Using the following thermochemical data, calculate ΔHf° of Yb2O3(s)?

Question

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

Naomi Aronson · Accepted Answer

Before we even do anything, we can usually look up #DeltaH_f^@# ahead of time to know what to expect... And we find that #(b)# should be it. (pg. 25). This is using Hess's Law. We can 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 #"Yb"_2"O"_3(s)#, i.e. forming #"1 mol"# of #"Yb"_2"O"_3(s)# starting from its elements in their elemental state at #25^@ "C"# and #"1 bar"#: #2"Yb"(s) + 3/2"O"_2(g) -> "Yb"_2"O"_3(s)# To achieve this,  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 #3/2"O"_2(g)# on the reactants' side, as it would be in the final reaction. 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")#

Wyatt Adkins · Answer

undefined To calculate ΔHf° of Yb2O3(s), we can use the following thermochemical dataTo calculate ΔHf° of Yb2O3(s), we can use the following thermochemical

2Yb(s) + 3/2O2(g) → Yb2O3(s) ΔH° = -1969.4 kJ/mol

ΔHf° of Yb2O3(s) can be calculated using the equation:

ΔHf°(Yb2O3) = ΣΔHf°(products) - ΣΔHf°(reactants)

Given that the only product is Yb2O3(s), and the reactants are Yb(s) and O2(g), we substitute the values:

ΔHf°(Yb2O3) = ΔH°(Yb2O3) - [ΔHf°(Yb) + ΔHf°(O2)]

ΔHf°(Yb2O3) = -1969.4 kJ/mol - [0 + 0]

ΔHf°(Yb2O3) = -1969.4 kJ/mol