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(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. |
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Answer» Solution :`{:(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` |
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