Question

The lattice energy of $NaCl{}_{(S)}$ is – 790 $kJmo{l^{ - 1}}$ and the enthalpy of hydration is - 785$kJmo{l^{ - 1}}$. Calculate the enthalpy of the solution of $NaCl{}_{(S)}$.

Answer 1

According to Hess’s Law, the change in enthalpies in any given chemical reaction, is not dependent on the path that is observed between the initial and the final states of the reaction. This law allows us to indirectly calculate the enthalpy change in a reaction, even when we cannot actually physically measure those changes.

Complete step by step answer:
The chemical reaction for the hydration of sodium chloride or NaCl can be given as follows:
$NaC{l_{(s)}} + aq\xrightarrow{{{\Delta _{sol}}H}}N{a^ + }_{(aq)} + C{l^ - }_{(aq)}$
From this reaction we can observe that the bond between the sodium and the chlorine atom is broken due to the introduction of an aqueous medium. The data that has been given to us is lattice energy of $NaCl{}_{(S)}$ is – 790 $kJmo{l^{ - 1}}$ and the enthalpy of hydration is -785$kJmo{l^{ - 1}}$.
In order to relate these to values, let us consider one case where the NaCl molecule dissociates into Na gas and Cl gas, since the lattice energy is given and this value is equivalent to the energy. Hydration of these gaseous ions would also yield similar results with the same hydration energy required. Hence,

$$NaC{l_{(s)}} + aq\xrightarrow{{{\Delta _{sol}}H}}N{a^ + }_{(aq)} + C{l^ - }_{(aq)} \\\ \downarrow \\\ \left[ N{a^ + }_{(g)} \\\ C{l^ - }_{(g)} \\\ \right]\xrightarrow{{hydration}}N{a^ + }_{(aq)} + C{l^ - }_{(aq)} \\\$$

Hence, ${\Delta _{sol}}H$ = 790 -785 = 5 $kJmo{l^{ - 1}}$

Note:
If a process is written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. Hess's law is valid because enthalpy is a state function.