(i) Determine `Delta_(f)H^(@) (NO, g)` at `25^(@)C`. Using the following information
`Delta_(f)H^(@) (CO_(2), g)= -393.5` kJ/mol
`{:(2NO(g)+O_(2)(g) rarr 2NO_(2)(g)(g),,,Delta_(r)H^(@)=-114.0" kJ/mol"),(2CO(g)+O_(2)(g) rarr 2CO_(2)(g),,,Delta_(r)H^(@)=-566.0" kJ/mol"),(4CO(g)+2NO_(2)(g) rarr4CO_(2)(g),+,N_(2)(g), Delta_(r)H^(@)=-1198.4" kJ/mol"):}`
(ii) Calculate the equilibrium pressure (in Pascal) for the conversion of grapgite to diamond at `25^(@)C`. The densities of graphite and diamond may be taken to be 2.20 and 3.40 g/cc respectively independent of pressure.
Given : `DeltaG^(@) (C("graphite") rarr C ("diamond"))=2900` J/mol.

Explanation

`{:(2CO(g)+O_(2)(g) rarr 2CO_(2)(g)," "Delta_(r)H^(@)=-566.0" kJ/mol"),(NO_(2)(g) rarr NO(g) +1//2O_(2)(g)," "Delta_(r)H^(@)=114//2" kJ/mol"),(2CO_(2)(g)+1//2N_(2)(g) rarr 2CO(g) +NO_(2)(g) ," "Delta_(r)H^(@)=1198.4//2" kJ/mol"),(bar(1/2 N_(2)(g)+1/2 O_(2)(g) rarr NO(g)," "Delta_(f)H^(@) (NO, g)=90.2" kJ/mol")):}`
(ii) `DeltaG_(2)-DeltaG_(1)=DeltaV [P_(2)-P_(1)]`
`DeltaV=12xx[1/3.4-1/2.2]xx10^(-6) m^(3) mol^(-1)`
`DeltaV=-14.4/(3.4xx2.2)xx10^(-6)rArr -1.925xx10^(-6) m^(3) mol^(-1)`
Let `P_(2)` is equilibrium pressure , `DeltaG_(2)=0, P_(2)=1" bar"=10^(5) Pa`
`0-DeltaG_(1)=-1.925xx10^(-6) [P_(2)-1]`
`2900=1.925xx10^(-6) [P_(2)-P_(1)]`
`P_(2)=2900/(1.925xx10^(-6))+P_(1)`
`rArr 1506.5xx10^(6)+10^(5)`
`P_(2)=1.50xx10^(9) Pa`