:

• ΔG = -1097.43 kJ (rounded to two decimal places)
• Spontaneity: The reaction is spontaneous at 298.15 K.

Would you like further details on any of these steps or have any questions?

Here are five related questions to deepen understanding:

1. How does the sign of ΔG relate to the spontaneity of a reaction?
2. Why is it necessary to convert ΔS to kJ/K in this calculation?
3. How would a positive ΔS impact the spontaneity of a reaction?
4. What would happen to ΔG if the temperature increased?
5. How does the enthalpy change (ΔH) affect the spontaneity of a reaction?

Tip: Remember that a negative ΔG indicates a spontaneous reaction, while a positive ΔG indicates non-spontaneity.