Hi TNT, you wouldnt happen to be studying through the uni of New England (Australia) would you?
Ive got the exact same question that im tackling now, the reactions you chose are not formation reactions which, if you have the same question is what there asking for, in other words the reactants have to appear as they would in there standard form, e.g, 2HCl is wrong because both chlorine and hydrogen gas are diatomic molecules (Cl2, H2).
my working on this question so far, which could be completely wrong, is:
the question states:
Using Hess's law, write out all of the formation reactions that add up to and calculate delta H (25C) for the following reaction:
2NaHCO3(s) --------> Na2CO3(s) + CO2(g) + H2O(l)
rxn 1: C(s) + O2(g) ------> CO2(g) Delta H rxn= -393.51 KJ
rxn 2: H2(g) + 1/2O2(g) -----> H2O(l) Delta H rxn= -285.83 KJ
rxn 3: 2Na(s) + C(s) + O2(g) + 1/2O2(g) ----->Na2CO3(s) Delta H rxn= -1130.77KJ
rxn 4: 2Na(s) + H2(g) + 2C(s) + O2(g) + 1/2O2(g)----> 2NaHCO3(s) Delta H rxn= -1901.62 KJ
so far all of the products we need are on the product side of the formation reactions, which is where we want them. all we need to do to get the sole reactant (the 2NaHCO3(s)) onto the reactant side of the formation reaction, this is done by reversing rxn 4 and changing its sign in the process.
so: Delta H rxn (4) = 1901.62 KJ
and by subtracting the sum of the enthalpy change of the products from the reactants, delta H for the final rxn can be calculated:
Delta Hrxn = [(-1130.77KJ)+(-393.51KJ)+(-285.83KJ)] - 1901.62KJ
the value i end up with seems a little high, so ive probably gone about this question the wrong way totally, if anyone can help me and TNT on this q it would be greatly appreciated.