1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Standard enthalpy of formation

  1. Oct 28, 2015 #1
    1. The problem statement, all variables and given/known data

    I have the following task:

    The Standard enthalpy of formation of gaseous H2O at 298K is -241.82 kJ mol-1. Estimate its value at 100 °C given the following values of the molar heat capacities at constant pressure: H2O (g): 33.58 JK-1mol-1, H2 (g): 28.84 JK-1mol-1, O2 (g): 29.37 JK-1mol-1. Assume that the heat capacities are independent of temperature.

    2. Relevant equations

    3. The attempt at a solution

    The only formula I can think of is this one:


    But I don't know how to calculate this exactly. Can some help me getting started?

  2. jcsd
  3. Oct 28, 2015 #2
    First, you must devise a pathway to get from reactants at 100C, through the reaction at 25C, and finally to the product at 100C.
  4. Oct 28, 2015 #3
    Ok, thx

    is it correct that the standard enthalpie of H2 and O2 are 0 because of the definition?

    Does this mean I that I can use my formula in this way?

  5. Oct 28, 2015 #4
    All correct.
  6. Oct 28, 2015 #5
    Therefore the ΔH(T1) is -241.82 kJ mol-1, the Cp(H2O) is 33.58 JK-1mol-1, the Cp(H2) is 28.84 JK-1mol-1, the Cp(O2) is 29.37 JK-1mol-1 and I just need to insert these values in the equation above and I get the result for ΔH at 373K and therefore my final result?
  7. Oct 28, 2015 #6
    Check all your units for consistency.
  8. Oct 28, 2015 #7
    I think everything is fine, I just need to transform ΔH(T1) from -241.82 kJ mol-1 to -241'820 Jmol-1 and then it should be consistent, I think
  9. Oct 28, 2015 #8
    Go for it! (Note: If you had left it as kJ, you'd have gotten a much different answer. Experience will tell you that heats of reaction don't change drastically with moderate temperature change, like here, so a "wild" answer is a tipoff that something's wrong.)
  10. Oct 28, 2015 #9
    Thanks a lot!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted