Question

The standard enthalpy change of the reaction:
C2H4(g)+H2O(g)=C2H5OH(l) can be calculated from standard enthalpy changes of combustion given below:
C2H4(g) DHc°= -1411Kjmol-1
C2H5OH(l). DHc°= -1367kjmol-1

Answer 1

Write the combustion reaction for C2H4. It is:
C2H4 + 3O2 ==> 2CO2 + 2H2O
Write the combustion reaction for C2H5OH. It is:
C2H5OH + 3O2 ==> 2CO2 + 3H2O
Now add the C2H4 combustion to the reverse of the C2H5OH combustion:

C2H4 + 3O2 + 2CO2 + 3H2O ==> 2CO2 + 2H2O+ C2H5OH + 3O2
Cancel the materials common to both sides to get the final:
C2H4 + H2O ==> C2H5OH
For dH for the rxn just add dHc for rxn 1 to the reverse of dHc for rxn 2.
-1411 kJ/mol +(+1367 kJ/mol) = ?