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=VdP-SdT " At 298 K, SdT = 0"`
`therefore dG=VdP`
`underset(1)overset(P)intdG=underset(1)overset(P)intVdPtherefore G-G^(@)=V(P-1)" "[because " Solids involved"therefore" V almost constant"]`
`therefore Delta_(r)G=[G_("diamond")^(@)+V_(d)(P-1)]-[G_("graphite")^(@)+V_(g)(P-1)]`
`0=2.9xx10^(3)+(P-1)10^(5)(-2xx10^(-6))" "thereforeP=14501 " bar"`