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Molar Heat of Neutralisation

  1. Dec 1, 2013 #1
    1. The problem statement, all variables and given/known data
    All solutions are (aq).

    In the reaction: HCl + NaOH → NaCl + H2O, ΔH = 60.4 kJ/mol
    In the reaction: CH3COOH + NaOH → CH3COONa + H20, ΔH = 51.9 kJ/mol
    In the reaction: CH3COOH + NH4OH → NH4CH3COO + H20, ΔH = 51.6 kJ/mol
    In the reaction: HCl + NH4OH → NH4Cl + H20, determine ΔH

    2. Relevant equations

    Hess's Law: ΔH(reaction) = ƩΔH(products) - ƩΔH(reactants)
    ΔH1 = ΔH(NaCl) + ΔH(H2O) - ΔH(HCl) - ΔH(NaOH)
    ΔH2 = ΔH(CH3COONa) + ΔH(H2O) - ΔH(CH3COOH) - ΔH(NaOH)
    ΔH3 = ΔH(NH4CH3COO) + ΔH(H2O) - ΔH(CH3COOH) - ΔH(NH4OH)
    ΔH4 = ΔH(NH4Cl) + ΔH(H2O) - ΔH(HCl) - ΔH(NH4OH)

    3. The attempt at a solution
    I've never quite understood how this works. All of our previous examples were like 2X + Y = 2B, and had a variation of all three variable in each equation. I think I understand it, and start the problem and 30 seconds later realise I don't understand it at all. I don't see how to make the connection between the first three reactions and the fourth. Please help.
     
  2. jcsd
  3. Dec 2, 2013 #2

    Borek

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    Staff: Mentor

    Apparently first equation must be taken as it is - that's the only way of having HCl on the left. It also looks like you should add third equation to the first - this way you will have both HCl and NH4OH on the left. Now you have to think how to use the third equation to get rid of the things you don't need.

    Could be writing them all as net ionic will help.

    Please remember you can subtract the reaction as well, you don't have to only add them.
     
  4. Dec 2, 2013 #3
    Hmmmm. I think I get it...but I've said that before. :)

    ΔH1-ΔH2+ΔH3 (Reactants): H + Cl + Na + OH - CH3CO - OH - Na - OH + CH3CO + OH + NH4 + OH
    ΔH1-ΔH2+ΔH3 (Reactants): H + Cl + NH4 + OH

    ΔH1-ΔH2+ΔH3 (Products): Na + Cl + 2H + O - CH3COO - Na - 2H - O + NH4 + CH3COO + 2N + O
    ΔH1-ΔH2+ΔH3 (Products): NH4 + Cl + 2H + O

    HCl + NH4OH → NH4Cl + H20

    Which proves that ΔH1-ΔH2+ΔH3 = ΔH4, so
    ΔH4 = 60.4 kJ/mol - 51.9 kJ/mol + 51.6 kJ/mol
    ΔH4 = 60.1 kJ/mol

    I looked it up and my results seems to be about 8 kJ/mol higher than it should, but we did this in lab.
     
  5. Dec 2, 2013 #4

    Borek

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    Staff: Mentor

    8 kJ/mol sounds like a reasonable experimental error to me :wink:
     
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