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Homework Help: Hess Law: Molar Enthalpy Change for Decomposition

  1. Apr 11, 2007 #1
    1. The problem statement, all variables and given/known data
    This is for a chemistry report. This is the task:
    "Determine the molar enthalpy change for the decomposition of sodium hydrogen carbonate into sodium carbonate, CO(2) and water.
    2NaHCO(3) -> Na(2)CO(3) + CO(2) + H(2)O
    This enthalpy change cannot be measured directly".

    Instead, we reacted NaHCO(3) and Na(2)CO(3) in separate containers with HCL (in excess (100 mL), about 1M).

    We measured the temperature changes, and the masses of each solid/liquid.

    For the NaHCO(3) reaction:
    Mass(NaHCO(3)): 0.001021 kg ± 0.000001 kg
    Mass(HCL): 0.100 ± 0.001 kg
    Change in T = 1 K

    For the Na(2)CO(3) reaction:
    Mass(Na(2)CO(3)): 0.000905 kg ± 0.000001 kg
    Mass(HCL): 0.100 ± 0.001 kg
    Change in T = -0.3 K

    2. Relevant equations
    Nr 1: 2NaHCO(3) -> Na(2)CO(3) + CO(2) + H(2)O
    Nr 2: NaHCO(3) + HCl -> NaCL + CO(2) + H(2)O
    Nr 3: Na(2)CO(3) + HCl -> 2NaCL + CO(2) + H(2)O

    Note: By multiplying equation 2 by two, reversing equation 3, and then adding them together, we get equation 1. This has to be done to the enthalpies I calculate in Part 3, too.

    3. The attempt at a solution
    My biggest problem is figuring out how to find the Molar Delta H in kJ/mol. By doing the following, I can calculate the approximate Delta H (only in kJ) for reaction 1:

    Following the equation of Heat = Mass * Specific Heat (c) * Delta T, we get:

    (Note, here I add the two masses of HCl and the solids, and then use the specific heat of water. I understand this holds quite a large uncertainty. If anybody has a better idea, please say so. I am also really unsure on what I do after calculating the heat, ie. when I invert the sign of the heat form + to - and vica versa. Is this correct?)

    For NaHCO(3):
    Mass = 0.101021 kg
    Specific Heat (water) 4186 J/(kg * K)
    Delta T = 1 K

    Heat = 422.87 J. As this is an exothermic reaction, it becomes -422.87 J

    For Na(2)CO(3):
    Mass = 0.100905 kg
    Delta T = -0.3 K

    Heat = -126.71 J. As this is an endothermic reaction, it becomes +126.71 J.

    Then, by doing the same to these numbers as is mentioned in Part 2 (the reactions), I end up with (-422.87 * 2) + (126.71*-1) = -970.71 J = -0.971 kJ!

    But this however, is just the enthalpy change for the reaction. Not the molar enthalpy change...
    Any comments on this would be absolutely wonderful, and greatly appreciated!
  2. jcsd
  3. Apr 12, 2007 #2


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    Gold Member

    The units for molar enthalpy change are kJ/mol. You have the KJ part for a reaction containing 0.001021 kg of NaHCO3. (the reaction where the delta T was 1K, -422.87J) Convert this to moles and you have the molar enthalpy of that part.
    You have the KJ part for a reaction consisting of 0.100905 kg of Na2CO3. (the reaction where delta T was -0.3K, 126.71J) Convert this to moles and you have the molar enthalpy of that part.


    It doesn't do you any good to subtract the enthalpies from two reactions that contain different amounts of materials (as you have done in your OP).
    I also assume that the CO2 gas was in thermal equilibrium with the solution before it bubbled away. That is, the evolved CO2 didn't carry away any heat energy ... You have also confounded the heats of reaction with the heats of solution. (a minor point)
  4. Apr 12, 2007 #3
    Thanks! I believe I've got it now. :)

    Anyway, it'll have to wait. I'm off to Kaliningrad in Russia with class now.

    Me and a couple of friends talked a bit, but your comment really cleared things up. Thank you very much!
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