The standard state Gibbs free energies of formation of C (Graphite) and C(diamond) at T = 298 K are
`Delta_(f)G^(@)["C(graphite)"]=0kJmol^(-1)`
`Delta_(f)G^(@)["C(diamond)"]=2.9kJmol^(-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 `2xx10^(-6)m^(3)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 [Use information : 1 J = `1kgm^(2)s^(-2),1Pa=1kgm^(-1)s^(-2),1"bar"=10^(5)Pa`]

Options

A. 29001 bar
B. 58001 bar
C. 14501 bar
D. 1450 bar

Answer

C