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Consider the following reaction at 25 °C: 5 SO3(g) + 2 NH3(g) → 2 NO(g) + 5 SO2(g) + 3 H2O(g) If AH° = 42.4 kJ./mol and AS° =
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Answer #1

At equilibrium,

ΔG° = 0

We know that

ΔG° = ΔH° – TΔS°

At equilibrium,

ΔG° = ΔH° – TΔS° = 0

ΔH° = TΔS°

Given that

ΔH° = 42.4 kJ/mol = 42400 J/mol

ΔS° = 562.3 J/(mol.K)

T = ΔH° / ΔS°

T = (42400 J/mol )/(562.3 J/(mol.K))

T = 75.40459 K

T = 75.40459 – 273.15 = – 197.7 °C = – 198 °C

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