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)` `Delt

Question

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)` `Delt

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. 58001 bar
D. 1405 bar

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
We know
`DeltaG_(2)-DeltaG_(1)=DeltaV (P_(2)-P_(1))`
`2.9xx10^(3)-0=2xx10^(-6)(P_(2)-1)`
`(P_(2)-1)=(2.9xx10^(3))/(2xx10^(-6))Pa`
`=1.45xx10^(9)Pa`
`=1.45xx10^(4)" bar "(because"1 b ar"=10^(5)Pa)`
`P_(2)=14501` bar

5 views Answered 2023-02-05 15:29