Formation of a solution from two componenets can be considered as :
(i) pure sovent `rarr` separated solvent molecules, `/_\H_(1)`
(ii) Pure solute `rarr`separated molecules,`/_\H_(2)`
(iii) separated sovent and solute molecules`rarr`solution, `/_\H_(3)`
solution so formed will be ideal if :

Options

A. `DeltaH_(soln)=DeltaH_-DeltaH_(2)-DeltaH_(3)`
B. `DeltaH_(soln)=DeltaH_(1)+DeltaH_-DeltaH_(3)`
C. `DeltaH_(soln)=DeltaH_(1)+DeltaH_(2)+DeltaH_`
D. `DeltaH_(soln)=DeltaH_(3)-DeltaH_(1)-DeltaH_(2)`

Answer

2

Explanation

For simplicity, we can imagine the solution process as taking place in three distinct steps: Step 1 involves the separation of solvent molecules, and step 2 entails the separation of solute molecules. These steps require energyinput to break attractive intermolecular forces, therefore, they are endothermic.
In step 3 the solvent and solute molecules mix. This process is exothermic. According to Hess's law, the heat of solution `(DeltaH_("soln"))` is given by
`DeltaH_("soln")=DeltaH_(1)+DeltaH_(2)+Delta_(3)`
In case of ideal solution, the solute-solvent attractive is identical with the solvent-solvent attraction and solute-solute attraction. Therefore, `DeltaH_("soln")` is zero. The solution process will be exothermic `(DeltaH_("soln")lt0)` if the solute-solvent attraction is stronger than the solvent-solvent attraction and solute-solute attraction. This happens in nonideal solution with negative deviation. The solution process will be endothermic `(DeltaH_("soln")gt0)`, if the solute-solvent interaction is weaker than the solvent-solvent and solute-solute interactions. This happens in case of nonideal solution with deviation.