Question

1. for the solutions formed below, calculate the standard enthalpy change, $delta h^{circ}$, according to the data provided in the table:

$kcl_{(s)} \to k^{1+}_{(aq)} + cl^{1-}_{(aq)}$
$delta h^{circ} = ___________ kj/mol$
$cacl_2 cdot 2h_2o_{(s)} \to ca^{2+}_{(aq)} + 2cl^{1-}_{(aq)}$
$delta h^{circ} = ___________ kj/mol$

Explanation:

Step1: Calculate $\Delta H^{\circ}$ for $KCl_{(s)}\to K^{+}_{(aq)}+Cl^{-}_{(aq)}$

Use the formula $\Delta H^{\circ}=\sum\Delta H^{\circ}_{formation}(products)-\sum\Delta H^{\circ}_{formation}(reactants)$.
For products: $\Delta H^{\circ}_{formation}(K^{+}_{(aq)})+\Delta H^{\circ}_{formation}(Cl^{-}_{(aq)})=-251.2+(-167.4)=-418.6\ kJ/mol$.
For reactant: $\Delta H^{\circ}_{formation}(KCl_{(s)})=-436.5\ kJ/mol$.
Then $\Delta H^{\circ}=-418.6-(-436.5)$.

Step2: Simplify the expression

$\Delta H^{\circ}=-418.6 + 436.5=17.9\ kJ/mol$.

Step3: Calculate $\Delta H^{\circ}$ for $CaCl_{2}\cdot2H_{2}O_{(s)}\to Ca^{2+}_{(aq)}+2Cl^{-}_{(aq)}$

For products: $\Delta H^{\circ}_{formation}(Ca^{2+}_{(aq)})+2\Delta H^{\circ}_{formation}(Cl^{-}_{(aq)})=-543.0+2\times(-167.4)=-543.0-334.8=-877.8\ kJ/mol$.
For reactant: $\Delta H^{\circ}_{formation}(CaCl_{2}\cdot2H_{2}O_{(s)})=-842.5\ kJ/mol$.
Then $\Delta H^{\circ}=-877.8-(-842.5)$.

Step4: Simplify the expression

$\Delta H^{\circ}=-877.8 + 842.5=-35.3\ kJ/mol$.

Answer:

$17.9\ kJ/mol$
$-35.3\ kJ/mol$