Question

Which one of the following statements is true?

The heat of solution in H2O for KOH is the same as that for KOH • H2O.
For a compound to spontaneously dissolve in H2O at 25 °C, the solution process must be exothermic.
Ne is less soluble in H2O at 2 atm pressure than at 1 atm pressure.
C2H4 is very soluble in liquid NH3.
None of the above

Answer 1

Answer:

All are false.

Explanation:

We can deal with each of these statements one by one.

Statement A is false. The heat of solution is the enthalpy change associated with the dissolution of a solute into a solution.

solute(s)−→−−−H2O(ℓ)solution(aq)ΔHsoln

We can write a process associated with each of the situations described.

KOH(s)−→−−−H2O(ℓ)KOH(aq)⟶K+(aq)ΔHsoln=ΔHKOH

KOH⋅H2O(s)−→−−−H2O(ℓ)KOH(aq)⟶K+(aq)ΔHsoln=ΔHKOH⋅H2O

The reason for the difference is because the hydrated KOH • H2O already has some water integrated in its crystal structure. This means there will be diminished KOH ionic interactions that need to be broken and some of the ion/H2O interactions are already formed in the hydrate prior to the dissolution process. Typically, the process of dissolution of the anhydrous form is more exothermic than the dissolution process of the hydrate.The final product is the same for both of these but since the initial state is different, we should expect there to be a difference in the solvation energy.

Statement B is false. For a process to be spontaneous, the system must move from a higher to a lower state of energy. While we often think of heat changes (enthalpy for example) as the main changes in energy, disorder is another form of an energy change that can and does occur in processes. The measure of disorder of a system is called the entropy and a change in disorder is ∆S, where S is the symbol used for entropy. A system will spontaneously move from a state of order to disorder. (To increase the order of a system we will have to put work, energy or effort of some kind into the system.)

So when a process (like dissolution) is exothermic, this only refers to the enthalpy change (∆H). However, when looking at the overall spontaneity of the reaction we also must look at the increase or decrease in disorder or entropy (∆S).  Therefore, just because a process is exothermic does not mean it will be spontaneous or lower in energy. In fact, if the increase in disorder is enough then the dissolution process can be endothermic (absorb heat) and still be spontaneous.

Statement C is false. At low concentrations, Henry’s Law can be used to determine the solubility of a gas.   Cg = PgkH   where Cg = solubility (concentration of) the gas, Pg = partial pressure of the gas in equilibrium with the solution, and kH is the Henry’s Law constant. In other words the solubility of a gas is directly proportional to the partial pressure of the gas above the solvent.  Therefore, this statement is not true because at 2 atm of pressure Ne will be twice as soluble as it would be if the  partial pressure were 1 atm.

Note: The Henry’s Law constant depends on the gas/solvent combination and must be determined experimentally.

Statement D is false. Qualitatively we can use the “likes dissolve likes” rule to answer this question.  First we have to figure out the solvent-solute interactions to determine if a solute will be readily soluble in a particular solvent. We need to know the structures of the solvents to be able to determine the important intermolecular forces.  (Draw Lewis structures, determine shape, polarity, etc.)

Lewis structures of ammonia and ethene

Liquid NH3 is a polar molecule that has intermolecular forces composed of hydrogen bonding, dipole-dipole interactions, and London dispersion forces.  Ethene is a non-polar covalent molecule that has LDF forces but no others.  Therefore, since the two molecules do not have similar intermolecular forces ethene will not be soluble in ammonia.

Since all 4 statements are false, only E is a true statement.