How do you calculate the energy released when 24.8 g #Na_2O# reacts in this reaction: #Na_2O(s) + 2HI(g) -> 2Nal(s) + H_2O(l)#?

1 Answer
Sep 16, 2017

#"48.0 kcal"# was released... at constant pressure.


This can't be done without data on the enthalpy of reaction... And #"Nal"# doesn't exist. #"NaI"# does. A brief google search gives conflicting results:

#DeltaH = -"120.0 kcal"# [Source 1] [Source 2]
#DeltaH = -"12.00 kcal"# [Source]

As this is a strong acid-strong base reaction, I would never believe a mere #-"12.00 kcal"#... The #-"120.0 kcal"# is more physically reasonable. Furthermore, this should be #"kcal/mol"#, i.e. per mol of #"Na"_2"O"# (since it has a stoichiometric coefficient of #1# and is on the reactants' side).

Now, since energy is an extensive quantity, it scales with the size of the system. Therefore, when we find the mols of #"Na"_2"O"#, we know the scaling factor.

#24.8 cancel("g Na"_2"O") xx "1 mol"/(61.979 cancel("g Na"_2"O"))#

#= "0.400 mols Na"_2"O"#

Reading the reaction itself gives us #"1 mol Na"_2"O"#, #"2 mols HI"#, etc. As a result, the energy released by #"0.400 mols"# is just #40%# of the #"120.0 kcal"# that was for every #"1 mol"#:

#nDeltaH_"rxn" = 0.400 cancel"mols" xx -"120.0 kcal"//cancel"mol"#

#= color(blue)(q_"rxn" = -"48.02 kcal")#

or we say that #"48.02 kcal"# was released. And in what scenario? At constant pressure.