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

Options

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

Answer

c

Explanation

`dG= V dP -SdT`
For isothermal process at 298 K, `DeltaT= 0 :.S d T =0`
`:. dG =VdP`
Starting from initial pressureof1 bar, if required pressure is P bar, then
`int_(1)^(P)dG =int_(1)^(P) VdP `or`DeltaG= V [P]_(1)^(P)`
`[V=` constant because solids are involved]
`= V(P-1)`
or`G-G^(@) = V( P-1)`
`:. `For the process , C ( graphite)`rarr C( `diamond)
`Delta_(r)G= [G_("diamond")^(@) +V_(d) (P-1)]-[G_("graphite")^(@) + V_(g)(P-1)]`
`= [ G_("diamond")^(@)-G_("graphite")^(@)]+ (P-1)(V_(d) -V_(g))`
WhenC ( graphite) and C ( diamond) are in equilibrium, `Delta_(r)G=0`
`0= ( 2.9 xx10^(3)-0)+( P-1) 10^(5) ( - 2 xx 10^(-6))`
or`P -1 = ( 2.9 xx 10^(3))/( 10^(5) xx ( 2 xx 10^(-6)))=14500 ` bar
or `P =14501 ` bar