posted by nicholas .
If heat of reaction, Ca(s) + 2H+(aq) Ca2+(aq) + H2(g) , is - 544 kJ/mol, heat of reaction, CaO(s) + 2H+(aq) Ca2+(aq) + H2O(l) , is -192 kJ/mol. and heat of formation of water is -286, what is heat of CaO?
i found that the formation of CaO is -638. However, I'm confused about the next step to find delta H for the end equation.
You're finished. That IS delta H for the end reaction which is the delta H formation for CaO.
Eqn1 is Ca(s)+2H+ ==> Ca^+2 +H2(g)
eqn2 reversed is
H2O + Ca+2 ==> 2H+ + CaO
eqn 3 is H2 + 1/2 O2 ==> H2O
I omitted most of the (aq) (g) and (s) to save time.
Note 2H+ on left in eqn 1 cancels with 2H+ in eqn 2 on right.
H2 in eqn 2 on right cancels with H2 on left in eqn 3.
H2O on left in eqn 2 cancels with H2O
on right in eqn 3.
Ca(II) ion on right in eqn 1 cancels with Ca(II) ion on left in equn 2. All of that leaves.
Ca(s) + 1/2 O2(g) ==> CaO(s) delta H = -638 kJ/mol
If all of that confuses you, and it's easy to get confused with so many different ions and solids, just add th three equations I wrote. I have already reversed eqn 2. Then note that there are some ions and gases that appear on both sides and in the same quantity. Those can be canceled leaving the final equation I have above. I find it easier, however, to look on the left side somewhere in the equations I have, then find one like it on the right side and cancel it. Then I look for the next one etc until I have canceled everything I can. Then I write what is left. Works every time although it gets a little confusing if one of the equations must be multiplied by some constant. Fortunately, that step wasn't necessary in this problem.