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Chapter 13 - Homework Solutions
Use the following diagrams for questions concerning solution process. )Hsoln > 0 (endothermic); product (solution) has a higher energy than reactants (solute & solvent).
Attractive forces between solute and solvent (unlike particles) are not as strong as solute-solute and solvent-solvent AF (like particles). More energy required to separate solute-solute and solvent-solvent particles than is released when form solute-solvent AF. ENDOthermic ()Hsoln > 0) A solution process with )Hsoln > 0 (+) tends to not be spontaneous (but can be). It also depends on the entropy of solution, )Ssoln. See page 3. 2 )Hsoln < 0 (exothermic); product (solution) has a lower energy than reactants (solute & solvent).
Attractive forces between solute and solvent (unlike particles) are stronger than solute-solute and solvent-solvent AF (like particles). Less energy required to separate solute-solute and solvent-solvent particles than is released when form solute-solvent AF. EXOthermic ()Hsoln < 0) A solution process with )Hsoln < 0 (-) tends to be spontaneous (but may not be). It also depends on the entropy of solution, )Ssoln. See page 3. 3 )Hsoln = 0 (ideal); product (solution) has same energy as reactants (solute & solvent).
Attractive forces between solute and solvent (unlike particles) are similar to solute-solute and solvent-solvent AF (like particles). Energy required to separate solute-solute and solvent-solvent particles is about the same as that released when form solute-solvent AF. Generally, particles with only LF are the most likely to form ideal solutions. IDEAL ()Hsoln = 0) For ideal solutions ()Hsoln = 0) and endothermic ()Hsoln > 0) to form the change in entropy (disorder) must be positive ()Ssoln > 0, disorder must inc.). Generally, )Ssoln > 0 for mixing. Remember, as given in class: )Hsoln = )H1 + )H2 + )H3 or )Hsoln = )Hsep solute + )Hsep solvent + )Hsolvation )G = )H - TC)S
If, and )G < 0 (negative) for a spontaneous process )G > 0 (positive) for a nonspontaneous process )H > 0 (+) or
then, )H = 0 MUST have )S > 0 (+) (an increase in disorder) in order to have )G < 0 (-), at some temperature, in order for a solution to form. A )S > 0 (+) tends to make a process spontaneous. If )H < 0 (-), )S can be negative (-) and still have )G < 0 (-), so a solution can form at some temp even with a decrease in disorder (inc. in order). 4
13.2) When an ionic compound dissolves in water the cation becomes surrounded by the H2O molecules (see figure below). Courtesy of Prentice Hall (BLB 10th ed.) This process is called hydration (specific term for solvation when the solute is an ion and the solvent is H2O). This is an Ion-Dipole AF between the ion and the polar solvent (H2O). This AF is quite strong. The attractive forces we’ve discussed have the relative order in terms of increasing strength: LF < DD < H-bonds < Ion-Dipole < Ionic & Covalent In terms of the energy diagrams depicted on pages 1-3 above (Figure 13.4 in the text, p. 532) to which step does this interaction correspond? This corresponds to step 3 in the diagrams. This is )H3 ()Hsolvation). For this specific case of an ion in H2O this is usually referred to as the heat of hydration, )Hhyd This step always releases energy and is exothermic. Whenever attractive forces are formed energy is released. Lattice energy is the main compoent of step 1, )H1, the enthalpy required to separate the solute particles (ions). If )H1 is too large the dissolving process is prohibitively endothermic and the substance won’t be very soluble. In other words, not enough energy is released in step 3, )H3 ()Hhyd), from the solutesolvent interactions (ion-dipole) to overcome the enthalpy required in the first two steps and give a small enough positive )Hsoln so any positive change in )Ssoln (inc in entropy) can overcome this and make the compound soluble. If you remember I said in class that ionic compounds with high lattice energies and thus high melting points would also tend to not be very soluble due to the fact that the ions are harder to pull apart. 5
13.5) Which of the following pictures best represents a saturated solution? Courtesy of Prentice Hall (BLB 10th ed.) Diagram (b) is the best representation of a saturate solution. A saturated solution has some undissolved solid in equilibrium with dissolved solute in the solution. As much solute has dissolved as can dissolve, leaving some undissolved solid in contact with a saturated solution. Liken it to a sponge that you are using to absorb a larger puddle of liquid. After awhile the sponge can’t absorb any more liquid (it won’t hold any more) and the sponge is saturated. A saturated solution won’t “hold” any more solute (no more will dissolve). An unsaturated solution is one in which the maximum amount of solute has not dissolved. More solute can dissolve. Again, this would be like using a sponge to absorb a drop of liquid. The sponge could still absorb more liquid and is not saturated (it is unsaturated). A supersaturated solution is one in which there is actually more solute dissolved at a given temperature than there should be (more than in a saturated solution). This is a metastable state. You generally get such a thing by heating the solution and more solute dissolves (which is usually the case). Then you do not allow the solute to crystalize out as you cool the solution (some solutes have trouble crystallizing). This “traps” the added heat in the system. When you do something to this metastable system (such as add a small crystal, scratch the side of the container, shake it, etc.) the excess solute (that doesn’t “belong” in the solution at the lower temperature) will crystalize out. As it does the system releases the “trapped” heat. This is one of the processes used in reusable hot packs. 6
13.7) The structures of Vitamin B6 and E are shown below. Courtesy of Prentice Hall (BLB 10th ed.) Vitamin B6 is a relatively small molecule consisting of a benzene-like 6-membered ring with a N atom in place of a C atom (5 C atoms and 1 N atom in a ring). It also has 3 polar -OH groups. These groups and the N atom make the molecule somewhat polar and the -OH groups can form H-bonds to water molecules in solution. The Vitamin B6 molecule is small enough that the H2O molecules can effectively surround the molecule and interact with it and keep it in solution. Vitamin E has one end that has a six-member and a 5-member ring fused together with only one -OH group (polar; DD & H-bonding) and one ether group, -O- (polar and H-bonding to water molecules). However, there is a long nonpolar hydrocarbon chain coming off the rings. The whole molecule has a slightly polar “head” and a long nonpolar “tail”. This makes the molecule behave more like a nonpolar molecule, with large London Forces (LF). The same thing happens in fats, which are largely nonpolar. The H2O molecules can not effectively interact with the long nonpolar end of Vitamin E and surround the molecules to keep them from coming back together and keep them in solution. 7
13.11) There is a balloon made of a semipermeable membrane containing 0.25 L of a 0.2 M solution of some solute which is submerged in a 0.1 M solution of the same solute (as shown below). Assuming the volume of solution outside the balloon is large, what would you expect to happen to the solution volume inside the balloon once the system has come to equilibrium through osmosis? Courtesy of Prentice Hall (BLB 10th ed.) Ideally, one would expect a volume in the balloon of 0.50 L (double what you started with). If the volume of solution outside the balloon, solvent will flow across the semipermeable membrane until the molarities of the solutions inside and outside the balloon are equal, 0.1 M. In order for the solution inside the balloon to decrease from 0.2 M to 0.1 M the volume inside the balloon must double (be twice as large as the initial volume, M2V2 = M1V1), to 0.50 L. In actuality, osmosis across the membrane is not perfect (non-ideal). The solution concentration inside the balloon will be slightly greater than 0.1 M and the volume of the balloon will thus be slightly less than 0.50 L. 8
13.14) a) )Hsoln = )Hsep ions + )Hsep H2O + )Hhyd ()Hsolvation is )Hhyd when solvent is water) You get the same general diagram for NaCl in benzene, C6H6, a nonpolar solvent. However, the energy released in step 3 ()Hsolvation) will be a lot less than for NaCl in H2O because you are trying to replace ionic AF with nonpolar LF between ions and the nonpolar C6H6 solvent molecules. )Hsoln = )Hsep ions + )Hsep C6H6 + )Hsolvation
Remember, )Hsoln > 0 when solute-solvent interactions (unlike forces) are less favorable than solute-solute & solvent-solvent interactions (like forces). This does NOT lead in general to a spontaneous dissolution (solution formation). )G = )H - TC)S and )G < 0 (negative) for a spontaneous process If )H > 0 (+) to get )G < 0, must have )S > 0 (+) so an increase in randomness (disorder) is an important factor. ** See next page ** 9
In these cases, )H1 ()Hsep ions) is the same. In step 2, the energy required to separate the H2O molecules is greater than that required to separate C6H6 molecules ()Hsep H2O > )Hsep C6H6). This is due to the fact that for H2O you have to break LF, DD and H-bonding to separate the molecules, while for C6H6 you only break LF (dispersion forces). In step 3 (involving solutesolvent interactions), the energy released when H2O surrounds the Na+ and Cl- ions (forming strong ion-dipole AF) is much greater than that when the NaCl tries to dissolve in the nonpolar C6H6 (which can form only LF with the ions). This large difference in )H3 in the two cases causes NaCl to be soluble in H2O but not C6H6. In both cases the solution process is endothermic ()Hsoln > 0) and an increase in disorder, entropy, ()S > 0) is necessary for a solution to form. In the case of NaCl in C6H6 the increase in disorder is not enough to overcome the positive enthalpy change ()Hsoln > 0) and make the solution form, whereas it is for NaCl in H2O). Lattice energy is the main compoent of step 1, )H1, the enthalpy required to separate the solute particles (ions). If )H1 is too large the dissolving process is prohibitively endothermic and the substance won’t be very soluble. In other words, not enough energy is released in step 3, )H3 ()Hhyd), from the solute-solvent interactions (ion-dipole) to overcome the enthalpy required in the first two steps and give a small enough positive )Hsoln so any positive change in )Ssoln (inc in entropy) can overcome this and make the compound soluble. If you remember I said in class that ionic compounds with high lattice energies and thus high melting points would also tend to not be very soluble due to the fact that the ions are harder to pull apart. b)
A cation (or anion) is strongly hydrated (strong ion-dipole AF) when the ion is highly charged and small, i.e. a large charge density. The ion-dipole energy is given by: Q: EI-D % ------d2 Q: charge on ion :: dipole moment of solvent molecule d: distance between centers of ions & usually determined by adding radii of the ion and the molecule. From this one can see that this energy (AF) increases as the charge on the ion increases and the size of the ion decreases. It also depends on the size of the dipole moment of the solvent molecule, i.e. how polar the solvent molecule is. The more polar the solvent molecule the more soluble will be the ionic solute. 10
13.15) Indicate the most important type of solute-solvent AF. Decide whether the solute and solvent are ionic, polar, covalent or nonpolar covalent (molecular). CCl4 nonpolar LF in C6H6 (benzene) nonpolar LF a) Both are nonpolar and have only LF so the only solute-solvent IAF will be LF (dispersion). b) CH3-OH in H2O polar polar LF LF DD DD H-bonding H-bonding The solute-solvent IAF are LF, DD and H-bonding. Since H-bonding is the strongest of these it is the most important. In a solution of CH3OH in H2O, the CH3OH can form H-bonds to H2O through the lone pair e& on the O atom of CH3OH to H atoms of H2O. Also, the -OH H-atom can from H-bonds to lone pair e& on the O atom of H2O. Note: Some molecules which can NOT form H-bonds in the pure liquid (or solid) CAN form H-bonds to the solvent molecules in solution (solvent must have a H bonded to N, O, F in the solvent molecule). An example would be CH3 - O - CH3. This can NOT form H-bonds as a pure substance. However, it can form H-bonds to H2O molecules through the lone pair e& on the O atom to the H atoms on the H2O solvent molecules.) 11
13.15) (cont.) c) KBr ionic ion-ion in H2O polar LF DD H-bonding H2O KBr (s) -----------> K+ (aq) + Br& (aq) Ionic solutes & polar solvents form Ion-Dipole IAF. Ion-Dipole (weaker than ionic bonds, but stronger than LF, DD & H-bonds)
QC: % ------d2 QC: (Att. Force, F % ------- ) d3
Q = chg on ion : = dipole moment of polar molecule d = distance between ion & polar molecule Eion-dipole Note: Strength of solute-solvent IAF: LF . DD < H-bonding < Ion-Dipole < Ionic & Covalent Remember, for ionic compounds the lattice energy and AF are given by the following:
Q+ Q& LE % ---------d Q+ Q& (Att. Force, F % ---------- ) d2 This is very similar to the Ion-Dipole energy and AF between ions and polar solvents. 12
13.15) (cont.) d) HCl polar LF DD in CH3C/N: polar LF DD The solute-solvent IAF are LF and DD. In this case, since both molecules are about the same size and not that large and fairly polar the Dipole-Dipole IAF are the most important. 13.16) 13
13.16) (cont.) 14
13.17) a) The energy diagrams for the solution process are given on pages 1-3. Equation 13.1 is: )Hsoln = )H1 + )H2 + )H3
or )Hsoln = )Hsep solute + )Hsep solvent + )Hsolvation
If the solvent is water ()Hsolvation = )Hhyd, the enthalpy of hydration) )Hsoln = )Hsep solute + )Hsep H2O + )Hhyd
When dissolving an ionic solute the lattice energy, LE, Q+ QLE % ------d
Q: charge on ions d: distance between centers of ions & usually determined by adding ionic radii. is the amount of energy required to completely separate a mole of ions from the solid to its gaseous ions. This corresponds to )H1 (solute-solute interactions). b) In eqn. 13.1, the 3rd step, )H3, is always exothermic. Formation of attractive forces, no matter how weak, always release energy (lowers the energy of the system), relative to the energy of the isolated particles. Step 1: Step 2: Step 3: separate solute particles separate solvent particles form solute-solvent AF )H1 > 0 (+), endothermic )H2 > 0 (+), endothermic )H3 > 0 (-), exothermic 15
13.20) (cont.) ...
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This note was uploaded on 04/20/2011 for the course CALC 103 taught by Professor Reinard during the Spring '11 term at Finger Lakes Community College.
- Spring '11