Saved Bookmarks
| 1. |
The standard state Gibbs free energies of formation of C (graphite) and C(diamond) at T=298 K areDelta_(f)G^(@)[C("graphite")]=0kJ mol^(-1)Delta_(f)G^(@)[C("diamond")]=2.9 kJ 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 with C (graphite) is in equilibrium with C (diamond), is[Useful information : 1J=1kg m^(2)s^(-2), 1Pa=1kg m^(-1)s^(-2)," 1 bar = "10^(5) Pa] |
|
Answer» <P>58001 BAR `therefore dG=VdP` `underset(1)overset(P)intdG=underset(1)overset(P)intVdPtherefore G-G^(@)=V(P-1)""[because " Solids involved"therefore" V ALMOST constant"]` `therefore Delta_(r)G=[G_("diamond")^(@)+V_(d)(P-1)]-[G_("graphite")^(@)+V_(g)(P-1)]` `0=2.9xx10^(3)+(P-1)10^(5)(-2xx10^(-6))""thereforeP=14501 " bar"` |
|