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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?

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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.

Determine ΔG at 298.15K for the reaction. Is it spontaneous? ΔH = -1141.1 kJ, ΔS = -146.5 J/K for the reaction 2NO(g) + O2(g) → 2NO2(g).

Learn how to calculate Gibbs Free Energy (ΔG) for the reaction 2NO(g) + O2(g) → 2NO2(g) at 298.15K. This problem involves thermodynamic principles, using ΔH = -1141.1 kJ and ΔS = -146.5 J/K to determine spontaneity.

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