Using the following thermochemical data, calculate ΔHf° of Yb2O3(s)?
Using the following thermochemical data, calculate ΔHf° of Yb2O3(s).
2YbCl3(s) + 3H2O(l) → Yb2O3(s) + 6HCl(g) ΔH° = 408.6 kJ/mol
2Yb(s) + 3Cl2(g) → 2YbCl3(s) ΔH° = -1919.6 kJ/mol
4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(l) ΔH° = -202.4 kJ/mol
a)-1713.4 kJ/mol
b) -1814.6 kJ/mol
c) 2530.6 kJ/mol
d) 1308.6 kJ/mol
e) -2125.8 kJ/mol
Using the following thermochemical data, calculate ΔHf° of Yb2O3(s).
2YbCl3(s) + 3H2O(l) → Yb2O3(s) + 6HCl(g) ΔH° = 408.6 kJ/mol
2Yb(s) + 3Cl2(g) → 2YbCl3(s) ΔH° = -1919.6 kJ/mol
4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(l) ΔH° = -202.4 kJ/mol
a)-1713.4 kJ/mol
b) -1814.6 kJ/mol
c) 2530.6 kJ/mol
d) 1308.6 kJ/mol
e) -2125.8 kJ/mol
1 Answer
Before we even do anything, we can usually look up
And we find that
This is using Hess's Law. We can
- multiply coefficients all the way through by a constant
#c# , giving#cDeltaH_i^@# for that step. - reverse a reaction so that
#DeltaH_i^@ -> -DeltaH_i^@# . - add arbitrary reaction steps together as long as all the intermediates cancel out and the resultant reaction is achieved.
It doesn't matter if it's using an obscure Lanthanide oxide, or some stratospheric ozone destruction, it works out the same way.
We begin with:
#2"YbCl"_3(s) + 3"H"_2"O"(l) → "Yb"_2"O"_3(s) + 6"HCl"(g)# ,#" "DeltaH_1^@ = "408.6 kJ/mol"#
#2"Yb"(s) + 3"Cl"_2(g) → 2"YbCl"_3(s)# ,#" "DeltaH_2^@ = -"1919.6 kJ/mol"#
#4"HCl"(g) + "O"_2(g) → 2"Cl"_2(g) + 2"H"_2"O"(l)# ,#" "DeltaH_3^@ = -"202.4 kJ/mol"#
We simply have to cancel out intermediates and form the reaction we want. We want the standard formation of
#2"Yb"(s) + 3/2"O"_2(g) -> "Yb"_2"O"_3(s)#
To achieve this,
- We can simply take
#3/2DeltaH_3^@# to achieve#3/2"O"_2(g)# on the reactants' side and cancel out the#6"HCl"(g)# from step 1. It also cancels out the#3"H"_2"O"(l)# from step 1, and the#3"Cl"_2(g)# from step 2. #2"Yb"(s)# is already on the reactants' side, so we do not touch#DeltaH_2^@# .#"Yb"_2"O"_3(s)# is already on the products' side, so we do not touch#DeltaH_1^@# .- The
#2"YbCl"_3(s)# cancels out on its own.
Everything else works itself out.
#" "cancel(2"YbCl"_3(s)) + cancel(3"H"_2"O"(l)) → "Yb"_2"O"_3(s) + cancel(6"HCl"(g))#
#" "2"Yb"(s) + cancel(3"Cl"_2(g)) → cancel(2"YbCl"_3(s))#
#ul(3/2(cancel(4"HCl"(g)) + "O"_2(g) → cancel(2"Cl"_2(g)) + cancel(2"H"_2"O"(l))))#
#" "2"Yb"(s) + 3/2"O"_2(g) -> "Yb"_2"O"_3(s)#
And all of this came out because we just searched for the easiest way to get
Thus:
#color(blue)(DeltaH_(f,"Yb"_2"O"_3(s))^@) = DeltaH_(rxn)^@#
#= DeltaH_1^@ + DeltaH_2^@ + 3/2DeltaH_3^@#
#= "408.6 kJ/mol" + (-"1919.6 kJ/mol") + 3/2(-"202.4 kJ/mol")#
#= color(blue)(-"1814.6 kJ/mol")#