Example 5.12: Writing Reaction Equations for \(\mathrm{ΔHf°}\)
Write the heat of formation reaction equations for:
and
Solution
Remembering that
\(\mathrm{ΔHf°}\) reaction equations are for forming
1 mole of the compound from its constituent elements under standard conditions, we have:
\(\ce{?~C(s,graphite)}\)\(\ce{ + }\)\(\ce{~?~H2(g)}\)\(\ce{ + }\)\(\ce{~?~O2(g)}\)\(\ce{->}\)\(\ce{C2H5OH(l)}\)\(\ce{ }\)
for the first substance,
\(\ce{C2H5OH}\)
To balance this, count how many of each element are found in the product, and adjust the coefficients of the elements on the reactant side, i.e. we need
2 carbons,
6 hydrogens, and
1 oxygen. For oxygen, the coefficient has to be less than one (a fraction) because one
\(\ce{O2}\) molecule already contains
2 atoms of oxygen, but we need only one.
\(\ce{2C(s,graphite)}\)\(\ce{ + }\)\(\ce{3H2(g)}\)\(\ce{ + }\)\(\ce{1/2O2(g)}\)\(\ce{->}\)\(\ce{C2H5OH(l)}\)\(\ce{ }\)
and
\(\ce{3Ca(s)}\)\(\ce{ + }\)\(\ce{1/2P4(s)}\)\(\ce{ + }\)\(\ce{4O2(g)}\)\(\ce{->}\)\(\ce{Ca3(PO4)2(s)}\)\(\ce{ }\)
Note: The standard state of carbon is graphite, and phosphorus exists as
\(\ce{P4}\) (you can use
appendix G to find the standard forms of elements).