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

`C_(2)H_(12)O_(6)(s) + 6O_(2)(g) to 6CO_(2)(g) + 6H_(2)O(l)`
`DeltaH^(@)_(comb) = Sigma DeltaH_(f)^(@)("products") = SigmaDeltaH_(f)^(@)("reactants")`
`[6 xx DeltaH_(f)^(@)(CO_(2)) + 6DeltaH_(f)^(@)(H_(2)O)]` - `[DeltaH_(f)^(@)(C_(6)H_(12)O_(6)) + 6DeltaH_(f)^(@)(O_(2))]`
` = [ 6 xx (-400) + 6(-300)] - [-1300 + 6(0)]`
` = -2400 - 1800 + 1300 = -2900 kJ "mol"^(-1)`
Standard enthalpy of combustion per gram` = (-2900)/(180) = -16.11 kJ "g^(-1)`