Practice Problems Week 9
Refer to your text in sections 8.12, 8.14, 8.15. All of them deal with material properties that are affected by intermolecular forces. If you think about how molecules can move, you can figure out how the properties will change with the strength of intermolecular forces.
(i) the strongest intermolecular forces
(ii) the highest vapor pressure at -25ºC
(iii) the greatest volatility
(iv) the highest boiling point
(v) the greatest surface tension
(vi) the highest DHf (heat of fusion)
(vii) the highest DHv (heat of vaporization)
(viii) the greatest viscosity
We recall that intermolecular forces can be
combinations of 1.) hydrogen bonding, 2.) electric dipole-dipole forces and 3.)
London dispersion forces, listed in order of decreasing strength. We also
recall that hydrogen bonding is about ten times stronger that dipole-dipole and
so, if it is present, it will dominate the properties. We also recall that ALL
(a) CO2 & SO2
CO2 is a linear molecule with two very polar C-O bonds. But they are equal and opposite, so they cancel out leaving us with an overall nonpolar molecule. SO2 , on the other hand, is a bent molecule. So the polar S-O bonds add up to an overall molecular polarity whose “arrow” points toward the oxygen atom end of the molecule. Neither molecule has hydrogen bonding possible.
So the intermolecular forces on SO2 are dipole-dipole and those in CO2 are L-D. Since dipole-dipole are much stronger than L-D, that will determine the differences between these two molecules.
So what’s the score for (i) through (viii) above?
SO2: (i), (iv), (iv), (vi), (vii), (viii)
CO2: (ii), (iii)
(b) CH3NH2 & CH3PH2
We observe that the two compounds are very similar with N in one and P in the other. Both N and P are group VA elements. But we notice that N is capable of supporting hydrogen bonding if it has an H bonded to it and there’s a lone pair available for the H to “hide behind.” Remember the H is naked. Since with CH3NH2 we have such a setup, we have hydrogen bonding. With the compound containing P, we don’t. So CH3NH2 has the strongest intermolecular forces and we can now compile the scores:
CH3NH2: (i), (iv), (iv), (vi), (vii), (viii)
CH3PH2: (ii), (iii)
(c) HF & NaF
HF and NaF are vastly different materials. HF is a polar covalent molecule and NaF is an ionic compound. What do we know about intermolecular forces between ionic compounds? Well, we know we don’t have “molecules” in the literal sense, because ionic compounds are made up of charged atoms (ions). So the notion of intermolecular forces must be interpreted as forces between ions instead. So in the comparison between HF and NaF, we will compare the intermolecular forces between polar molecules (in HF) with those between ions an ionic compound like NaF. There is no contest. The electrostatic forces between ions is so much greater than the forces between polar molecules, that NaF wins. So the score must be:
HF : (ii), (iii)
NaF: (i), (iv), (iv), (vi), (vii), (viii)
We have to imagine the molecules being attracted to one another and how does the strength of that attraction affect the property we are looking at? If the process involves pulling molecules apart, stronger attraction will retard the process (such as melting, for example). If the attraction is less (weaker intermolecular forces), the the process will be enhanced. Boiling pulls molecules apart. Melting pulls molecules apart. Higher heat of vaporization and fusion means it’s harder to pull molecules apart. Viscosity is a measure of how difficult it is to move one molecule over another which is a form of pulling molecules apart. So higher intermolecular forces means higher viscosity (thickness). Vaporization is a process of pulling molecules apart in the liquid and sending one out to the vapor. So higher intermolecular forces retard that process and require higher temperatures. And so forth.
If the strength if intermolecular forces decreases, the boiling point decreases, the vapor pressure increases, the viscosity decreases, the heat of vaporization decreases, the rate of evaporation increases, the heat of fusion decreases, and volatility increases.
(a) More volatile: A because it has higher vapor pressure at all temperatures
(b) Stronger forces of attraction: B because of lower vapor pressure
(c) Greater viscosity: B because of higher intermolecular forces
(d) Normal boiling point of A: about 158 looking at the temperature where the “A” curve crosses the 760 mmHg line.
(e) Vapor pressure of “B” at 120ºC: about 500 mmHg looking to the left-hand axis from the point where the “B” curve crosses 120ºC.
(f) Boiling point of “A” at 650 mmHg: about 150ºC because that’s about the temperature where the vapor pressure of “A” crosses the 650 mmHg line.
· DHf = 6.48 kcal/mole
· DHv = 28.5 kcal/mole
· Boiling point = 975ºC
· Melting point = 438ºC
· Specific heats:
¨ Csolid = 0.168 cal/(gºC)
¨ Cliquid = 0.275 cal/(gºC)
¨ Cgas = 0.0975 cal/(gºC)
¨ Molar mass, M = 125 g/mole
How many calories of heat are required to heat 50.0 grams of Dunhamonium from 325ºC to 1025ºC?
We will be heating up 50.0 g, , of this here Dunhamonium, from below its melting point (so it starts out as a solid) to above its boiling point. That means that it will melt, once it reaches its melting point, and it will boil, once it reaches its boiling point, and then the gas will get superheated to 1025ºC. So we have the heating curve with its five regions:
1. Warming 50.0 g of the solid from 325ºC to 438ºC (Q1 = mc1DT1)
2. Melting 0.400 moles at 438ºC with DHf = 6.48 kcal/mole (Q2 = nDHf)
3. Warming 50.0 g of the liquid from 438ºC to 975ºC (Q3=mc3DT3)
4. Boiling 0.400 moles at 975ºC with DHv = 28.5 kcal/mole (Q4 = nDHv)
5. Warming 50.0 g of the gas from 975ºC to 1025ºC (Q=mcDT) (Q5=mc5DT5)
Q is the quantity of heat in each step which we can now calculate and add up for the total. We’ll convert the heats in steps 1, 3 and 5 to kcal after the calculation of the heat. We will adjust the significant figures for each heat based on the individual steps. Then we will add up the heats, in kcal, observing the number of decimal places. Note that step 4 had a temperature difference of 50ºC, so we only got 2 SF. On adding, we saw that the number with the fewest decimal places was 11.4 kcal, and that limited our answer to one decimal place.