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Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g) delta H1?

MgO(s) + 2HCl(aq) ->MgCl2(aq) + H2O(l) delta H2?

To make these equations add up to the formation reaction of MgO, you will need to include the following:

H2(g) + 1/2O2(l) delta H =-241.8 kJ

Using Hess's Law and manipulating these three equations and their respective enthalpy values you can solve for deltaH formation of MgO.

The equations above are already balanced so so I do not understand what they mean by manipulating the equations. To find the delta H value for the first two is this when I use the sum of products - the sum of reactants?

  • Chemistry - ,

    The only problem is that when I looked up the thermodynamic values I could not find MgCl2(aq) so would this just be 0?

  • Chemistry - ,

    For the first equation I did
    0 + 2(-167.159) -> 0 + 0
    0 - 334.318 = deltaH1 = -334.318

    Second equation I did
    -601.24 + (-334.318) -> -285.83
    -285.83 -(-935.558) = deltaH2 = 649.728

    Then deltaH3 = -241.8

    On my lab report it says to find the deltaH formation of MgO and it has two columns, one for Mg and MgO. So is deltaH1 my Mg and deltaH2 my MgO?

  • Chemistry - ,

    I will refer to the three equations you have above as eqn 1, eqn 2 and eqn 3.
    Write equn 1 as is/
    Reverse eqn 2.
    Eqn 3 as is
    Add the three equations together and you see that you get
    Mg(s) + 1/2 O2(g) ==> MgO(s)
    The delta H1 for eqn 1 is the delta H you measured.
    Delta H2 for eqn 2 is the delta H you measured but since you reversed the equation you want to change the sign. Eqn 3 has dH given.
    Add those 3 as I've shown and you obtain the dH formation for MgO.

  • Chemistry - ,

    you are correct

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