Question

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are
$\Delta_fG^0$ [C(graphite)] = 0 kJ mol⁻¹
$\Delta_fG^0$ [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 m²s⁻²; 1 Pa = 1 kg m⁻¹ s⁻²; 1 bar = 10⁵ Pa]
(A) 14501 bar
(B) 58001 bar
(C) 1450 bar
(D) 29001 bar

Answer 1

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