The standard enthalpies fo formation of `CO_(2) (g), H_(2) O (1)`, and glucose (s) at `25^(@)C` are `- 400 kJ mol^(-1), - 300 kJ mol^(-)`, and `- 1300 kJ mol^(-1)`, respectively. The standard enthalply of combustion per gram of glucose at `25^(@)C` is

Options

A. `+2900 kJ`
B. `- 2900 kJ`
C. `- 16.11 kJ`
D. `+ 16.11 kJ`

Answer

C

Explanation

`Delta_(C )H^(@` (standard heat of combustion) is the standard enthalpy change when one mole of the substance is completely oxidised.
Also standard heat of formation can be taken as the standard of that substance.
`H_(CO_2)^(@) = Delta_(f)H^(@)(CO_2) = -400 kJ moL^(-1)`
`H_(H_2O)^(@) = Delta_(f)H^(@)(H_2O) = -300 kJ mol^(-1)`
`H_("glucose")^(@) = Delta_(f)H^(@)("glucose") = -1300 kJ mol^(-1)`
`H_(O_2)^(@) = Delta_(f) H^(@)(O_2) = 0.00`
`C_(6)H_(12)O_(6)(s) + 6(O_2) (g) rarr 6CO_(2)(g) + 6H_(2)O(l)`
`Delta_(C )H^(@)("glucose") = 6[Delta_(f) H^(@)(H_2O)] - [ Delta_(f) H^(@)(C_6H_12 O_3) + 6Delta_(f) H^(@)(O_2)]`
`= 6[-400 - 300] - [-1300 + 6 xx 0] = -2900 kJ mol^(-1)`
Molar mass of `C_(6)H_(12)O_(6) = 180 g mol^(-1)`
Thus, standard heat of combustion of glucose per gram ` = (-2900)/(180) = -16.11 kJ g^(-1)`
To solve such problem, students are advised to keep much importance in unit conversion. As here value of R `(8.314 Jk^(-1) mol^(-1))` in `JK^(-1)mol^(-1)` must be converted into `kJ` by dividing the unit by 1000.