posted by Vince on .
I've attempted this problem but not sure if i'm on the right track.
Use the enthalpy diagram provided above and apply Hess’s Law to determine the standard enthalpy of formation for C12O36H20N12 (s) using the results from part (a) and the following values:
The standard enthalpy of formation of gaseous carbon dioxide is -393.5 kJ/mol
The standard enthalpy of formation of liquid water is -286 kJ/mol
(Strictly speaking, standard enthalpy is defined for a reaction that occurs at 25 °C, but for the purpose of this question you may ignore the difference in temperature.)
So i got that part a which asked for the heat evolved during the decomposition reaction of C12O36H20N12 (s) → 12CO2 (g) + O2 (g) + 6N2 (g) + 10H2O (l) the molar enthalpy change is -756.5kJ.
I use delta h(rxn)=delta H(products-reactants)
The product is C12O36H20N12 (s) which is part a and is -756.5
the reactants are 12( -3935.5kJ)+10(-286kJ) with N2(g) and O2(g) being in their standard which is just zero.
Therefore we have -756.5kJ-(12( -3935.5kJ)+10(-286kJ)) which equals 6822.5kJ for the standard enthalpy of formation for C12O36H20N12. hints/tips are greatly appreciated