dHrxn = (n*dHo products) - (n*dHo reactants)
dHrxn = (2*dHo CO2 + 2*dHo H2O) - (1*dHo C2H4 + 3*dHo O2)
dHrxn = 564.4 kJ
You have dHo for all but O2 and that is zero. I would recognize that the 564.4 kJ is for 11.22 grams C2H4 so for a mole of C2H4 that will be
564.4 kJ x (28/11.22) = ? kJ/mol. Make that substitution for dHrxn and solve for dHo C2H4. Post your work if you get stuck.
C2H4(g)+ 3O2(g) --> 2 CO2 (g) + 2H2O (l)
The complete combustion of 11.22g of ethylene, C2H4 (g), via the given equation, produces 564,4 KJ of heat. Calculate the standard enthalpy of formation for C2H4(g), using the standard enthalpies of formation given in the table.
In the table,CO2(g) -393.5 kJ/moL, H2O (l) -285.8 kJ/mol
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