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The standard state Gibbs free energies o...

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]`

A

14501 bar

B

29001 bar

C

1450 bar

D

58001 bar

Text Solution

Verified by Experts

The correct Answer is:
a
`p= (DeltaG)/(DeltaV) = (2.9 kJ mol^(-1))/(2 xx 10^(-6) m^(3) mol^(-1)) = (2.9 xx 10^(3) Jmol^(-1))/(2 xx 10^(-6) m^(3) mol^(-1)) = 1.45 xx 10^(9) Nm^(-2)`
`P = (1.45 xx 10^(9))/10^(5) "bar" = 1.45 xx 10^(4)` bar =14500 bar
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