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?
Ca(s) etc is equation 1.
CaO + 2H^+ etc is equation 2.
H2 + 1/2 O2 ==> H2O is equation 3.
Write equation 1 as is.
Reverse equation 2 and change sign of delta H.
Write equation 3 as is.
Add all the equations (1, reversed 2, and 3) and see if that won't give you the formation of CaO; i.e., Ca(s) + 1/2 O2 ==> CaO.
Then add delta H1 + reversed delta H2 with sign changed + delta H 3 for H2O as is. The total should be the delta H for the end equation.
Post your work if you get stuck.
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.
To find the ΔH for the end equation, you will need to add up the values of ΔH from equation 1 (ΔH1), reversed equation 2 with the sign changed (ΔH2), and equation 3 (ΔH3).
Given that the formation of CaO is -638 kJ/mol, this means that ΔH1 = -638 kJ/mol.
From equation 2, by reversing it and changing the sign of ΔH, we have:
Ca2+(aq) + H2O(l) → CaO(s) + 2H+(aq)
Therefore, ΔH2 = -(-192 kJ/mol) = +192 kJ/mol.
From equation 3, H2 + 1/2 O2 → H2O, we have:
ΔH3 = -286 kJ/mol (given).
Now, add up all three values:
ΔH(end equation) = ΔH1 + ΔH2 + ΔH3
ΔH(end equation) = -638 kJ/mol + 192 kJ/mol + (-286 kJ/mol)
ΔH(end equation) = -732 kJ/mol
So the ΔH for the end equation, Ca(s) + (1/2)O2 → CaO(s), is -732 kJ/mol.