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?
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
Yes, you are correct. To find the enthalpy change, or delta H value, for a reaction, you can use Hess's Law, which states that the overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps of the reaction.
In this case, you have three equations with their respective enthalpy values. You can manipulate these equations by multiplying them by appropriate coefficients to make them add up to the formation reaction of MgO. In this way, you can use the known enthalpy changes to calculate the enthalpy change for the formation of MgO.
However, in this specific case, since the equations are already balanced, you do not need to manipulate them. You can directly use the enthalpy change values for the reactions to find the enthalpy change for the formation of MgO.
For the first equation: Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g), you can calculate the enthalpy change (delta H1) by subtracting the sum of the enthalpies of the reactants from the sum of the enthalpies of the products.
delta H1 = [enthalpy of MgCl2(aq) + enthalpy of H2(g)] - [enthalpy of Mg(s) + enthalpy of 2HCl(aq)]
Similarly, for the second equation: MgO(s) + 2HCl(aq) -> MgCl2(aq) + H2O(l), you can calculate the enthalpy change (delta H2) in the same way.
delta H2 = [enthalpy of MgCl2(aq) + enthalpy of H2O(l)] - [enthalpy of MgO(s) + enthalpy of 2HCl(aq)]
By using experimental data or tabulated values for the enthalpies of the individual species involved in the reactions, you can compute the enthalpy changes delta H1 and delta H2. Then, you can use these values along with the known enthalpy change for the formation of H2(g) + 1/2O2(l) to calculate the enthalpy change for the formation of MgO.