Question

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?

Answer 1

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

Answer 2

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?

Answer 3

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.

Answer 4

you are correct