Question

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are
ㅿ????G⁰ [C (graphite)]= 0 kJmol⁻¹
ㅿ????G⁰ [C (daimond)]= 29 kJmol⁻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 x 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⁻¹ s⁻²; 1 bar = 10⁵ Pa]
[A] 14501 bar [B] 58001 bar
[C] 1450 bar [D] 29001 bar

Answer 1

Answer:

this is which class question