Question

Using the following thermochemical equations, calculate the standard enthalpy
of combustion for one mole of liquid acetone (C 3 H 6 O).
3C(s) + 3H 2 (g) + 1 ⁄ 2 O 2 (g) ---> C 3 H 6 O(ℓ) ΔH° = −285.0 kJ
C(s) + O 2 (g) ---> CO 2 (g) ΔH° = −394.0 kJ
H 2 (g) + 1 ⁄ 2 O 2 (g) ---> H 2 O(ℓ) ΔH° = −286.0 kJ

Answer 1

From Hess law; The enthalpy of a given chemical reaction is constant, regardless of the reaction happening in one or many steps

$\therefore$ $\Delta H_{(reaction)}=\sum\Delta H_{(reactants)}H-\sum\Delta H_{(products)}$

$\Delta H_{(combustion)}=[3(-285kJ)+3(-394kJ)-1/2(-286kJ)]$

$=[(-855kJ)+(-1182kJ)-(-143kJ)]$

$=-1894kJ$