Question
Explain the enthalpy of solution of an ionic compound.
✓
Answer
An ionic compound dissolves in a polar solvent like water in two steps as follows :
Step-I : The ions are separated from the molecule involving crystal lattice enthalpy $\Delta_L H$.
$MX _{( s )} \rightarrow M _{( g )}^{+}+ X _{( g )}^{-} \quad \Delta_{ L } H$
$\Delta_{ L } H$ is always positive.
Step-II : The gaseous ions are hydrated with water molecules involving hydration energy, $\Delta_{\text {hyd }} H$.
$\begin{aligned}
& M _{( g )}^{+}+ xH _2 O _{( l )} \rightarrow\left[ M \left( H _2 O \right)_{ x }\right]^{+} \\
& X _{( g )}^{-}+ yH _2 O _{(l)} \rightarrow\left[ X \left( H _2 O \right)_{ y }\right]^{-}
\end{aligned}$
$\Delta_{\text {hyd }} H$ is always negative.
The enthalpy change $\Delta$ solnH of solution is given by,
$\Delta_{\text {soln }} H=\Delta_L H +\Delta_{\text {hyd }} H$
For example, consider enthalpy of solution of $NaCl ( s )$.
$\Delta_{ L } H _{ NaCl }=790 kJ mol ^{-1}$
$\Delta_{\text {hyd }} H _{ NaCl }=-786 kJ mol ^{-1}$
Hence enthalpy change for solution of $NaCl ( s )$ is,
$\begin{aligned}
& \Delta_{\text {soln }} H =\Delta_L H _{ NaCl }+\Delta_{\text {hyd }} H _{ NaCl } \\
& =790+(-786) \\
& =+4 kJ mol ^{-1}
\end{aligned}$
Therefore dissolution of $NaCl$ in water is an endothermic process.
Step-I : The ions are separated from the molecule involving crystal lattice enthalpy $\Delta_L H$.
$MX _{( s )} \rightarrow M _{( g )}^{+}+ X _{( g )}^{-} \quad \Delta_{ L } H$
$\Delta_{ L } H$ is always positive.
Step-II : The gaseous ions are hydrated with water molecules involving hydration energy, $\Delta_{\text {hyd }} H$.
$\begin{aligned}
& M _{( g )}^{+}+ xH _2 O _{( l )} \rightarrow\left[ M \left( H _2 O \right)_{ x }\right]^{+} \\
& X _{( g )}^{-}+ yH _2 O _{(l)} \rightarrow\left[ X \left( H _2 O \right)_{ y }\right]^{-}
\end{aligned}$
$\Delta_{\text {hyd }} H$ is always negative.
The enthalpy change $\Delta$ solnH of solution is given by,
$\Delta_{\text {soln }} H=\Delta_L H +\Delta_{\text {hyd }} H$
For example, consider enthalpy of solution of $NaCl ( s )$.
$\Delta_{ L } H _{ NaCl }=790 kJ mol ^{-1}$
$\Delta_{\text {hyd }} H _{ NaCl }=-786 kJ mol ^{-1}$
Hence enthalpy change for solution of $NaCl ( s )$ is,
$\begin{aligned}
& \Delta_{\text {soln }} H =\Delta_L H _{ NaCl }+\Delta_{\text {hyd }} H _{ NaCl } \\
& =790+(-786) \\
& =+4 kJ mol ^{-1}
\end{aligned}$
Therefore dissolution of $NaCl$ in water is an endothermic process.
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