|
|
 |
Reply
| |
Hi, How would you go about setting this up? Thanks A 50.0-mL sample of a 1.00 M solution of CuS0 4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both solutions was 20.2 °C before mixing and 26.3°C after mixing. The heat capacity of the calorimeter is 12.1 J/K. From these data calculate �?/EM> H for the process CuSO 4( l M) + 2KOH (2 M) �?Cu(OH)2(s) + K2SO4 (0.5 M) Assume the specific heat and density of the solution after mixing are the same as those of pure water. |
|
Reply
| | From:  ·Steve· | Sent: 3/1/2006 11:12 PM |
Yep, but what about the sign of DH?  |
|
Reply
| |
| This message has been deleted by the author. |
|
Reply
| |
forgot! - sign of course, exothermic. <o:p> </o:p> <o:p> </o:p> OK how about another problem<o:p></o:p> <o:p> </o:p> On a visit to the <st1:place w:st="on">Midwest</st1:place>, you find yourself in a freak hailstorm and get hit by a 6 pound piece of hail You really want to keep this ice chunk as a souvenir so you decide to pack it in dry ice, put it on the plane with you, and stick it in your freezer. Unfortunately, you only have a frost-free freezer at home. How long will the piece of hail last (start the timer from when you get bonked on the head in <st1:State w:st="on"><st1:place w:st="on">Iowa</st1:place></st1:State>). There are lots of different variables to consider so you can do a lot with this problem!<o:p></o:p> The molar heat capacity of ice is 37.6 J/mol -*C<o:p></o:p> I figured this to be 6 lb ice ball to be 2721.54 g and 151.06 moles of water. I am saying the outside temp is -10 C and it took me an hour to pack it in dry ice so I guess it would still be ok?<o:p></o:p> I think I can get it home without losing any of it to melting.? <o:p></o:p> so 151.06 mol x 37.6 J/mol - *C = 5679.86 J/*C ????<o:p></o:p> How would I calculate the kJ needed to melt it without a temp change? I need to know that don't I?<o:p></o:p> I am still pondering. <o:p></o:p> I am thinking the frost free freezer is the key. Any insight so far? <o:p></o:p> I will keep thinking about it and post back.....<o:p></o:p> Thanks<o:p></o:p> |
|
Reply
| | | From: ·Steve· | Sent: 3/4/2006 12:49 AM |
Well of course without specific information we can't calculate an actual time. Even the shape of the chunk of hail would be a factor (because of the effect of surface area). For the time being I would assume as you did that there is no loss of ice all the way home, until it is placed in the freezer. We also need to know the temperature of the freezer, but I guess we can assume 0°C for simplicity. I've noticed that ice cubes in a frost-free freezer slowly disappear if they are just left in the freezer for a long time. During the defrost cycle they may melt on the surface just slightly and slowly evaporate that way, or if no actual melting occurs the ice may just slowly sublime. Either way, you are converting ice at 0°C to water vapor at 0°C overall.
Or, the problem could be asking how long it will take the piece of hail simply to melt when it asks how long it will "last", in which case we would have a shorter calculation for H2O (s, 0°) --> H2O (l, 0°). In this case we don't have to consider the conversion to vapor, H2O (l, 0°) --> H2O (g, 0°).
Using the heat of fusion of ice (6.01 kJ/mol) you can calculate the joules of heat needed for melting the moles of ice in the piece of hail at 0°C. There is no temperature change here. If you want to also include the heat needed to evaporate the liquid, add moles of ice X 44.01 kJ/mol (this number is the standard DH° for the process H2O(l) --> H2O (g) which should be about right at 0° also).
Now you have a value for the total heat, but what about the time? That depends on lots of things, such as how well the ice contacts the defroster coils in the freezer, if the freezer is "standard" or "automatic" frost-free, how long the defroster is on and off (automatic defrost), and the wattage of the defroster. Typical defroster times are 20 minutes every six hours.
Again for simplicity I guess you could assume that all of the heat from the defroster coils are transferred to the ice during the defrosting cycle (I'm assuming automatic defrost). If you know how many joules of heat are transferred to the ice in each 20 minute heating period, you can calculate the total time it would take to melt, or melt and vaporize, the ice.
This amount of heat would be the watts of power used by the defroster coils times the time in seconds (20 min = 1200 s). A watt is a joule per second, J/s. The defroster coils use perhaps 350 watts of power (estimated from here). The joules of heat transferred to the ice would be (350 J/s)(1200 s) = 420,000 J or 420 kJ. From this, calculate the number of 20 min heating periods needed to evaporate all the ice and add the times between the periods when the defroster is off (say, cycles of 6 hours off, 20 min on) and that should give you the total time for your precious chunk of hail to disappear! This is definitely a " " kind of problem! (And a " " analysis! Hope it gets you on the right track.)
Steve
How frost-free freezers work -
|
|
Reply
| |
Thanks Steve. You know how to explain things perfectly, now it all seems easy!!! Curious where you got the values for 44.01 kJ/mol (this number is the standard DH° for the process H2O(l) --> H2O (g) which should be about right at 0° also. It is probably in CRC Handbook, right, but is there a web site you use also? If I used DH° sublimation that would just equal 44.01 kJ/mol for fusion + the 6.01 kJ/mol for vaporization? Hess' Law? Thanks |
|
Reply
| |
also where would I find the molecular mass of ice - would it be that much different than water? for calculating moles in ice ball. |
|
Reply
| |
Seem right? So I first figured moles in the ball: <o:p> </o:p> 1 lb = 453.59g <o:p> </o:p> 6 lb hail ball = 2721.54 g. <o:p> </o:p> 2721.54g H20 x (1 mol H2O/ 18.016g H2O) = 151.06 mol H2O<o:p></o:p> <o:p> </o:p> 151.06 mol H2O (6.01 kJ/mol + 44.01 kJ/mol) = 7556.02 kJ<o:p></o:p> <o:p> </o:p> So I need 7556.02 kJ of heat to melt the ball and then evaporate it. <o:p></o:p> |
|
Reply
| | | From: ·Steve· | Sent: 3/4/2006 6:39 PM |
>> Curious where you got the values for 44.01 kJ/mol <<
I think that should have been 44.1 kJ. I used standard heats of formation values of H2O (l) and H2O (g) from the appendix of one of my general chemistry textbooks. DHf° for H2O (l) = -285.85 kJ/mol and DHf° for H2O (g) = -241.8 kJ/mol. Therefore DH° for the reaction H2O(l) --> H2O (g) will be
DH° reaction = [(1 mol)(-241.8 kJ/mol)] - [(1 mol)(-285.85 kJ/mol) = 44.05 kJ
(This is just the "products minus reactants" method for calculating DH° of reactions using the standard enthalpies of formation of the substances.)
That is DH°vap at standard temperature, 25°C. But meanwhile I found some better values. According to one of my textbooks it is 44.94 kJ/mol at 0°C and 40.67 kJ/mol at 100°C, so it does vary with temperature somewhat. They are in the CRC Handbook also, under "Steam Tables" in my edition; at "T = 32°F, Evap = 1075.5 Btu/lbm" which converts to 45.07 kJ/mol). Here's a web site that has some of this information too: http://www.lsbu.ac.uk/water/data.html. It gives DHvap = 45.051 kJ/mol at 0°C. As you can see, the values vary a little from source to source. I'll just go with 45.05 kJ/mol.
>> If I used DH° sublimation that would just equal 44.01 kJ/mol for fusion + the 6.01 kJ/mol for vaporization? Hess' Law? <<
That's right, here's what you have:
H2O (s) --> H2O (l) DH = 6.01 kJ at 0°C
H2O (l) --> H2O (g) DH = 45.05 kJ at 0°C
H2O (s) --> H2O (g) DH = 51.06 kJ at 0°C
>> also where would I find the molecular mass of ice - would it be that much different than water? <<
Since ice is made of water, that's the molecular weight to use, 18.0153 g/mol!
>> So I need 7556.02 kJ of heat to melt the ball and then evaporate it. <<
That's right; it will be a little more if you use the higher DH of 51.06 kJ/mol.
Steve |
|
Reply
| | | From: ·Steve· | Sent: 3/6/2006 2:29 AM |
I did some keyword searching to try to find some kind of estimate of the rate of evaporation of ice in a freezer. It's actually a popular experiment - Freeze a weighed amount of water in your freezer in a bowl and leave it in the freezer for a week, then weigh it again, to find that it now weighs less. But there are a number of variables - vapor pressure of water in the freezer (humidity), the temperature, the amount of surface area exposed to the air, the air movement over the surface, how often the freezer door is opened, how often the defroster runs, etc. Well, if you have a few days you could do this experiment and get an actual estimate of how long it will take the ice to evaporate! An example problem here gives a rate of loss of six grams per week which sounds reasonable, so you might want to use that. This way, it's not necessary to try to answer the question just from the heat absorbed perspective, defrost cycle, and so forth. If we assume all of the heat is transferred to the ice, it won't last very long! But, that would not be the case anyway. But your teacher may like all the calculations - moles of ice, heat of fusion, heat of vaporization, and all that good stuff! It's certainly a good "thinking" problem! Steve |
|
Reply
| |
Thanks Steve, I just said 10% of the heat got to the ice, and it was something like 485 days. 100%, it disappeared too fast. Ok how about this: We did with a seperation funnel, In this part of the experiment, you will be extracting the red dye Sudan IV also known as Scarlet Red, from a solution that is 50:50 water and ethanol. The solvent chosen for this extraction is hexane. Here are some of the answers to questions on this, how do they look? Thanks On 6 I thought about gas chromatography or using denities after evaporating hexane? |
| | sep.jpg |
Reply
| | | From: ·Steve· | Sent: 3/7/2006 5:48 AM |
In question 4 ("Identify the components of the two layers formed in step 2. How did you make this determination?") did the hexane extract the dye? You have the upper hexane layer marked "clear". I assumed this was the extraction of the dye from the water-ethanol solution.
In question 5 and 6, you are asked to confirm the contents of three flasks:
a) water and ethanol
b) Sudan IV and hexane
c) mostly hexane with trace Sudan IV
Yes, I think GC could be used to identify the solvent or solvent mixture after distillation. You could evaporate the remaining solvent and confirm the identities of the dry solid residues of flask b and flask c using IR spectroscopy (best) or by their melting points (may be less reliable, as the reported melting range of even pure Sudan IV is rather broad, 181-188°C).
Although I don't know the steps of the procedures, your answers look reasonable!
Steve
|
|
Reply
| |
oops, the steps were: 1 . Take a ring support for the separators funnel and attach it to the lab stand. Place the separators funnel in it as shown in figure 1. Make sure that the stopcock is closed (horizontal position). Pour in 20ml of Sudan IV solution. Record your observations. 2. Add 10ml of hexane to the mixture in separators funnel. Record your observations. 3. Put the stopper in the top of the funnel securely. Securely hold the funnel and invert it a few times as shown in fig. 2. With the stopcock pointing UP and away from everyone, carefully open it (the handle should be parallel with funnel.) to release any pressure. IF THIS IS UNCLEAR ASK YOUR INSTUCTOR OR LAB STAFF BEFORE ATTEMPTING IT. There will probably be a gentle hiss of escaping gas. CLOSE THE VALVE. GENTLY swirl the mixture with the funnel still inverted and VENT again. Repeat process until no more gas is released. 4. Place the funnel in the support ring. Wait until the two liquids separate. If the interface between the two liquids isn't sharply defined, it means that the system (liquids; solute; impurities) is prone to form an emulsion, droplets of one liquid trapped in the other as in salad dressing. Shake the two liquids together as vigorously as possible without forming an emulsion. Record your observations. 5. Let the mixture rest in the support ring, until the liquids separate. 6. Take three 125 mL Erlenmeyer flask. Label them "A", "B" and "C". 7. Place Erlenmeyer flask(A) under funnel. Remove stopper from separatory funnel. stopcock slowly and drain bottom layer. Stop just after you get to the interface. 8. Now pour the top layer out of the top into flask(B). 9. Put the contents of flask (A) into the funnel. Add 10ml hexane and repeat steps 3 through 10. This time put the bottom layer back into flask (A) and the top layer into flask(C). Compare the color of each flask. Record your observations. |
|
Reply
| | | From: ·Steve· | Sent: 3/7/2006 9:01 PM |
OK, that's clear. In question 4, I think it's mainly asking which layer is the hexane layer (the upper layer). Since the mixture has not been shaken yet, no dye was extracted and the hexane layer is still colorless. Questions 5 & 6 make sense now too! Steve |
|
Reply
| |
I guess I am not totally clear: Why is the Sudan IV in a mixture of ethanol and water instead of just plain water? What if it were in just water or just ethanol? I think I was misled in lab so not totally sure. I see Sudan IV has both nonpolar and polar regions so I am thinking ethanol kind of bridges the two (Sudan IV and water). But the hexane draws out the Sudan so it can also be highly attracted to nonpolar hexane. One question asked what the dominant intermolecular force present in Sudan IV was and I said London Dispersion since it interacts with nonpolar hexane and got that wrong? |
|
Reply
| | | From: ·Steve· | Sent: 3/10/2006 6:02 PM |
>> Why is the Sudan IV in a mixture of ethanol and water instead of just plain water? << I think you answered that already - Sudan IV is insoluble in pure water, but has some solubility in ethanol, but... >> ...What if it were in ... just ethanol? << Then the extraction would not work, because ethanol and hexane are miscible (so you would not get two layers). >> I am thinking ethanol kind of bridges the two (Sudan IV and water). << That's right - ethanol is polar, but it has nonpolar character as well, and as a result it is able to dissolve a wide variety of organic compounds. >> One question asked what the dominant intermolecular force present in Sudan IV was and I said London Dispersion since it interacts with nonpolar hexane << Right again - "like dissolves like" so that makes sense. Steve |
|
|
|
Free Forum Hosting
Important Announcement
First
0 recommendations 
Return to Inorganic