Question

If the enthalpy of formation of $HCl_{(g)}$ and $C{l^ - }_{(aq)}$ are $ - 92kJ\;mo{l^{ - 1}}$ and $ - 167.44kJ\;mo{l^{ - 1}}$, find the enthalpy of solution of hydrogen chloride gas.
(A) $ - 75.14\;kJ\;mo{l^{ - 1}}$
(B) $ + 75.14\;kJ\;mo{l^{ - 1}}$
(C) $ - 260.7\;kJ\;mo{l^{ - 1}}$
(D) None of these

Answer 1

Hint: As we know that the enthalpy of formation is the change in enthalpy that takes place when one mole of product is formed from its element which are present in their standard state in its most stable allotropic form.

Complete answer:
As we know that standard heat of formation involves the formation of a product from the elements which are present in their standard forms like some elements which are present in their standard state of formation includes the graphite, white phosphorus, rhombic sulphur and oxygen in gaseous state. Similarly chlorine and hydrogen in their standard states forms compounds like hydrochloric acid.
We also know that hydrochloric acid is formed by the reaction of hydrogen gas and chlorine gas which we can show with the help of the following equation as below:
${H_2} + C{l_2} \to 2HCl$ and we are given that the enthalpy of formation of hydrochloric acid gas is $ - 92kJ\;mo{l^{ - 1}}$.
We are also given in the question that formation of aqueous chloride have the enthalpy of formation as $ - 167.44kJ\;mo{l^{ - 1}}$ and we can show this formation of chloride as given below:
$C{l_2} \to 2C{l^ - }$

Now, we have to find the enthalpy of solution of hydrogen chloride and we know that hydrogen chloride is formed as shown in the given reaction below:
$\Delta {H_{{H^ + }}} + \Delta {H_{solution}} \to \Delta {H_{chloride}}$
So putting the values of enthalpies of formation we can get the enthalpy of solution as:
$
\Rightarrow ( - 92) + \Delta {H_{solution}} \to - 167.44 \\
\Rightarrow \Delta {H_{solution}} = - 167.44 + 92 \\
\Rightarrow \Delta {H_{solution}} = - 75.14 kJ\;mo{l^{ - 1}} \\
 $

Therefore, from the above explanation we can say that the correct answer is, ‘(A) $ - 75.14\;kJ\;mo{l^{ - 1}}$’.

Note: Always remember that the enthalpy of solution which is also called the change in enthalpy of dissolution is basically the change in enthalpy associated with the dissociation of any substance which is present in solvent. It can be exothermic as well as endothermic.