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Δ G° = -RT ln K = -8.314 × 300 × ln(100) = -8.314 × 300 × 4.605 = -11488 J/mol = -11.49 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
The standard state of an element is its most stable form at 1 bar and 298 K. For carbon, this is graphite.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
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