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Enthalpy of a Reaction NH4Cl

  1. Apr 11, 2008 #1
    [SOLVED] Enthalpy of a Reaction

    1. The problem statement, all variables and given/known data
    2. Relevant equations

    So we are studying the reaction NH4Cl (s) ---> NH3 (g) + HCl (g) and we have it in 4 steps:

    1) NH3 (aq) + HCl (aq) ---> NH4Cl (aq) + H2O (l)

    enthalpy of reaction per mole NH3: 1569.54 J/mol
    enthalpy of reaction per mole HCl: 3358.73 J/mol

    2) NH4Cl (s)+ H2O (l)---> NH4Cl (aq) + H2O (l)
    enthalpy of reaction per mole NH4Cl: -16129.01 J/mol

    3) NH3 (g) ---> NH3 (aq)
    delta H=-34640 J/mol

    4) HCl (g) ---> HCl (aq)
    delta H= -75140 J/mol

    3. The attempt at a solution

    I understand that you need to add up the enthalpies, but the sign part confuses me. I am not sure how to pick a sign at each step. Also, for the first step, do I add those two values together to get an overall value for that equation? That's what I though, but I'm not sure.

    My idea was add the two from 1) and make it negative, add to it 2) (no sign change), and add to it 3) and 4), both made negative. This results in 88722.75 J/mol, but I'm not sure if I picked the signs correctly.

    Thank you.
  2. jcsd
  3. Apr 11, 2008 #2
    What you want to do is rearrange the equations so that things cancel out and you're left with the overall equation. When flipping a particular equation, you have to flip the sign of the change in enthalpy with that equation and then add all the resulting enthalpies.
  4. Apr 12, 2008 #3
    Hess's law of constant heat summation.
  5. Apr 12, 2008 #4
    Thank you for stating the obvious.....
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