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The standard state Gibbs free energies of formation of C ( graphite) and C( diamond) at T=298 K are Delta_(f) G^(@) [C ( graphite) = 0 kJmol^(-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 ata given temperature. The conversionofgraphite [C(graphite ) ]to diamond [ C ( diamond)] reducesits volumeby 2 xx 10^(-6) m^(2) mol^(-1). If C( graphite0 is convertedto C( diamond) isothermally at T= 298 K , the pressure at which C( graphtie) is in equilibriumwith C ( diamond) is [ Useful information : 1 J = 1kgm^(2) s^(-2), 1 Pa=1 kg m^(-1) s ^(-2), 1 "bar" = 10^(5) Pa] |
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Answer» 58001 bar For isothermal process at 298 K, `DeltaT= 0 :.S d T =0` `:. dG =VdP` Starting from initial pressureof1 bar, if required pressure is P bar, then `int_(1)^(P)dG =int_(1)^(P) VdP `or`DELTAG= V [P]_(1)^(P)` `[V=` constant becausesolids are involved] `= V(P-1)` or`G-G^(@) = V( P-1)` `:. `For the process , C ( graphite)`rarr C( `DIAMOND) `Delta_(r)G= [G_("diamond")^(@) +V_(d) (P-1)]-[G_("graphite")^(@) + V_(g)(P-1)]` `= [ G_("diamond")^(@)-G_("graphite")^(@)]+ (P-1)(V_(d) -V_(g))` WhenC ( graphite) and C ( diamond) are in equilibrium, `Delta_(r)G=0` `0= ( 2.9 xx10^(3)-0)+( P-1) 10^(5) ( - 2 xx 10^(-6))` or`P -1 = ( 2.9 xx 10^(3))/( 10^(5) xx ( 2 xx 10^(-6)))=14500 ` bar or `P =14501 ` bar |
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