Question

15. Given the reaction of diamond converting to graphite: 2C(s, diamond) ? 2C(s, graphite) Determine ?G° at 298 K and predict whether the reaction is spontaneous or not. ?Hf° (C, diamond) = 1.9 kJ/mol S°(C, diamond) = 2.38 J/mol S°(C, graphite) = 5.74 J/mol A) ?G° = +2.90 kJ/mol; Non spontaneous reaction B) ?G° = -5.8 kJ/mol; Spontaneous reaction C) ?G° = -2.90 kJ/mol; spontaneous reaction D) ?G° = +5.8 kJ/mol; Spontaneous reaction

          15. Given the reaction of diamond converting to graphite:
2C(s, diamond) ? 2C(s, graphite)
Determine ?G° at 298 K and predict whether the reaction is spontaneous or not.
?Hf° (C, diamond) = 1.9 kJ/mol
S°(C, diamond) = 2.38 J/mol
S°(C, graphite) = 5.74 J/mol
A) ?G° = +2.90 kJ/mol; Non spontaneous reaction
B) ?G° = -5.8 kJ/mol; Spontaneous reaction
C) ?G° = -2.90 kJ/mol; spontaneous reaction
D) ?G° = +5.8 kJ/mol; Spontaneous reaction

15. Given the reaction of diamond converting to graphite:
2C(s, diamond) ? 2C(s, graphite)
Determine ?G° at 298 K and predict whether the reaction is spontaneous or not.
?Hf° (C, diamond) = 1.9 kJ/mol
S°(C, diamond) = 2.38 J/mol
S°(C, graphite) = 5.74 J/mol
A) ?G° = +2.90 kJ/mol; Non spontaneous reaction
B) ?G° = -5.8 kJ/mol; Spontaneous reaction
C) ?G° = -2.90 kJ/mol; spontaneous reaction
D) ?G° = +5.8 kJ/mol; Spontaneous reaction

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Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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We are given the reaction of diamond converting to graphite, with the enthalpy change (ΔH°) of 1.9 kJ/mol for diamond and the standard entropy changes (ΔS°) for diamond and graphite as 2.38 J/(mol·K) and 5.74 J/(mol·K), respectively. We need to calculate the Show more…

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Given the reaction of diamond converting to graphite: 2C(diamond) → 2C(graphite) Determine ΔG° at 298 K and predict whether the reaction is spontaneous or not. ΔHf° = 1.9 kJ/mol (diamond) S° = 2.38 J/mol (diamond) S° = 5.74 J/mol (graphite) A) ΔG° = +2.90 kJ/mol: Non spontaneous reaction B) ΔG° = -5.8 kJ/mol: Spontaneous reaction C) ΔG° = -2.90 kJ/mol: spontaneous reaction D) ΔG° = +5.8 kJ/mol: Spontaneous reaction