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
A. `+ 16.11 kJ`
B. `- 16.11 kJ`
C. `+ 2900 kJ`
D. `- 2900 kJ`
A. `+ 16.11 kJ`
B. `- 16.11 kJ`
C. `+ 2900 kJ`
D. `- 2900 kJ`
Correct Answer – B
Thermochemical equation for the combusion of glucose is
`C_(6) H_(12) O_(6) (s) + 6O_(2) (g) rarr 6 CO_(2) (g) + 6 H_(2) + 6 H_(2) O (1)`
According to thermodynamics,
`Delta_(r) H^(@) = sum a_(i) Delta_(f) H^(@)` (products) `- sum b_(i) Delta_(f) H^(@)` (reactants)
`Delta_(C ) H^(@) = [(6 mol) Delta_(f) H^(@) (CO_(2), g) + (6 mol) Delta_(f) H^(@) (H_(2) O, 1)] – [(1 mol) Delta_(f) H^(@) (C_(6) H_(12) O_(6), s) + (6 mol) Delta_(f) H^(@) (O_(2), g)]`
`= [6(-400)+6(-300)]-(-1300)`
`= – 2900 kJ mol^(-1)`
` = (- 2900 kJ mol^(-1))/(180 g mol^(-1))`
(molar mass of glucose = `180 g mol^(-1))`
`= 16.11 kJ g^(-1)`
Note that `Delta_(f) H^(@) (O_(2)) = 0`
Here students often make the mistake of not calculating the standard enthalpy of combustion per gram. In the examination hall, they just read the question as ….. standard enthalpy of combustion of glucose…