The combustion of ethene, C ₂H ₄, is given by the equation C ₂H ₄ + 30 ₂ → 2CO₂ + 2H₂ O; ∆H = - 1428 KJ. If the molar heats of formation of water and carbon (IV) oxide are - 286 kJ and - 396 kJ respectively, calculate the molar heat of formation of ethene in KJ

Efranklyn

5 years ago

C2H4 +3O2 ---------> 2CO2 + 2H2O ∆H= -1428kJ ---------(1)
H2 + 1/2O2 ---------> H2O ∆H = -286kj -----------(2)
C + O2 ----------> CO2 ∆H = -396kj ----------(3)
The formation of ethene is given by
2C + 2H2 ---------> C2H4 ∆H = xkj -----------(4)
Now applying Hess law
From eqn (4) we see that the formation must be on the r.h.s and also there are 2 moles of carbon and H2 in the formation of ethene.
Multiply eqn (2) and eqn (3) by 2 including their energies and rewrite eqn (1) to have the ethene on the r.h.s doing all that we have
2CO2 + 2H2O ----------> C2H4 + 3O2 ∆H = +1428kj
2C + 2O2 ----------> 2CO2 ∆H = -572kj
2H2 + O2 ----------> 2H2O ∆H = -792kj
Finally add the whole epression we have that
X = +1428 + ( -572-792)
X = 1428- 1364
X = +64 QED