The enthalpy of combustion of glucose, `C _(6) H _(12) O _(6)(s)` is `-2816 kJ mol^(-1)` at `25^(@)C.` Calculate `Delta H_(f) ^(@)` for glucose. The `Delta H _(f) ^(@)` values for `CO_(2)` (g) and `H_(2)`O(l) are -393.5 and -285.9 `kJ mol^(-1),` respectively.

Explanation

`C _(6) H _(12) O _(6) (s) + 6O _(2) (g) to 6 CO _(2) (g) + 6 H _(2) O (l), Delta H =-2816kJ`
`Delta H = sum Delta H _(f) ^(@)` (products) `-sum Delta H _(f) ^(@)` (reactions) ,
`-2816kJ = 6 ( - 393.5 kJ mol ^(-1)) + 6 (- 285 . 9 kJ mol ^(-1)) - Delta H _(f) ^(@) ( C _(6) H _(12) O _(6)) - 6 Delta H _(f) ^(@) (O _(2))`
On simplification and recalling that `Delta H _(f) ^(@) (O _(2)) = 0,`
`Delta H _(f) ^(@) (C _(6) H _(12) O _(6)) =- 1260 . 4 kJ mol ^(-1).`