The Standard Gibb's free energy for the transformation of diamond to graphite at 298 k is -2.9 kJ/mol.
Standard Gibb's free energy of the reaction will be calculated as:
ΔG° = ΔG°(product) - ΔG°(reactant)
For this reaction tandard Gibb's free energy will be calculated as:
ΔG° = ΔG°(graphite) - ΔG°(diamond)
ΔG° for diamond = 2.9 kJ/mol
ΔG° for graphite = 0 kJ/mol
On putting these values on the above equation we get:
ΔG° = 0 - 2.9 kJ/mol = -2.9 kJ/mol
Hence, required value is -2.9 kJ/mol.
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Answer:
standard Gibbs energy for transformation of diamond to graphite at 298K is -2.89828 = -2.90 kJ/mol
Explanation:
**I have included images of the formulas to use + the table of products/reactants below!!**
1) calculate delta_rH using the formula for enthalpy of reaction:
delta_rH = (sum of products) - (sum of reactants)
delta_rH = (0 kJ/mol) - (1.897 kJ/mol) = -1.897 kJ/mol
2) calculate the delta_rS using the formula for entropy of reaction:
delta_rS = (sum of products) - (sum of reactants)
delta_rS = (5.740 J/mol*K) - (2.38 J/mol*K) = 3.36 J/mol*K
(this answer should be positive because graphite is more disordered than diamond)
3) change the above answers to the same unit (all either to kJ, or all either to J - I will do kJ):
3.36 J/mol*K = 0.00336 kJ/mol*k
4) use the Gibbs energy formula to calculate the Gibbs energy for transformation:
delta_rG = delta_rH - (T)(delta_rS)
delta_rG = (-1.897 kJ/mol) - (298K)(0.00336 kJ/mol*K)
delta_rG = (-1.897 kJ/mol) - (1.00128 kJ/mol) = -2.89828 = -2.90 kJ/mol
