For which of the following reactions is ΔH∘rxn equal to ΔH∘f of the product(s)?
2Na(s)+Cl2(g)→2NaCl(s)
H2(g)+12O2(g)→H2O(g)
Na(s)+12Cl2(g)→NaCl(s)
H2O2(g)→12O2(g)+H2O(g)
2H2(g)+O2(g)→2H2O(g)
Na(s)+12Cl2(l)→NaCl(s)
2Na(s)+Cl2(g)→2NaCl(s)
H2(g)+12O2(g)→H2O(g)
Na(s)+12Cl2(g)→NaCl(s)
H2O2(g)→12O2(g)+H2O(g)
2H2(g)+O2(g)→2H2O(g)
Na(s)+12Cl2(l)→NaCl(s)
1 Answer
Jan 14, 2018
Well, which of these are proper formation reactions that make one mol of product?
First of all, many of these are imbalanced. They should be:
#1)# #2"Na"(s) + "Cl"_2(g) -> 2"NaCl"(s)#
#2)# #"H"_2(g) + 1/2"O"_2(g) -> "H"_2"O"(g)#
#3)# #"Na"(s) + 1/2"Cl"_2(g) -> "NaCl"(s)#
#4)# #"H"_2"O"_2(g) -> 1/2"O"_2(g) + "H"_2"O"(g)#
#5)# #2"H"_2(g) + "O"_2(g) -> 2"H"_2"O"(g)#
#6)# #"Na"(s) + 1/2"Cl"_2(l) -> "NaCl"(s)#
- As mentioned, there is by convention
#"1 mol"# of product made. So we can eliminate#(1)# and#(5)# immediately. - They must also use elements in their elemental state, and chlorine is a gas. Thus, we strike out
#(6)# . - They obviously must form a compound to be a nontrivial formation reaction, so that eliminates
#(4)# , which is the opposite of a formation, AND water is not an element in its elemental state.
Why are