Elemental carbon usually exists in one of two forms: graphite or diamond. It is

Question

Elemental carbon usually exists in one of two forms: graphite or diamond. It is generally believed that diamonds last forever. The table shows the standard enthalpy of formation ( Constants | Periodic Table Gibbs free energy is a measure of the spontaneity of a chemical reaction. It is the chemical potential for a reaction, and is minimized at equilibrium. The change in Gibbs free energy can be calculated by Hi, ) and the standard molar entropy (S*) values for diamond and graphite Substance AH (kJ/mol) S° (J/mol-K) 5.740 graphite Cdiamond 1.897 2.38 where Grn is the standard Gibbs free energy change for a reaction, n is the standard enthalpy change for a reaction, T is the temperature in kelvins, and &n; is the standard entropy change for a reaction. Part A For a reaction at equilibrium, Grxn is equal to zero. For negative values, the reaction is spontaneous, and for positive values the reaction is nonspontaneous What is the standard Gibbs free energy for the transformation of diamond to graphite at 298 K? Cdiamond Cgraphite Express your answer to three significant figures and include the appropriate units View Available Hint(s) 2.90 kJ mol SubmitPrev evious Answer Incorrect, Try Again: 6 attempts remaining

Explanation / Answer

Given:
Hof(C(diamond)) = 1.897 KJ/mol
Hof(C(graphite)) = 0.0 KJ/mol

Balanced chemical equation is:
C(diamond) ---> C(graphite)

Ho rxn = 1*Hof(C(graphite)) - 1*Hof( C(diamond))
Ho rxn = 1*(0.0) - 1*(1.897)
Ho rxn = -1.897 KJ

Given:
Sof(C(diamond)) = 2.38 J/mol.K
Sof(C(graphite)) = 5.74 J/mol.K

Balanced chemical equation is:
C(diamond) ---> C(graphite)

So rxn = 1*Sof(C(graphite)) - 1*Sof( C(diamond))
So rxn = 1*(5.74) - 1*(2.38)
So rxn = 3.36 J/K
Ho = -1.897 KJ/mol
So = 3.36 J/mol.K
= 0.00336 KJ/mol.K
T = 298 K

use:
Go = Ho - T*So
Go = -1.897 - 298.0 * 0.00336
Go = -2.90 KJ
Answer: -2.90 KJ

-2.90 is correct
Please try unit as KJ

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