1.

Calculate the standard heat of formation of propane, if its heat of combustion is-2220.2 kJ "mol"^(-1) , the heats of formation of CO_(2(g)) and H_2O_((l)) are - 393.5 and -285.8 kJ "mol"^(-1) respectively.

Answer»

SOLUTION :`C_(3)H_(8)+5O_(2) rarr2CO_(2)+4H_(2) O`
`DeltaH_(C)^(@) = -2220.2 "kJ .mol"^(-1)""…(1)`
`C+O_(2) rarrCO_(2)`
`DeltaH_(f)^(@)=-393.5 "kJ.mol"^(-1)""..(2)`
`H_(2)+1/2O_(2)rarrH_(2)O`
`DeltaH_(f)^(@) = -285.8 " kJ mol"^(-1)""..(3)`
`3C+4H_(2)rarrC_(3)H_(8)`
`DeltaH_(f)^(@)=? `
`(2)xx3rArr3C+3O_(2)rarr3CO_(2)`
`DeltaH_(f)^(@)=-1180.5kJ ""...(4)`
`(3)xx4rArr4H_(2)+2O_(2)rarr4H_(2)O`
`DeltaH_(f)^(@)=-1143.2 KJ""....(5)`
`(4)+(5)-(1)rArr3C+3O_(2)+4H_(2)+2O_(2)+3CO_(2)+4H_(2)Orarr3CO_(2)+4H_(2)+C_(3)H_(8)+5O_(2)`
`DeltaH_(f)^(@)=-11805-1143.2-(-2220.2)kJ`
`3C+4H_(2)rarrC_(3)H_(8)`
`DeltaH_(f)^(@)=-103.5kJ`
STANDARD heat of formation of PROPANE is
`DeltaH_(f)^(@)(C_(3)H_(8))=-103.5 kJ`.


Discussion

No Comment Found