The standard state Gibbs energies of formation of C (graphite) and C (diamond) at T=298 K are Delta_(f)G^(@)[C("graphite")]=0"kJ mol"^(-1) Delta_(f)G^(@)[C("diamond")]=2.9"kJ mol"^(-1) The standard state means that the pressure should be 1 bar and substance should be pure at a given temperature. The conversion of graphite to dimond reduces its volume by 2xx10^(-6) m^(3) mol^(-1) . If C (graphite) is converted to C isothermally at T=298 K, the pressure at which C (graphite) is in equilibrium with C (diamond) is (1J=1 kg m^(2)s^(-2),1 Pa=1kg m^(-1)s^(-2), 1bar=10^(5) Pa)

The standard state Gibbs energies of formation of C (graphite) and C (

1.	The standard state Gibbs energies of formation of C (graphite) and C (diamond) at T=298 K are Delta_(f)G^(@)[C("graphite")]=0"kJ mol"^(-1) Delta_(f)G^(@)[C("diamond")]=2.9"kJ mol"^(-1) The standard state means that the pressure should be 1 bar and substance should be pure at a given temperature. The conversion of graphite to dimond reduces its volume by 2xx10^(-6) m^(3) mol^(-1) . If C (graphite) is converted to C isothermally at T=298 K, the pressure at which C (graphite) is in equilibrium with C (diamond) is (1J=1 kg m^(2)s^(-2),1 Pa=1kg m^(-1)s^(-2), 1bar=10^(5) Pa)
Answer» `58001` BAR `1450` bar `14501` bar `29001` bar Solution :`DeltaG=VDeltap-SDeltaT,AsDeltaT=0,DeltaG=VDeltap` `:.G_(gr)-G_(gr)^(@)=V_(gr)DeltaP "and"G_(dim)-G_(dim)^(@)=V_(dim)DeltaP` `:.(V_(dim)-V_(gr))DeltaP=(G_(dim)-G_(gr))+(G_(gr)^(@)-G_(dim)^(@))` at eqb. `G_(dim)=G_(gr)` `:.(V_(dim)-V_(gr))DeltaP=G_(dim)^(@)-G_(gr)^(@)=2.9xx10^(3)` CALCULATE `Deltap,i.e.,p-p_(0)`,"then p"i.e.,(p_(0)=1).`