dHrxn = (n*dHf products) - (n*dHf reactants)
-43 = (1*52.3) - [(1*-285.6) + (dHf C2H4)]
Solve for dHf C2H4. Post your work if you get stuck.
Calculate the enthalpy of formation of ethene C2H4 from the given information
C2H4+H2O------ C2H5OH DH=-43kJmol
Enthalpy of formation of water And ethanol are -285.6kJmol and 52.3kJmol respectively
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