1.

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 barB. 58001 barC. 1450 barD. 29001 bar

Answer» Correct Answer - A


Discussion

No Comment Found

Related InterviewSolutions