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Topic: intermolecular forces and evaporative cooling  (Read 19869 times)

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Offline Kyle1990

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intermolecular forces and evaporative cooling
« on: March 31, 2008, 03:30:09 PM »
Ok, so i recently did a lab on intermolecular forces and evaporative cooling. Here is my question.
The data i have is

Initial temperature: (t1)
n-hexane 24.28 deg C
n-pentane 24.82 deg C

The temperature probe recorded the following temperatures as the two solvents evaporated on the probe:

Lowest temperature (t2)

n-hexane 5.89 deg C
n-pentane 9.56 deg C

Thus, delta T for the two would be
for n-hexane delta t = 18.39 deg C
for n-pentane delta t= 15.26 deg C

ok so here's the question. Because hexane has a higher molecular weight than pentane, shouldn't it have a lower delta t because it would have stronger intermolecular forces? This data does not coincide with this principle. Could it be an error in the lab or is this how it's supposed to be?
"Theories are nets cast to catch what we call 'the world': to rationalize, to explain, and to master it. We endeavor to make the mesh ever finer and finer."
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Offline Rabn

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Re: intermolecular forces and evaporative cooling
« Reply #1 on: April 01, 2008, 05:31:13 AM »
What happens when something turns into a gas? Think of what needs to occur for both liquids to evaporate.  The intermolecular forces are low between straight chain organic molecules....especially in comparison to water.  To help you visualize this. Imagine having just one molecule of n-Heptane on a surface.  What do you have to do to it in order to turn it into a gas? Now think about the n-pentane in the same way. That should help. If not post back and I'll help guide you along some more.

Offline Kyle1990

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Re: intermolecular forces and evaporative cooling
« Reply #2 on: April 01, 2008, 01:11:36 PM »
Well, on the surface, the molecules are only bound by the molecules beneath and beside them, thus they experience less of a "pull" than the other molecules in the liquid. When a liquid vaporizes these molecules overcome these attractive forces and evaporate into the gaseous phase. And since nonpolar molecules only have london dispersion forces, they are relatively easy to overcome, which accounts for their high volatility. Ah, the thing that i'm still not quite clear about is this: polar molecules, such as alcohols, the strength of their molecular forces increases with molecular weight-because there is increased hydrogen bonding. Does the same principle apply to nonpolar molecules? But instead of hydrogen bonding, do the strength of london forces increase as molecular weight increases?
"Theories are nets cast to catch what we call 'the world': to rationalize, to explain, and to master it. We endeavor to make the mesh ever finer and finer."
-Karl Popper

Offline Arkcon

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Re: intermolecular forces and evaporative cooling
« Reply #3 on: April 01, 2008, 01:21:46 PM »
Well, on the surface, the molecules are only bound by the molecules beneath and beside them, thus they experience less of a "pull" than the other molecules in the liquid. When a liquid vaporizes these molecules overcome these attractive forces and evaporate into the gaseous phase. And since nonpolar molecules only have london dispersion forces, they are relatively easy to overcome, which accounts for their high volatility.

Pretty logical so far.

Quote
Ah, the thing that i'm still not quite clear about is this: polar molecules, such as alcohols, the strength of their molecular forces increases with molecular weight-because there is increased hydrogen bonding. Does the same principle apply to nonpolar molecules?

Hmm... higher molecular weight, perhaps derived from the hydrocarbon chain, CH2-CH2-CH2, etc -- that is what increases hydrogen bonding strength?  That does not follow logically.

Quote
But instead of hydrogen bonding, do the strength of london forces increase as molecular weight increases?

Don't stop at pentane, look up higher alkanes, what happens, and can you see a trend, that you can then work backwards to hexane and pentane.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Rabn

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Re: intermolecular forces and evaporative cooling
« Reply #4 on: April 01, 2008, 03:00:41 PM »
Go back to my suggestion of imagining one molecule laying on the surface of a table. In order to turn it into a gaseous molecule what determines how much energy needs to be transferred to it? Think of two bowling balls, how do you determine how hard to lift it off the ground to make it airborn? When in doubt, looking at the energy involved is always going to lead you in the right direction. 

Offline Kyle1990

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Re: intermolecular forces and evaporative cooling
« Reply #5 on: April 02, 2008, 12:21:39 AM »
ok so there must be something wrong with my data because hexane should have a smaller delta t because there are more atoms which = more forces which should = a small drop in temperature because the molecules don't evaporate as readily as molecules with a lower molecular weight. Is this correct?
"Theories are nets cast to catch what we call 'the world': to rationalize, to explain, and to master it. We endeavor to make the mesh ever finer and finer."
-Karl Popper

Offline Rabn

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Re: intermolecular forces and evaporative cooling
« Reply #6 on: April 02, 2008, 01:27:40 AM »
Your data is correct...think about where the energy to evaporate the liquid comes from.  You are standing in front of El Dorado with your back turned to it; if you want to see it all you have to do is turn around..

Offline Arkcon

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Re: intermolecular forces and evaporative cooling
« Reply #7 on: April 02, 2008, 07:13:55 AM »
You are standing in front of El Dorado with your back turned to it

 ;D

You rock, Rabn.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Rabn

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Re: intermolecular forces and evaporative cooling
« Reply #8 on: April 02, 2008, 09:15:27 PM »
Thanks Arkcon, I thought it would be a very fitting metaphor that those who know what I mean would greatly appreciate.  I wonder if Kyle1990 has had any luck.

Offline Kyle1990

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Re: intermolecular forces and evaporative cooling
« Reply #9 on: April 02, 2008, 10:50:12 PM »
i just don't understand why hexane has a high change in temp because it's supposed to be the molecule with the stronger molecular forces, thus not as many molecules would evaporate, so there should be a lower cheange in temp. ugh something is just not clicking.
"Theories are nets cast to catch what we call 'the world': to rationalize, to explain, and to master it. We endeavor to make the mesh ever finer and finer."
-Karl Popper

Offline Rabn

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Re: intermolecular forces and evaporative cooling
« Reply #10 on: April 03, 2008, 02:37:36 AM »
O.k. I think I know why you're getting hung up on this. let's quickly review what these attractive forces are:

H-bond: this occurs between an O, N, or F bonded to an H and another O, N, or F. It is caused by the delocalisation of the electrons on the H, increasing the frequency that the nuceus is exposed; this exposure of the positively charged nucleus allows an O, F or N which all have free electron pairs to be attracted to it. Let's give H-bond a value 10

London-Dispersive: the electrons around two atoms approaching each other repel each other, this destabilizes the electrons which expose one of the atoms nucleus to the electrons around the other atom.  This however does not last long and the exposure is much, much smaller than that which occurs with hydrogen bonding. let's give L-D attraction a value of 2 relative to h-bond

Now, let's consider pentane.  C-C-C-C-C  as another pentane approaches the electrons around the hydrogen get displaced a little by the electrons around a hydrogen of the incoming molecule. This produces a small attraction.  so if two pentanes are lined up next to each other the bonding force is a 10 because there are 5 L-D interactions. To put those two molecules in the gaseous state you have do two things: 1) break each L-D interaction and 2) give the molecules enough energy to enter the gaseous state. What determines how energy is required to send this molecule zooming around as a gas? Its molecular weight. Now let's make some more extreme simplifications. let's say that each attraction unit, H-Bond has 10 and L-D has 2, requires 1 unit of energy for it to be broken. THat takes care of the energy requirement from 1. Let's also say that for each unit of molecular weight it also takes 1 energy unit for it to be sent into a gaseous form, i.e. if you have 2 pentanes and pentane has a molecular weight of 30, 6 units per carbon(1 unit for each proton), you would need 10 energy units to break the L-D interactions and 60 energy units to send both molecules into the gaseous form. So a total of 70 energy units is required to make both pentane molecules gaseous. Let's do the same thing for 2 hexanes. 6 L-D interactions and 12 carbons give a value of 78 energy units to send the two molecules away as a gas. to recap:

energy units required to make molecules gaseous:
pentane: 70
hexane : 78

now put the two molecules of pentane onto the thermocouple and in a special box without any other atoms as a gas and vapor pressure doesn't exist.  That the only way for the molecules to become a gas is to add energy to it.  The only source for the energy to get to molecules is the thermocouple.  Can you speculate which delta T would be greater if you only had 2 of each molecule to turn into a gas?

edited to add some clarity...

Offline Kyle1990

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Re: intermolecular forces and evaporative cooling
« Reply #11 on: April 03, 2008, 01:28:30 PM »
OMG IT ALL MAKES PERFECT SENSE!!! thank you so much Rabn for all your *delete me* I'm so sorry it took me so long to figure this out!  :)
"Theories are nets cast to catch what we call 'the world': to rationalize, to explain, and to master it. We endeavor to make the mesh ever finer and finer."
-Karl Popper

Offline Rabn

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Re: intermolecular forces and evaporative cooling
« Reply #12 on: April 03, 2008, 02:17:11 PM »
Understanding is journey.  I don't think it took you that long to grasp the concept. Clarifying this now will undoubtedly make many things easier for you in the future. cheers

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