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

  1. Nov 23, 2008 #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

    help?
     
  2. jcsd
  3. Nov 23, 2008 #2

    Borek

    User Avatar

    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. Nov 23, 2008 #3
    dang...ouch 38% error

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