# Question c63e6

Mar 2, 2017

By using Hess's Law.

#### Explanation:

We are given the two equations

1. $\text{BaCl"_2"(s)" → "BaCl"_2"(aq)";color(white)(mmmmmmmmml) Δ_text(sol)H = "-20.6 kJ/mol}$
2. $\text{BaCl"_2·"2H"_2"O(s)" → "BaCl"_2"(aq)" + "2H"_2"O(l)"; Δ_text(sol)H = color(white)(l)"8.8 kJ/mol}$

From these, we must derive the target equation

$\text{BaCl"_2"(s)" + "2H"_2"O(l)" → "BaCl"_2·"2H"_2"O"; color(white)(mmm)Δ_text(hyd)H = ?}$

The target equation has ${\text{BaCl}}_{2}$ on the left, so we start with equation 1.

It also has $\text{BaCl"_2·2"H"_2"O}$ on the right, so we reverse equation 2 to get equation 3.

When we reverse an equation, we change the sign of its ΔH.

Finally, we add equations 1 and 3, canceling species that appear on opposite sides of the reaction arrow.

When we add two equations, we add their ΔH values.

This gives the target equation.

1. $\text{BaCl"_2"(s)" → color(red)(cancel(color(black)("BaCl"_2"(aq)"))); color(white)(mmmmmmmmml)Δ_text(sol)H =color(white)(l) "-20.6 kJ/mol}$
3. color(red)(cancel(color(black)("BaCl"_2"(aq)"))) + "2H"_2"O(l)" → "BaCl"_2·"2H"_2"O(s)"; Δ_text(sol)H =color(white)(ll) "-8.8 kJ/mol"
$\text{BaCl"_2"(s)" + "2H"_2"O(l)" → "BaCl"_2·"2H"_2"O"; color(white)(mmmll)Δ_text(hyd)H = "-29.4 kJ/mol}$

Thus, for the hydration of ${\text{BaCl}}_{2}$, Δ_text(hyd)H = "-29.4 kJ/mol"#.