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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:

    HVSEQp6.png

    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?

    e9af3459ba109af49490024c1ccdc164.gif
     
  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!
     
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