posted by Hannah on .
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?
The only problem is that when I looked up the thermodynamic values I could not find MgCl2(aq) so would this just be 0?
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?
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.
you are correct