Question #cd6d2
2 Answers
To calculate
Explanation:
Let us see it with an example:
We must know enthalpy of formation of all compounds and elements:
-
#Delta H_"form" ("CH"_4) = -17.9 " kJ/mol"# , from the reaction:
#"C" + 2 "H"_2 rightarrow "CH"_4# -
#Delta H_"form" ("H"_2"O") = -241.82 " kJ/mol"# , from the reaction:
#"H"_2 + 1/2 "O"_2 rightarrow "H"_2 "O"# -
#Delta H_"form" ("CO"_2) = -393.5 " kJ/mol"# , from the reaction:
#"C" + "O"_2 rightarrow "CO"_2# -
#Delta H_"form" ("O"_2) = 0 " kJ/mol"# , like every element at its natural state.
So, now:
where
Note: albeit there are other methods to obtain enthalpy (such as Hess's law, for example), I think this is the one which better adjusts to what you are looking for.
Comment - the link wouldn't show up in the Comments section.
Explanation:
Handy tables of compounds here:
http://bilbo.chm.uri.edu/CHM112/tables/thermtable.htm