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Homework Help: Thermodynamics: internal energy change

  1. Oct 14, 2011 #1
    1. The problem statement, all variables and given/known data

    Enthalpy of formation of a mole of atomic hydrogen = 218kJ. Enthalpy changes when a mole of atomic hydrogen is formed by dissociating half a mole of molecular hydrogen. Calculate ΔU of the process of molecular hydrogen dissociation.

    2. Relevant equations

    ΔH = ΔU + Δ(PV)

    3. The attempt at a solution

    ΔU = ΔH - Δ(PV)
    ΔU = ΔH - Δ(nRT)
    ΔU = 218kJ - (1/2 mol)(8.31 J/Kmol)(298K)
    ΔU = 216kJ

    I really don't think my answer is correct.
  2. jcsd
  3. Oct 15, 2011 #2
  4. Oct 16, 2011 #3
  5. Oct 16, 2011 #4

    Andrew Mason

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    You will need to know the temperature at which the disassociation occurs. So I assume you are referring to the standard enthalpy of formation, which occurs T = 25C.

    dU = dH - d(PV) = dH - PdV -VdP

    If the process occurs at constant pressure (eg. at atmospheric pressure) the VdP term is zero. Alternatively, if it occurs at constant volume, the PdV term is zero. Let's assume that it occurs at constant pressure of one atmosphere (and temperature of 25C = 298K).


    dU = dH - PdV

    Since there are double the number of molecules after disassociation, the volume must double, as you note. I think you should be able to take it from there.

  6. Oct 16, 2011 #5
    so then

    ΔU = ΔH -P(dv)
    ΔU = ΔH - ∫ nRT/V dv
    ΔU = ΔH - nRT ln(vf/vi)
    ΔU = 218kJ - 1 mol * 8.21 J/k Mol *298 K * ln2
    ΔU = 216kJ ?
    Last edited: Oct 16, 2011
  7. Oct 16, 2011 #6
  8. Oct 16, 2011 #7
    ΔH formation is +'ve

    shouldn't my answer be negative since the system has less potential energy.
    1/2 H2 -> H
  9. Oct 16, 2011 #8

    Andrew Mason

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    Since pressure is constant, ∫PdV = PΔV

  10. Oct 16, 2011 #9

    Andrew Mason

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    The system has greater potential energy since it takes energy to separate the hydrogen atoms. (In forming H2 from H atoms, heat would be given off).

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