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

Answer

C

Explanation

`C("graphite") to C("diamond")`
`DeltaG^(@) = Delta_(f)G^(@)("diamond") - Delta_(f)G^(@)("graphite")`
`DeltaG^(@) = 2.9 - 0 = 2.9 kJ "mol"^(-1) = 2900 J "mol^(1)`
`DeltaV = 2 xx 10^(-6)m^(3)mol^(-1)`
`DeltaG^(@) = DeltaV.Deltap`
`2900 J mol^(-1) = (2 xx 10^(-6)m^(3) mol^(-1))(p_(e) -1)`
`p_(e) -1 = (2900)/(2 xx 10^(-6))= 1450 xx 10^(6)`
or `p_(e) - 1 = (1450 xx 10^(6))/(10^(5)) = 14500` bar
`p_(e) = 14500 + 1 = 14501` bar.