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Standard enthalpy of formation water

  1. Apr 17, 2015 #1
    1. The problem statement, all variables and given/known data
    The standard enthalpy of formation for liquid water is:
    H2 = 1/2O2→H2O ΔHf=-285.8 kJ/mol
    Which of the following could be the standard enthalpy of formation for water vapor?
    (A) -480.7 kJ/mol
    (B) -285.8 kJ/mol
    (C) -241.8 kJ/mol
    (D) +224.6 kJ/mol

    2. Relevant equations
    NA

    3. The attempt at a solution
    The equation above shows that condensation occurs, which is an exothermic process giving a negative ΔH. Perhaps I am not understanding the question because how can the standard enthalpy of formation of water vapor be determined from the above equation?
     
  2. jcsd
  3. Apr 17, 2015 #2
    Is the enthalpy of water vapor higher or lower than the enthalpy of liquid water? Is the heat of vaporization of water higher or lower than the heat given off when H2 and O2 react to form water?

    Chet
     
  4. Apr 17, 2015 #3
    Ok, so the enthalpy of fusion is lower in energy than the enthalpy of water vapor. ΔHvap is higher than ΔHfus. When I look at a heating curve, this true. Well, wouldn't we expect the standard enthalpy of formation of water vapor to be higher than the standard enthalpy of formation of liquid water based on a heating curve?
     
  5. Apr 17, 2015 #4
    Who said anything about fusion? How did ice get into the discussion? The standard heat of formation of either liquid water or water vapor is the result of forming chemical bonds between the hydrogen atoms and the oxygen atoms.

    Chet
     
  6. Apr 17, 2015 #5
    Alright, this is where I think my confusion is. My book does not clearly define these different ΔH's that well. We discussed standard heat of formation in a previous thread which defines as the change in enthalpy for a reaction that creates one mole of that compound from its raw elements in their standard state. We know that if the ΔHf is negative, then the products are more stable than the reactants. To see this more clearly, I wrote down the equations for what is known and what is being asked:

    H2(g) + 1/2O2(g) →H2O (l) ΔHf = -285.8 kJ/mol

    H2(g) + 1/2O2(g) →H2O(g) ΔHf = ?

    For the first equation, condensation occurs, which releases heat (this makes sense). But the second equation describes a gas going to a gas, so how can the sign for the ΔHf be determined? Would we assume that it is lower since it is not technically going through a phase change?
     
  7. Apr 17, 2015 #6
    The easiest way to do this is to use Hess' Law. What is ΔH for:

    H2O(l)-->H2O(v)

    Chet
     
  8. Apr 17, 2015 #7
    Well, I would write it down, but the question does not provide that information. That's why I'm so confused! They just give multiple choice answer to choose from so I assume that they want you to try to estimate the ΔHf.
     
  9. Apr 17, 2015 #8
    Do you think that more heat is given off when you react hydrogen and oxygen to form one mole of water, or when you condense one mole of water vapor to form one mole of liquid water? At least tell me what you think the sign of ΔH in post #6 is.
     
  10. Apr 17, 2015 #9
    Response for post #6: Condensation releases heat so ΔH <0 and vaporization requires energy so ΔH>0.
     
  11. Apr 17, 2015 #10
    Response to post #8: I would predict that condensation would be less since the water vapor is already in the H2O molecular structure.
     
  12. Apr 17, 2015 #11
    Both this and your previous post are correct. Now, can you apply Hess' law to the first reaction equation in post #5 and the phase change reaction in post #6 to determine the answer to the question posed in post #1?

    Chet
     
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