Question

The enthalpy of dissolution (or solution) of sodium chloride is $4KJmo{l^{ - 1}}$ and its enthalpy of hydration of ions is $- 784KJmo{l^{ - 1}}$ . What will be the lattice enthalpy of sodium chloride?
A. $+ 780KJmo{l^{ - 1}}$
B. $+ 394KJmo{l^{ - 1}}$
C. $+ 788KJmo{l^{ - 1}}$
D. $+ 398KJmo{l^{ - 1}}$

Answer 1

The term ‘Enthalpy’ is a thermodynamic function. It is introduced to study the total heat changes at constant pressure. Enthalpy is a sum of internal energy and a product of pressure and volume. It is represented by H. Enthalpy of hydration $\left( {{\Delta _{Hyd}}H} \right)$ is defined as the quantity of heat released or absorbed. Here it is the heat absorbed or released when one mole of anhydrous salt combines with more moles of water.

Complete step by step answer:
Write the given quantities from the question in standard form.
${\Delta _{Sol}}H = + 4KJmo{l^{ - 1}},{\Delta _{Hyd}}H = - 784KJmo{l^{ - 1}}$
Enthalpy of Solution $\left( {{\Delta _{Sol}}H} \right)$ is defined as the change in heat during the dissolution of one mole of a substance above a solvent such that further addition of solvent will not produce any change in heat. After that change in heat will be constant.
$NaCl\left( s \right) + aq \to NaCl\left( {aq} \right);{\Delta _{Sol}}H = + 4KJmo{l^{ - 1}}$ $NaCl\left( s \right) + {H_2}O\left( l \right) \to N{a^ + }\left( {aq} \right) + C{l^ - }\left( {aq} \right);{\Delta _{Hyd}}H = - 784KJmo{l^{ - 1}}$

Now we need to calculate the lattice enthalpy of sodium chloride but what it means. Lattice enthalpy of Sodium chloride ( ${\Delta _{lattice}}H$ ) is defined as the lattice enthalpy of an ionic compound is the enthalpy change one mole of compound dissociates in its gaseous state
$N{a^ + }C{l^ - }\left( s \right) \to N{a^ + }\left( g \right) + C{l^ - }\left( g \right);{\Delta _{lattice}}H = ?$

Formula used ${\Delta _{Sol}}H = {\Delta _{lattice}}H + {\Delta _{Hyd}}H,$
We got the above equation as $\left( {{\Delta _{Sol}}H} \right)$ can be obtained in two process enthalpy of lattice ${\Delta _{lattice}}H$ Followed by the enthalpy of hydration $\left( {{\Delta _{Hyd}}H} \right)$ .
Now for 1 mole of $NaCl$,
We have, ${\Delta _{Sol}}H = + 4KJmo{l^{ - 1}},{\Delta _{Hyd}}H = - 784KJmo{l^{ - 1}}$
Using the above formula we get
${\Delta _{Sol}}H = {\Delta _{lattice}}H + {\Delta _{Hyd}}H,$
After arranging we get,
$\Rightarrow {\Delta _{lattice}}H = {\Delta _{Sol}}H - {\Delta _{Hyd}}H,$
$\Rightarrow {\Delta _{lattice}}H = + 4KJmo{l^{ - 1}} - \left( { - 784KJmo{l^{ - 1s}}} \right),$
$\Rightarrow {\Delta _{lattice}}H = + 788KJmo{l^{ - 1}}$
The lattice enthalpy of sodium chloride is ${\Delta _{lattice}}H = + 788KJmo{l^{ - 1}}$

So, the correct answer is Option C.

Note: Symbol (aq) represents the excess of water.
The heat of solution $\left( {{\Delta _{Sol}}H} \right)$ of hydrated salts like $CuS{O_4}.5{H_2}O,CaC{l_2}.6{H_2}O$ or the salt which do not form hydrates like $NaCl,KCl,N{H_4}Cl$ is generally positive while for anhydrous salts which form hydrates like $CuS{O_4}$ is negative.