# 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

Jan 18, 2018

Before we even do anything, we can usually look up $\Delta {H}_{f}^{\circ}$ ahead of time to know what to expect...

This is using Hess's Law. We can

• multiply coefficients all the way through by a constant $c$, giving $c \Delta {H}_{i}^{\circ}$ for that step.
• reverse a reaction so that $\Delta {H}_{i}^{\circ} \to - \Delta {H}_{i}^{\circ}$.
• 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:

$2 \text{YbCl"_3(s) + 3"H"_2"O"(l) → "Yb"_2"O"_3(s) + 6"HCl} \left(g\right)$, $\text{ "DeltaH_1^@ = "408.6 kJ/mol}$

$2 {\text{Yb"(s) + 3"Cl"_2(g) → 2"YbCl}}_{3} \left(s\right)$, $\text{ "DeltaH_2^@ = -"1919.6 kJ/mol}$

$4 \text{HCl"(g) + "O"_2(g) → 2"Cl"_2(g) + 2"H"_2"O} \left(l\right)$, $\text{ "DeltaH_3^@ = -"202.4 kJ/mol}$

We simply have to cancel out intermediates and form the reaction we want. We want the standard formation of ${\text{Yb"_2"O}}_{3} \left(s\right)$, i.e. forming $\text{1 mol}$ of ${\text{Yb"_2"O}}_{3} \left(s\right)$ starting from its elements in their elemental state at ${25}^{\circ} \text{C}$ and $\text{1 bar}$:

$2 {\text{Yb"(s) + 3/2"O"_2(g) -> "Yb"_2"O}}_{3} \left(s\right)$

To achieve this,

• We can simply take $\frac{3}{2} \Delta {H}_{3}^{\circ}$ to achieve $\frac{3}{2} {\text{O}}_{2} \left(g\right)$ on the reactants' side and cancel out the $6 \text{HCl} \left(g\right)$ from step 1. It also cancels out the $3 \text{H"_2"O} \left(l\right)$ from step 1, and the $3 {\text{Cl}}_{2} \left(g\right)$ from step 2.
• $2 \text{Yb} \left(s\right)$ is already on the reactants' side, so we do not touch $\Delta {H}_{2}^{\circ}$.
• ${\text{Yb"_2"O}}_{3} \left(s\right)$ is already on the products' side, so we do not touch $\Delta {H}_{1}^{\circ}$.
• The $2 {\text{YbCl}}_{3} \left(s\right)$ 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))
$\underline{\frac{3}{2} \left(\cancel{4 \text{HCl"(g)) + "O"_2(g) → cancel(2"Cl"_2(g)) + cancel(2"H"_2"O} \left(l\right)}\right)}$
${\text{ "2"Yb"(s) + 3/2"O"_2(g) -> "Yb"_2"O}}_{3} \left(s\right)$

And all of this came out because we just searched for the easiest way to get $\frac{3}{2} {\text{O}}_{2} \left(g\right)$ on the reactants' side, as it would be in the final reaction.

Thus:

$\textcolor{b l u e}{\Delta {H}_{f , {\text{Yb"_2"O}}_{3} \left(s\right)}^{\circ}} = \Delta {H}_{r x n}^{\circ}$

$= \Delta {H}_{1}^{\circ} + \Delta {H}_{2}^{\circ} + \frac{3}{2} \Delta {H}_{3}^{\circ}$

= "408.6 kJ/mol" + (-"1919.6 kJ/mol") + 3/2(-"202.4 kJ/mol")

$= \textcolor{b l u e}{- \text{1814.6 kJ/mol}}$