Answer:
-138.9 kJ/mol
Explanation:
Step 1: Convert 235.8°C to the Kelvin scale
We will use the following expression.
K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K
Step 2: Calculate the standard enthalpy of reaction (ΔH°)
We will use the following expression.
ΔG° = ΔH° - T.ΔS°
ΔH° = ΔG° / T.ΔS°
ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K
ΔH° = -3.583 kJ (for 1 mole of balanced reaction)
Step 3: Convert -9.9°C to the Kelvin scale
K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K
Step 4: Calculate ΔG° at 263.3 K
ΔG° = ΔH° - T.ΔS°
ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K
ΔG° = -138.9 kJ/mol
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