CHEM 1A Lecture Notes - Lecture 2: London Dispersion Force, Dynamic Equilibrium, Chemical Energy

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cyansardine547

21 Dec 2020

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2 free energy and chemical equilibrium limits of solubility solvent dissolves the solute saturated : when solvent has dissolved all the solute that it can and some undissolved solutes left. Solid solute concentration at greatest ( no more can dissolve) molar solubility : the numerical value of the molar concentration of a saturated solution of a substance. If principle cohesive forces between solute molecules are london forces , best solvent mimics the force the key is to replace the strong solute - solute attractions with strong solute - solvent attractions example : soap. Pressure: solubility of the gas increases as its pressure increases. If gas above liquid is a mixture , then solubility of each component depends on that component"s partial pressure. Henry "s law : the solubility of a gas in a liquid is proportional liquid to its partial pressure above the. Increase in pressure gas molecules strike the surface of solvent more.

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Question about mixed solvents for recrystallization.

I am having a hard time conceptualizing how a mixed solvent would work in a scenario where P and O nitronitroacetanilide are recrystallized with a small amount of hot Ethanol followed by a larger volume of water.

From the structure of the molecules I assumed that the Para form of nitroacetanilide is more polar than the Ortho form. Second, I know that water>ethanol for polarity. Finally, I know that water and ethanol can hydrogen bond with one-another and with itself. Also I know that water and ethanol will hydrogen bond together to form a more densely packed liquid. I imagine that this would look somewhat like this: [ Eth-OH] --[O-H2]---[OH-Eth]

My understanding of dissolving is that the solute gets surrounded by solvent molecules and separates from other solute molecules. This happens to all the solute molecules until no more solvent molecules are left to dissolve the solute.

Having said that, does the following make sense? If not, please correct me

1) the O/P mixture of crude nitroacetanilide dissolves in hot ethanol because ethanol hydrogen bonds with the nitroacetanilide

2) when water is added, ethanol and water hydrogen bond with one another creating a denser solvent because the ethanol hydrogens are attracted to the oxygen in water and vice versa

3) because there's less hydrogens "available" to form new hydrogen bonds after the water is added, the solvent can't make new hydrogen bonds as easily

4) because the P-nitroacetanilide needs more hydrogens to dissolve due to it being more polar, it is now less soluble in the ethanol/water mixture. this causes it to precipitate out while leaving the O-nitroacetanilide that's less polar still in solution

Is this correct? if not please explain how the mixed solvent helps precipitate the p-nitroacetanilide but maintain the o-nitroacetanilide in solution.

For an ideal solution, we discussed in class that the chemical potential is given by: Î¼=Î¼_0+RT "ln" (c/c_0 ) where the units of Âµ written in this form are Joules/mole. The concentration c_0 is a reference concentration, which we will take to be 1 Molar for this problem. Suppose solution 1 with solute concentration of 1 mM (i.e., 1 x 10-3 Molar) is in contact with solution 2 with solute concentration of 0.1 mM (i.e., 1 x 10-4 Molar). If a differentially small amount of solute, dn, moves from the higher concentration to the lower concentration, the differential change in free energy is calculated as follows: dG = [-((âG_1)/dn)_(P,T) dn + ((âG_2)/dn)_(P,T) dn] where G_1 is the free energy of solution 1 (the higher concentration solution), G_2 is the free energy of solution 2 (the lower concentration solution), and the relative signs indicate that the number of moles of solute in solution 1 is decreased by dn, while the number of moles of solute in solution 2 is increased by dn. Recalling the definition of chemical potential, we can write: dG = [-Î¼_1 dn + Î¼_2 dn] = [Î¼_2-Î¼_1 ] dn

2a. Show that for the case described above dG = [RT "ln" (c_2/c_1 )] dn and insert the values of the high (c_1) and low (c_2) concentrations to confirm that, indeed, dG < 0 for transfer of solute from a high concentration solution to a low concentration solution. This result confirms that such transfer will be a spontaneous process if the solutions are placed in contact with one another.

2b. Without worrying about how it might be done, at T = 298 K, by what amount (in kJ/mole) would the chemical potential of the solute in the low concentration solution need to be increased in order for the system to be at equilibrium, even though the two solutions remain at their different concentrations of c_1=1 "mM" and c_2=0.1 "mM" ? (Note: In the case where the solute is an ion, and therefore charged, the most common way of changing the chemical potential of one solution versus the other is to apply a different electrical potential (in Volts) to one solution versus the other. Depending on the signs of the charge on the ion and of the electrical potential, the free energy of a charged solute can be either increased or decreased. This situation of maintaining different ionic concentrations in equilibrium is common in electrochemistry and in living cells.)

CHEM 1A Lecture Notes - Lecture 2: London Dispersion Force, Dynamic Equilibrium, Chemical Energy

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