The heat evolved on combustion of 1g starch `(C _(6) H _(10) O _(5)) _(n)` into `CO _(2) and H _(2) O _((l)) is 17.49 kJ g ^(-1).` Compute the enthalpy of formation of 1g starch. Given `Delta H _(f) ^(@) of H _(2) O _((l)) =- 285.85 kJ mol ^(-1), Delta H _(f) ^(@) of CO _(2) = - 392.5 kJ mol ^(-1).`

Options

A. `-5.9 kJ mol ^(-1)`
B. `-11. 8 1 kJ mol ^(-1)`
C. `- 2. 95 kJ mol ^(-1)`
D. `+ 11 . 81 kJ mol ^(-1)`

Explanation

`H _(2) + (1)/(2) O _(2) to H _(2) O (1) : Delta H _(f) ^(@) =- 285. 85 kJ mol ^(-1) : C + O _(2) to CO _(2) (l) Delta H _(f) ^(@) =- 293.7 kJ mol ^(-1)`
Since the heat evolved on combustion of 1 g sarch is `17.49 kJ,` the heat evolved on combustion of 162 n g starch is `17.49 xx 162 n = 2833 . 38 n kJ mol ^(-1).` The molecular mass of starch `( C _(6) H _(10) O _(5)) _(n) = 162 n.` Therefore, for the reaction
`(C _(6) H _(10) O _(5)) _(n) + 6 n O _(2) to 6 N CO _(2) + 5n H _(2) O , Delta H _(f) ^(@) = - 2833 . 3 8 n kJ mol ^(-1) " "...(1)`
For the reaction `5n H_(2) + ( 5n )/ (2)O_(2) + 6n C + 6n O _(2) to 5n H _(2) O + 6n CO _(2)`
The heat of reaction can be clculated from the given heat of formation of `H _(2) O and CO _(2)` as
`Delta _(r) H = 5n Delta H _(f) (H_(2)O) + 6n Delta H _(r) (CO _(2))`
`Delta _(r) H = 5xx ( - 285.85) + 6 ( - 393. 5) =- 3790. 25 n kJ mol ^(-1) " "...(2)`
The heat of formation of starch can be calculated by ubstacting Eq. (1) from Eq. (2)
`6n C + 5n H _(2) + ( 5n )/(2) O _(2) to (C _(6) H _(10) O _(5)) , Delta H _(I) ^(@) =- 956.87 n KJ mol ^(-1)`
Heat of formation of 1 g of starch `= - ( 956. 87n )/(162 n ) = - 5. 9 kJ mol ^(-1)`