Hess's Law question...



    H2O(g) → H2O(l) ∆H (kJ/mole) = –44
    C(s) + O2(g) → CO2(g) –394
    H2(g) + 1/2O2(g) → H2O(l) –286
    C2H5OH(l)+ 3O2(g) → 2CO2(g) + 3H2O(l) –1367

    Calculate the enthalpy change for the reaction:
    2C(s) + 2H2(g) + H2O(l) → C2H5OH(l)
    A. –226 kJ/mole
    B. +7 kJ/mole
    C. +109 kJ/mole
    D. +344 kJ/mole
    E. +687 kJ/mole


    The answer is B, but in the answer the first rxn where H2O(g) goes to H2O(l) is not involved in the calculation to arrive at choice B. Why not?



    Don't know what i'm doing
    10+ Year Member
    5+ Year Member
    Jul 11, 2005
      You only have to use the rxns that will give you the final rxn with the appropriate coefficients. You don't need to use the first rxn to get the final rxn given. If you had used the first rxn, there's no other reaction that has H2O(g) so you wouldn't be able to cancel it out to get the final reaction.
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        2 C(s) + 2 O2(g) → 2 CO2(g) 2(–394)
        H2(g) + O2(g) → 2 H2O(l) 2(–286)
        2 CO2(g) + 3 H2O(l) → C2H5OH(l)+ 3 O2(g) –(–1367)
        2 C(s) + 2 H2(g) + H2O(l) → C2H5OH(l) ∆H= 2(–394) + 2(–286) – (–1367)
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