the standard state gibbs free energy of the formation of graphite and diamond at temperature 298 Kelvin are 0 kilo joule per mole and 2.9 kilo joule per mole. standard state means the pressure should be 1 bar and the substance should be pure at the given temperature .the conversion of graphite to Diamond reduces the volume by 2 ?10 to the power -6 m cube per mole of carbon graphite is converted to Carbon diamond isothermally at temperature 298 Kelvin then the what is the pressure at which carbon graphite is in equilibrium with Carbon diamond ques no. 1 in image

C(graphite)C(diamond)Go=fGo(diamond)-fGo(graphite)Go=2.9-0=2.9KJ/molV=2×10-6 m3 mol-1Isothermally at T=0E=0At equilibrium,G=0Go=PV2.9×103= P(2×10-6)P-1=2.9×1032×10-6=1.45×109 Pascal1 bar=105PaP=1.45×109 bar+1P=14501 bar
Hence option A is correct

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