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.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Identify which of the following pairs has larger entropy? Why?
(a) $He _{( g )}$ in a volume of $1 L$ or $He _{( g )}$ in a volume of $5 L$ both at $25^{\circ} C$.
(b) $O _{2( g )}$ at $1 atm$ or $O _{2( g )}$ at $10 atm$ both at the same temperature.
(c) $C _2 H _5 OH _{( l )}$ or $C _2 H _5 OH _{( g )}$
(d) $5 mol$ of $Ne$ or $2 mol$ of $Ne$.
→(a) $He _{( g )}$ in a volume of $1 L$ or $He _{( g )}$ in a volume of $5 L$ both at $25^{\circ} C$.
(b) $O _{2( g )}$ at $1 atm$ or $O _{2( g )}$ at $10 atm$ both at the same temperature.
(c) $C _2 H _5 OH _{( l )}$ or $C _2 H _5 OH _{( g )}$
(d) $5 mol$ of $Ne$ or $2 mol$ of $Ne$.
Describe the manufacturing of $H_2SO_4$ by the contact process.
→Write the structure of the products obtained from the following ketones by action of hydrazine in presence of (1) slightly acidic medium (2) strong base KOH.
Explain the construction and working of a dry cell (or Leclanche’s cell).###Write a note on dry cell.
Explain the origin of magnetic properties in solids.
What is the action of Grignard reagent, $CH _3- Mg - I$ on : (1) formaldehyde (2) acetone?
→Explain and define the term vapour pressure of a liquid.
###
What is vapour pressure of a liquid?
###
What is vapour pressure of a liquid?
Write a note on Clemmensen reduction.Explain Clemmeusen’s reduction.Explain the reduction of carbonyl group into methylene group.
Predict whether the following
reactions would occur spontaneously
under standard state conditions.
a. $Ca_{(s)} + Cd^{2+}_{(aq)} \rightarrow Ca^{2+}_{(aq)} + Cd_{(s)}$
b. $2 Br-_{(s)} + Sn^{2+}(aq) \rightarrow Br_2_{(l)} + Sn_{(s)}$
c. $2Ag_{(s)} + Ni^{2+}_{(aq)} \rightarrow 2 Ag^+_{(aq)} + Ni_{(s)}$
→reactions would occur spontaneously
under standard state conditions.
a. $Ca_{(s)} + Cd^{2+}_{(aq)} \rightarrow Ca^{2+}_{(aq)} + Cd_{(s)}$
b. $2 Br-_{(s)} + Sn^{2+}(aq) \rightarrow Br_2_{(l)} + Sn_{(s)}$
c. $2Ag_{(s)} + Ni^{2+}_{(aq)} \rightarrow 2 Ag^+_{(aq)} + Ni_{(s)}$
The following redox reaction occurs in a galvanic cell.
$2 Al _{( s )}+3 Fe ^{2+}(1 M ) \rightarrow 2 Al ^{3+}(1 M )+3 Fe _{( s )}$
(a) Write the cell notation.
(b) Identify anode and cathode
(c) Calculate $E _{\text {cell }}^0$ if $E _{\text {anode }}^0=-1.66 V$ and $E _{\text {cathode }}^0=-044 V$
(d) Calculate $\Delta G^0$ for the reaction.
→$2 Al _{( s )}+3 Fe ^{2+}(1 M ) \rightarrow 2 Al ^{3+}(1 M )+3 Fe _{( s )}$
(a) Write the cell notation.
(b) Identify anode and cathode
(c) Calculate $E _{\text {cell }}^0$ if $E _{\text {anode }}^0=-1.66 V$ and $E _{\text {cathode }}^0=-044 V$
(d) Calculate $\Delta G^0$ for the reaction.