Question

The molar enthalpy of combustion of carbon, hydrogen and buta-l ,
3-diene (CH2 = CHCH = CH2) are - 394, - 288 and - 2542 KJ/mol respectively. By
using a suitable thermochemical diagram or otherwise find the enthalpy of formation of buta-1, 3-diene.

Answer 1

C + O2 = CO2 $\Delta$H1 = -394 kJ/mol

H2 + 1/2O2 = H2O $\Delta$H2 = -288 kJ/mol

C4H6 + O2 = 4CO2 + 3H2O $\Delta$H3 = -2542 kJ/mol

4C + 3H2 = C4H6 $\Delta$H4 = ?

$\Delta$H4 = - $\Delta$H3 + 4*$\Delta$H1 + 3*$\Delta$H2 = 2542 - 4*394 - 3*288 = 102 kJ/mol