The standard state Gibba free energies of formations of C("graphite") and ("diamond") at ``T =298k are `DeltafG^(@) [ C("graphite") = " O "kJ mol^(-1)``DeltafG^(@) [ C("graphite") = " O "kJ mol6^(-1)` The standed state means that the presses should be 1 bar, and substance of graphite [C(graphite)] to diamond [C(diamond )] reduces its volume by `2xx 10^(10) m^(-1)` If C(graphite) is converted to C(diamond ) isothemally at T= 298k the pressure at which C( graphite) is in equilibrium with C(diamond) ,is [ useful infromation : `1 J - Kgm^(2) s^(-2) ` ` 1pa = 1kg m^(-1) s^(-2) : 1^(-) = 10 ^(5) pa]`A. 58001 barB. 1450 barC. 14501 barD. 29001

The standard state Gibba free energies of formations of C("graphite")

1.	The standard state Gibba free energies of formations of C("graphite") and ("diamond") at ``T =298k are `DeltafG^(@) [ C("graphite") = " O "kJ mol^(-1)``DeltafG^(@) [ C("graphite") = " O "kJ mol6^(-1)` The standed state means that the presses should be 1 bar, and substance of graphite [C(graphite)] to diamond [C(diamond )] reduces its volume by `2xx 10^(10) m^(-1)` If C(graphite) is converted to C(diamond ) isothemally at T= 298k the pressure at which C( graphite) is in equilibrium with C(diamond) ,is [ useful infromation : `1 J - Kgm^(2) s^(-2) ` ` 1pa = 1kg m^(-1) s^(-2) : 1^(-) = 10 ^(5) pa]`A. 58001 barB. 1450 barC. 14501 barD. 29001
Answer» Correct Answer - c

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