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 (ΔH∘f) and the standard molar entropy (S∘) values for diamond and graphite. Substance ΔH∘f (kJ/mol) S∘ (J/mol⋅K) Cgraphite 0 5.740 Cdiamond 1.897 2.38 Part A 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. Hints ΔG∘rxn = −2.89 kJ mol

Answer 1

Answer=
The equation that you're utilizing for

​G=H - TS

Hf = 5.470 kj/mole
Now we have to calculate the S which is the entropy sum.
1.897-2.38 = -0.483 J/mol ( S)
Take a second look at the units - they are in "j/Mol" whereas the enthalpy is given in "kJ/mol." Our final answer is going to be in kJ, so it would make sense to convert our answer to kJ.
-0.483 J/mol =- 0.000483 kJ/mol.
We are given the temperature in Kelvin which fits this equation so no further conversions have to be completed. Now we can plug our answers into the equation...
-5.470-(298 x 0.000483)
and we get our final answer of +5.61 kJ.

====>Same another example use for the = when H=1.897

G(rxn)= H-T

Delta H = Enthalpy sum of products - enthalpy sum of reactants. So in this case, it would be:
0 - 1.897 = -1.897 kJ/Mol. (delta H)
Now we have to calculate the Delta S which is the entropy sum.
5.740-2.38 = 3.36 J/mol (delta S)
Take a second look at the units - they are in "j/Mol" whereas the enthalpy is given in "kJ/mol." Our final answer is going to be in kJ, so it would make sense to convert our answer to kJ.
3.36 J/mol = 0.00336 kJ/mol.
We are given the temperature in Kelvin which fits this equation so no further conversions have to be completed. Now we can plug our answers into the equation...
-1.897-(298 x 0.00336)
and we get our final answer of -2.90 kJ.