The standard state Gibb's free energies of formation of C (graphite) and C (diamond) at...
The standard state Gibb's free energies of formation of C (graphite) and C (diamond) at T=298 K are
ΔfGo[C (graphite]=0 kJ mol-1
ΔfGo[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 [C (graphite)] to diamond [C (diamond)] reduces its volume by 2×10-6 m3 mol-1 . If C (graphite) is converted to C (diamond) isothermally at T=298 K, the pressure at which C (graphite) is in equilibrium with C (diamond), is
[Useful information: 1 J=1 kg m2 s-2, 1 Pa=1 kg m-1s-2; 1bar=105Pa]
ΔfGo[C (graphite]=0 kJ mol-1
ΔfGo[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 [C (graphite)] to diamond [C (diamond)] reduces its volume by 2×10-6 m3 mol-1 . If C (graphite) is converted to C (diamond) isothermally at T=298 K, the pressure at which C (graphite) is in equilibrium with C (diamond), is
[Useful information: 1 J=1 kg m2 s-2, 1 Pa=1 kg m-1s-2; 1bar=105Pa]
Solution:
C(graphite)→C(diamond); ΔGo=ΔfGodiamond-ΔfGographite=2.9 kJ/mol at 1 bar
As dGT=V.dP
ΔG2∫ΔG1d(ΔGT)= P2∫P1ΔV.dP
ΔG2-ΔG1=ΔV. (P2-P1)
(2.9×103-0)=(-2×10-6) (1-P2)
P2-1=2.9×1032×10-6Pa=1.45×104 bar
P2=14501 bar.
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