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

Added by Heidi N.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Sri K

01/10/2023

Step 1

This is done by subtracting the entropy of the reactants from the entropy of the products. In this case, it would be: ΔS° = 2*(5.74 J/(mol·K)) - 2*(2.38 J/(mol·K)) = 6.72 J/(mol·K) Show more…

Show all steps

Thanks for your feedback!

Given the reaction of diamond converting to graphite: 2C(s, diamond) → 2C(s, graphite) Determine ΔG° at 298 K and predict whether the reaction is spontaneous or not. ΔHf° (C, diamond) = 1.9 kJ/mol S° (C, diamond) = 2.38 J/mol S° (C, graphite) = 5.74 J/mol 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