Determination of the enthalpy of combustion of Magnesium using Hess's law

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

    ok so i'm trying to find the change in enthalpy for this reaction:
    Mg(s) + (1/2)O2--> MgO(s)

    I've done the lab to find the temperatures of certain reactions, these were:
    9.70 Celsius for the reaction Mg(s) + 2HCl(aq)--> MgCl2(aq) + H2(l)

    7.60 Celsius for the reaction MgO(s) + 2HCl --> MgCl2(aq) + H2O(l) reaction

    (and we know the enthalpy of change for the reaction
    H2(g) + (1/2)O2(g) --> H2O(l) is 285.8 kJ)

    2. Relevant equations

    change in enthalpy = mass (100ml/1000ml) * specific heat (4.18) * change in temp (above)

    3. The attempt at a solution

    using the (above) equation i found that the enthalpy of change for the first reaction was -405 kJ per mol and the second was -318 kJ per mol.

    so i built a Hess's law chart and found that [Mg(s) + 2HCl(aq)--> MgCl2(aq) + H2(l)] + [H2(g) + (1/2)O2(g) --> H2] = [MgO(s) + 2HCl --> MgCl2(aq) + H2O(l)] + [Mg(s) + (1/2)O2--> MgO(s)]

    so basically the equation look like this: -405 + (-285.8) = -318 + x

    therefore x should be -372.8 kJ per mol

    but all the answers i've found on the net so far say that the enthalpy of combustion is approximately 602 kJ per mol

  2. jcsd
  3. Borek

    Staff: Mentor

    -372.8 kJ/mol agrees with your experimental data, so if it doesn't agree with tables - you have to look for experimental error.
  4. dang...ouch 38% error

    well...i guess thats what you get when you use Styrofoam insulators...
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