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. `+16.11kJ`

Answer

C

Explanation

`C_6H_12 O_6(s) + 6O_2(g) to 6CO_2(g) + 6H_2O(l)`
`Delta_c H = 6 Delta_f H(CO)_2 + 6 Delta_fH (H_2O) - Delta_f H (C_6H_12O_6)- 6Delta_f H (O_2,g)`
`= 6 xx (-400 - 300)- (-1300)-0`
` = -4200 + 1300`
` =-2900 kJ //mol`
For one gram of glucose, enthalpy of combustion
` = -(2900)/(180) = -16.11 kJ //g`.