The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298K are Δf G∘ [C(graphite)]=0 kJ mol−1 Δf G∘ [C(diamond)]=2.9KJ 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 2×10−6m3 mol−1. If C(graphite) is converted to C(diamond) isothermally at T = 298K, the pressure at which C(graphite) is in equilibrium with C(diamond), is [Useful information: 1J=1kg m2 s−2], 1 Pa=1 kgm−1s−2; 1 bar=105 Pa

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

1.	The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298K are Δf G∘ [C(graphite)]=0 kJ mol−1 Δf G∘ [C(diamond)]=2.9KJ 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 2×10−6m3 mol−1. If C(graphite) is converted to C(diamond) isothermally at T = 298K, the pressure at which C(graphite) is in equilibrium with C(diamond), is [Useful information: 1J=1kg m2 s−2], 1 Pa=1 kgm−1s−2; 1 bar=105 Pa
Answer» The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298K are $Δ_{f} G^{\circ} [C (g r a p h i t e)] = 0 k J m o l^{- 1}$ $Δ_{f} G^{\circ} [C (d i a m o n d)] = 2.9 K J m o l^{- 1}$ 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 $2 \times 10^{- 6} m^{3} m o l^{- 1}$ . If C(graphite) is converted to C(diamond) isothermally at T = 298K, the pressure at which C(graphite) is in equilibrium with C(diamond), is [Useful information: $1 J = 1 k g m^{2} s^{- 2}$ ], $1 P a = 1 k g m^{- 1} s^{- 2}; 1 b a r = 10^{5} P a$