Question

The integral enthalpy of solution in ${\rm{kJ}}$of one mole of
${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$ is dissolved in ${\rm{n}}$ mole of water is
given by:

\+ {\rm{1}}{\rm{.8}}}}$$ Calculate $$\Delta {\rm{H}}$$for $${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$$dissolved in $${\rm{2}}\,{\rm{mole}}$$of $${{\rm{H}}_{\rm{2}}}{\rm{O}}$$. (A) $$ - {\rm{48}}{\rm{.54}}\,{\rm{kJ}}$$ (B) $$39.79\,{\rm{kJ}}$$ (C) $${\rm{48}}{\rm{.54}}\,{\rm{kJ}}$$ (D) $${\rm{75}}{\rm{.60}}\,{\rm{kJ}}$$

Answer 1

As we know that the solution is made from solute and solvent. When we dissolve solute then the heat is either absorbed or heat will be released it depends upon the interaction of solute-solvent and the nature of solute particles.

Step by step answer:
Now, let’s first talk about the Integral heat which is the heat released or absorbed when one mole of solute is dissolved in solvent to make a solution at constant pressure and temperature.
The value of ${\rm{\Delta }}{{\rm{H}}_{{\rm{soln}}}}$ depends upon the final concentration of solute in the solution. If the solvent is in large excess so that further dilution does not change the enthalpy of the solution, it is called an enthalpy of solution at infinite dilution. This is because at this state, the interaction between the ions of the solution becomes negligible, hence, no energy is absorbed or released.
These are the theory parts about integral heat, now come to our calculation.
So, here we are given a formula and values of variables so we can easily get our answer by simply putting the values in the formula.
Given:
${\rm{n}} = {\rm{2}}\,{\rm{mole}}$
Applying on the formula ${\rm{\Delta }}{{\rm{H}}_{\rm{s}}}{\rm{ = }}\dfrac{{{\rm{75}}{\rm{.6}} \times {\rm{n}}}}{{{\rm{n}} + {\rm{1}}{\rm{.8}}}}$
We get as;

{\rm{\Delta }}{{\rm{H}}_{\rm{s}}} = \dfrac{{{\rm{75}}{\rm{.6}} \times {\rm{2}}}}{{{\rm{2}} + {\rm{1}}{\rm{.8}}}}\\\ = 39.79\,{\rm{kJ}} \end{array}$$ So, the enthalpy change is $$39.79\,{\rm{kJ}}$$. **Therefore, the correct answer is option (B).** **Note:** There are two types of enthalpy: a) Lattice enthalpy ($${\rm{\Delta }}{{\rm{H}}_{{\rm{lattice}}}}$$)- the change in enthalpy when one mole of solid breaks up into ions in the gaseous state is known as lattice enthalpy. b) Hydration enthalpy ($${\rm{\Delta }}{{\rm{H}}_{{\rm{hyd}}}}$$)- the enthalpy change when the ions are solvated or surrounded with solvent molecules is known as hydration enthalpy. Hence, the enthalpy of solution is $${\rm{\Delta }}{{\rm{H}}_{{\rm{sol}}}}{\rm{ = \Delta }}{{\rm{H}}_{{\rm{lattice}}}} + {\rm{\Delta }}{{\rm{H}}_{{\rm{hyd}}}}$$