The h Bar

user228700

9:00 AM

8

Q: Justification for Freezing Point Depression & Boiling Point Elevation in Solutions?

I was wondering if the following justification for freezing point depression and boiling point elevation are conceptually correct. The reason why I ask this question is because I have been self studying chemistry for a course I will be taking this fall, and I don't have a human reference to check...

user228700

And in this ^, the O.P has argued that:

user228700

$$\Delta G_{impure} = \Delta H - T \Delta S$$

user228700

Where $\Delta S = S_{liquid} - S_{solid}$

John Rennie

OK

I guess the $\Delta G$ there is the free energy of melting ...

user228700

The O.P's argument is that since ${S_{liquid}}_{impure} > {S_{liquid}}_{pure}$, we see that temperature must drop, since $\Delta H$ doesn't significantly change.

user228700

9:05 AM

@JohnRennie But that's not what the O.P was going for, anyway.

John Rennie

I would take a very similar though slightly different approach

Kenshin

The OP is saying a process is spontanous if DeltaG is +ve (which is ok using his definition of DeltaG)

user228700

@JohnRennie Okay...

Kenshin

where I define OP = person who answered the question

John Rennie

At the melting point the free energy of the solid and liquid phases must be the same. Yes?

Kenshin

9:07 AM

why?

user228700

You know, I still don't quite understand that. The two phases are in equilibrium with each other?

John Rennie

> The two phases are in equilibrium with each other

Yes.

And we get equilibrium when $\Delta G = 0$

That is the free energy of both phases is equal.

user228700

Hmm, and then we increase the temperature and then it starts properly melting?

Kenshin

yes

but the temperature at which DeltaG is 0 is the freezing/melting point

John Rennie

@Kaumudi.H well if you increase the temperature a bit then some solid will melt and absorb heat as latent heat of fusion so the temperature falls back down to the melting point.

And you re-establish the equilibrium but with a bit less solid and a bit more liquid.

user228700

9:11 AM

Are u telling me that all phase transitions happen little by little, each state being an equilibrium state? Like one of those quasi-static processes?

Kenshin

He's saying the phase transiation happens at the freezing/melting point

not higher, not lower

if you add more heat, the heat is used not to increase temp, but to melt

at the freezig/melting point

John Rennie

Well an equilibrium state is by definition a state in which nothing happens. We need to push the system slightly out of equilibrium for anything to happen.

user228700

Yeah, but it keeps falling back down to an equilibrium state, doesn't it? And then we keep screwing with that. Okay...

John Rennie

@Kaumudi.H yes, exactly.

user228700

@JohnRennie Okay, yes, now.

John Rennie

9:13 AM

We add a bit of heat and push the system out of equilbrium. The system then re-equilibrates and in the process uses the heat to convert a bit of the solid to liquid.

user228700

Right.

John Rennie

I find I'm not keen on the Vader hat. I think I'll go back to the hairy one ...

user228700

Wait, what, @JohnRennie: u were done?

John Rennie

@Kaumudi.H Yes. What remains to be said?

Or do you want to back to the original question no?

user228700

11 mins ago, by John Rennie

I would take a very similar though slightly different approach

John Rennie

9:17 AM

OK.

Start with a pure solvent. At the melting point the solid and liquid are in equilibrium.

user228700

Yeah...

John Rennie

If we add something to the liquid the we increase the entropy of the liquid and therefore decrease the free energy. Is that statement OK or should I justify it?

user228700

U've justified that statement like, 2 times already. I wonder how come u don't remember having spent all these hours :-P

John Rennie

I do other things as well as chatting on the PSE :-)

Anyway ...

If the free energy of the liquid has decreased the system is now out of equilibrium because the free energy of the solid is now higher than the free energy of the liquid.

user228700

@JohnRennie It was literally hours, one day :-)

John Rennie

9:21 AM

Solids don't change entropy (much) because, well, they're solid and their molecules are fixed in place.

user228700

Right, okay...

Kenshin

yes, adding slt to a solid doesn't increase the number of microstates

John Rennie

So free energy changes in solids are mostly due to changes in the enthalpy $H$.

Kenshin

because there is no additional degrees of freedom being added

John Rennie

And the change in enthalpy is basically due to a change in temperature. Specifically: $$\Delta H \approx C \Delta T$$ where $C$ is the specific heat.

So the only way to lower the free energy of a solid is to reduce its temperature.

Can you see where I'm going with this?

user228700

9:24 AM

Yep.

John Rennie

And that's why adding a solute to the liquid reduces the melting point.

I think that's pretty much what that question says as well.

user228700

Right.

user228700

Okay, thank you :-) @Kenshin: I was terribly confused yesterday. I'm sorry for not believing u :-P It's just that u type way too fast and that doesn't give me the proper amount of time to process everything and then I get super confused.

Kenshin

@Kaumudi.H I don't mind if you don't believe me, but you should know i'm always right

John Rennie

Anything more? I have a job to do but it can wait if you wanted to ask more.

koolman

9:25 AM

Why the coordinates of centre are same

@JohnRennie yes

John Rennie

:: John waits expectantly ::

user228700

I realize that the freezing point temperature is probably the same as the melting point but it's not immediately clear to me, so gimme a sec to figure that out...

John Rennie

@Kaumudi.H They are both the temperature at which the solid and liquid are in equilibrium.

In practice the temperature at which the liquid freezes is lower than the temperature at which the solid melts, because we need to push the system out of equilibrium to get anything to happen.

Slereah

@BernardoMeurer $T \in T\mathcal{M}$

user228700

@JohnRennie Riight...

John Rennie

9:30 AM

Technically this is known as superheating and supercooling.

user228700

Wokay, I think I'm onto the last and final page of this wretched chapter and have a "to be clear..." question about reverse osmosis.

John Rennie

OK

Kenshin

@JohnRennie do you relaly look like your display pic?

John Rennie

@Kenshin Without the white hair you mean? :-)

Kenshin

lol yea

John Rennie

9:33 AM

That picture is about 25 years old. I no longer have the beard and I have a lot more wrinkles. But I still look basically the same.

user228700

U apply even a little pressure at the solution side...and some of the solvent particles move to the other side...is that correct?

Kenshin

solvent particles are always moving back and forth

so technically even without pushing you have solvent particles moving to the other side

John Rennie

@Kaumudi.H Remember my explanation from yesterday (or was it Saturday) talking about the partial pressure of the water on the two sides of the membrane?

user228700

No, I mean how u can disturb the dynamic equilibrium.

user228700

@JohnRennie Yeah...

John Rennie

9:35 AM

At equilibrium we have to increase the pressure on the solution side until the partialpressure of the water matches the pressure of the pure water on the other side.

At that point the water flow is the same in both directions.

If we now increase the pressure on the solution side even more the flow rate from the solution into the pure water is higher than the flow rate from the pure water into the solution.

And that's how reverse osmosis works.

user228700

Huh.

Kenshin

that's how you remove toxic fluoride from the water supply

John Rennie

@Kaumudi.H What?

Kenshin

wat

user228700

@JohnRennie Okay, this isn't an equilibrium state. We need to keep applying that pressure, yeah?

user228700

9:38 AM

@Kenshin Hmm, I see.

John Rennie

@Kaumudi.H Yes. If we increase the pressure a bit then water flows out of the solution, so its volume goes down, and the pressure falls again.

So then we need to do some more work on it to increase the pressure again.

user228700

Oh, wow, yeah, okay...

John Rennie

That's why reverse osmosis takes energy of course, because we need to keep doing work to keep the pressure up.

user228700

Although, I didn't get this: (I'm verry sorry :-/)

user228700

> "until the partial pressure of the water matches the pressure of the pure water on the other side."

John Rennie

9:41 AM

Suppose it was ideal gases (because water is complicated).

user228700

OK... (:-P)

John Rennie

On one side we have pure oxygen and on the other a 50/50 mixture of oxygen and nitrogen.

user228700

Okay...

John Rennie

And we'll assume we have a membrane that only lets through oxygen.

user228700

Wokay...

Kenshin

9:42 AM

we already did this bit yesturday

John Rennie

And suppose the whole system is at a pressure of 1 atm.

user228700

@Kenshin What bit is that, exactly?

Kenshin

"until the partial pressure of the water matches the pressure of the pure water on the other side."

John Rennie

@Kaumudi.H Then the partial pressure of oxygen on the oxygen side is 1 atm. Yes?

user228700

Okay, hang on...

user228700

9:44 AM

I do remember u talking about this pressure stuff and I did understand that but I don't want to make u explain all over again, so one sec...

user228700

@JohnRennie Why? (::slowly backs away for fear of outburst::)

John Rennie

Well, go back to the ideal gas system I mentioned.

user228700

Okay... (I deeply apologize on behalf of my dumb brain _/\ _)

John Rennie

Are you happy that the partial pressure of the oxgen is 1 atm on the pure side and 0.5 atm on the mixture side?

user228700

Yep.

John Rennie

9:48 AM

And pressure is due to collisions of the gas molecules with whatever they are in. Yes?

user228700

Yeah...

John Rennie

If the temperature is constant then the gas molecule velocities are constant.

So the pressure is proportional to the number of molecules per second bouncing off the surface.

user228700

The average velocity anyway, okay, yeah...

John Rennie

Yes?

@Kaumudi.H yes :-)

user228700

Yeah (Gosh, I feel terrible. I'm really sorry)

John Rennie

9:50 AM

So on the oxygen side there are twice as many molecules hitting the membrane per second than there are on the mixture side. Yes?

user228700

Yeah.

John Rennie

Becuase the (partial) pressure is twice as high.

Kenshin

so the net flow rate is from the oxygen side to the mixture side

John Rennie

Suppose there is some probability $X$ that if an oxygen molecule hits the membrane it will go through.

This probability $X$ might be as simple as the fraction of the membrance area made up by the holes.

Kenshin

oh I should have said spoiler alert

user228700

9:52 AM

No, no, no, yeah, we can skip to how the net flow rate is from the oxygen side to the mixture side.

John Rennie

Good, because the rest is really simple.

Suppose we keep the pressure on the pure side at 1 atm but raise the pressuire on the mixture side to 2 atm.

Now what are the (partial) pressures of oxygen on the two sides?

user228700

1 and 1.

user228700

(Assuming that the mole fraction of oxygen is 0.5)

John Rennie

And how does this affect the flow rate through the membrane?

user228700

It, um, no net flow now?

Kenshin

9:54 AM

correct

John Rennie

Correct.

Kenshin

dynamic equillibrium

John Rennie

Because the same number of oxygen molecules are hitting the membrane per second on both sides.

user228700

Yeah, OK...

John Rennie

It is as simple as that.

Kenshin

9:55 AM

and flow rate right to left = flow rate left to right

John Rennie

Now suppose I raise the pressure on the mixture side to 3 atm but keep the pressure on the pure side at 1 atm. Would you like to speculate on what happens and why?

user228700

Sure. So the partial pressure of oxygen in the mixture side is now 1.5 atm, that's 0.5 atm greater than the pressure of oxygen on the other side.

user228700

(Still typing)

John Rennie

Yes.

Kenshin

net flow from mixture to pure

user228700

9:58 AM

^ Dood stole my punchline. So yeah, now that more molecules of oxygen are hitting the membrane on the mixture side, there's a net flow from mixture to pure.

John Rennie

And that's how reverse osmosis works.

Kenshin

how?

user228700

Okay. One last terminology question.

user228700

@Kenshin -__-

Kenshin

lol

user228700

9:59 AM

Osmosis is defined as the movement from solventbto solution before attaining equilibrium?

user253310

@SwapnilDas I never studied them separately. I spent the first 2 months in learning trigonometry and inverse trigonometry.

Kenshin

one could argue the "osmotic pressure" is present even in equillibrium

John Rennie

@Kaumudi.H Yes ...

Kenshin

but there is no movement because it is balanced by the opposing "Hydrostatic pressure"

Swapnil Das

@Doraemonドラえもん After Boards?

user228700

10:01 AM

@JohnRennie OK, that's about it, I think. Unless I decide to revisit thermodynamics, that's the last u'll hear about it from for at least two months. This calls for a celebration :-P

John Rennie

:-)

Kenshin

@JohnRennie one question about "osmosis"

user253310

@SwapnilDas Yeah. But you can learn them before boards also if you have free time...

John Rennie

@Kaumudi.H OK I need to get back to work for about an hour. Let me know if/when the customs give you any news.

Kenshin

so it only applies in disequillibrium?

user228700

10:02 AM

@JohnRennie U can count on it :-) Thanku!

Kenshin

isn't there osmotic flow in dynamic equillibrium

Swapnil Das

@Doraemonドラえもん So should I start from sets/n/relations or trig/algebra?

John Rennie

@Kenshin the word osmosis tends to be used ina rather vauge way to mean the whole phenomenon. I'd be cautious about attaching too specific a meaning to it.

Kenshin

k

user228700

^ That. Okay.

user228700

10:03 AM

@Kenshin Is that a word? O_o

user253310

@SwapnilDas Sets and relations isn't very important. You can learn it in 2-3 days.

Kenshin

@Kaumudi.H for sure

investopedia.com/terms/d/disequilibrium.asp

user228700

Cool.

Swapnil Das

@Doraemonドラえもん Theory of equations?

John Rennie

@Kaumudi.H yes dis equilibrium is the one here while dat equilibrium is the one over dere.

4

Kenshin

10:05 AM

lol

good one

user228700

x'D

user253310

@SwapnilDas It is better to do that after learning trigonometry. (You can leave out solution of triangles for the time being )

Swapnil Das

@Doraemonドラえもん Thanks for the valuable advice.

Qmechanic

@BernardoMeurer : Out of curiosity, what was the intended travel route? :)

Swapnil Das

@Qmechanic Out of curiosity, in which field of Theoretical Physics do you work in, Sir? :)

Kenshin

10:14 AM

let me guess

condensed matter

user253310

@SwapnilDas Out of curiosity, how are you sure that Qmechanic is "Sir" and not "Madam" ? :P

Swapnil Das

Oh so many curious people lol

Kenshin

it's very rare for Qmechanic to join us here

Swapnil Das

@Kenshin yeah. He's a active researcher perhaps

Kenshin

he doesn't like associating with underlings

Swapnil Das

10:16 AM

@Kenshin My guess, Quantum Electrodynamics :P

user253310

I used to think that Qmechanic is a bot. Qmechanic seems to edit every new question almost instantaneously !

Swapnil Das

lol.

That's his experience and expertise.

user228700

@Doraemonドラえもん: I vaguely remember that a lot of ionic equilibrium is involved in Electrochemistry.

user253310

@Kaumudi.H Ummm, yeah it is

user228700

But I'm not at all comfortable with IE yet. Will it be OK if I went ahead and did Electrochemistry anyway?

user228700

10:21 AM

Dang it.

user253310

@Kaumudi.H I don't think so...

user228700

Awesome. I'm super screwed.

user253310

Electrochemistry is the most complicated chapter in the syllabus

user253310

You need to master thermo and ionic

user253310

before that

user253310

10:22 AM

i guess

user253310

I am still doing ionic

user228700

Yeah. And Ionic...just, nope.

Swapnil Das

@Doraemonドラえもん and @Kaumudi.H You guys doing Irodov in Physics/

user253310

will start electro in january

user228700

Yeah, I was doing before I gave up for the moment and moved on to other chaps that I need to finish.

user253310

10:23 AM

@SwapnilDas well, stay away from that book for the first one year...its just for show

user228700

@SwapnilDas No, I don't have the time to do everything. My reso material is pretty good :-)

Swapnil Das

@Doraemonドラえもん Why?

@Kaumudi.H Fine :)

user253310

@SwapnilDas It is very time consuming. You will be better off completing the whole syllabus and then return to irodov at the end

Swapnil Das

Oh, hope I did know that earlier :(

user253310

Anyway physics galaxy covers most of the irodov problems which are needed for jee

Swapnil Das

10:25 AM

And wth I am doing the book bit by bit.

user228700

@Doraemonドラえもん: Have u done Surface Chem. yet?

user253310

@Kaumudi.H Oh, yeah i did it some time ago...not a very imp chapter

user253310

you can complete in one day

user228700

Yeah :-P One day is an exaggeration but yeah.

user228700

Solid state?

user253310

10:27 AM

@Kaumudi.H I studied and forgot..got to revise

user228700

Yeah that chapter is almost completely to do with memorizing crap >.<

user228700

CK?

user253310

@Kaumudi.H oh yes...just finished it yesterday :D

user228700

Nice :-)

user253310

This week i will do chemical and ionic equilibrium i suppose

user253310

10:29 AM

Okay..good luck

user228700

U too. Bye :-)

user253310

Bye @SwapnilDas and @Kaumudi.H

user253310

:)

Kenshin

@Doraemonドラえもん bye

user253310

@Kenshin :)

user228700

10:36 AM

@Kenshin: For how much longer will u be around?

user228700

Dang it, are u gone already?

Slereah

fuark

the SE christmas thing has begun

I'd better answer some questions

Kenshin

@Kaumudi.H wat up

user228700

For how much longer are u gonna be around..?

Kenshin

ages y

user228700

10:42 AM

Cool. I was thinking I'd start and finish a really quick chapter now and was hoping there'd be people around in case my brain effs up.

Kenshin

lolk

user228700

U're down to discuss right about any subject, yes? This one is Chemical Kinetics.

Kenshin

easy peasy

my favourite

user228700

@Kenshin I love it when u say that :-P

user228700

To be clear tho, ages = ?

Kenshin

10:44 AM

3 hrs or so

Swapnil Das

@Doraemonドラえもん Goodbye :)

user228700

@Kenshin Perfect. So may I ping u?

Kenshin

no problemo

user228700

Thanks!

koolman

10:56 AM

@Kenshin could you update the system to get notification of edited answers as soon as possible

It is very necessary

Kenshin

@koolman no worries dude, i'll make that the top of my to do list on the site

koolman

Thanks

Aswin Madhu

11:16 AM

@EmilioPisanty Can you please answer the following question. physics.stackexchange.com/questions/299051/…

user228700

11:28 AM

Thus is what my textbook says with regard to the term "rate of a reaction":

user228700

> "The rate of change of concentration with time of different chemical species taking part in a chemical reaction is known as rate of reaction of that species"

user228700

^ Of that species? Isn't rate defined for the whole reaction? @Kenshin

John Rennie

The term rate of reaction isn't precisely defined.

user228700

Huh?

Kenshin

@Kaumudi.H there are two rates

(1) the rate of the entire reaction

and (2) the rate of reaction for each species

user228700

11:34 AM

Right.

John Rennie

Typically if you have some reaction like $A + B \to C + D$ the rate might be $d[A]/dt$ or $d[B]/dt$.

Kenshin

@Kaumudi.H normally you will pick a species in the reaction, and say the reaction is consuming/producing 3 moles per second of these species for example

user228700

* $-(d[A]/dt)/a$

user228700

@Kenshin Yes, yes, then u divide by the stoichiometric coefficient to get the rate of the reaction...

Kenshin

that's right

very gud

user228700

11:37 AM

Alright, thanks, guys :-) I'll be back with more, do not fret :-P

Kenshin

lol better be

@koolman i've made an adjustment to the code

let's test to see if it works

user228700

Hmm, actually, no, never mind. I gots to finish making notes about osmosis and all first and my mom is compelling me to go cycle for a bit now.

user228700

I've done too much (:-P) physical chemistry in the past month. I think I'll give it a rest for awhile and then come back to this chap.

user228700

Sorry for raising up ur hopes, @Kenshin :-P

user228700

Since I need to go cycling and all, I do have a very important question: does anybody here listen to any podcasts?

user228700

11:49 AM

Anybody? No? Wokay.

Kenshin

back

hang on

@koolman your request has been granted

notifications have been implemented when an answer is edited

user228700

@Kenshin: Podcasts?

Kenshin

no

wat do yu mean

user228700

Are u asking what the word "podcast" means?

Kenshin

no why did you think i was talking about podcast

user228700

11:56 AM

I asked this:

Kenshin

oh

user228700

8 mins ago, by Kaumudi. H

Since I need to go cycling and all, I do have a very important question: does anybody here listen to any podcasts?

Kenshin

i see sorry I hadn't read the text wall yet

na i don't :p

user228700

Oh :-) Okay.

Kenshin

wat ones do u listen to

user228700

11:58 AM

Well, I'm a regular listener of:

user228700

itunes.apple.com/us/podcast/dear-hank-and-john/…

user228700

> "It's a comedy podcast about death in which John and Hank Green answer questions, give u dubious advice and discuss the latest news from Mars, the cold dead rock and AFC Wimbledon, the greatest football team in England"

user228700

It sounds very weird but it is very insightful :-)

user228700

I occasionally used to listen to:

user228700

m.soundcloud.com/earbiscuits

Kenshin

12:01 PM

lol

those guys look weird

user228700

Yeah x'D

Kenshin

wat's that one about?

just random stuff?

user228700

No, those two are Rhett and Link. They have a channel on YouTube called "Good Mythical Morning".

user228700

They've been best friends since 1st grade. Sometimes they talk about themselves/random stuff but most of the time they invite other guests and interview them brutally (←Deep questions and all)

Kenshin

sounds fun

user228700

12:04 PM

Yeah :-P

koolman

@Kenshin thanks

@Kenshin its working :)

Kenshin

12:23 PM

awesome dood

heather

hello

Kenshin

hi @heather

heather

yay hats!

wait...why isn't it showing up?

Kenshin

i dunno is mine showing?

heather

yes

Kenshin

12:29 PM

did u just put your hat on?

heather

mine is the same hat as yours

yes

Kenshin

k it takes a few min to display that's all

heather

phew

okay

Kenshin

@koolman if you click on the "users" tab on the new site, can you see "blocked users"

oh

Emilio Pisanty

@AswinMadhu The fact that I edited it for clarity but didn't answer should tell you plenty. I do not particularly appreciate being pinged for this, and certainly not twice for the same question.

Kenshin

12:32 PM

@heather I see in the "to do" list on the site, it says hide the "blocked users" tab

was this ur idea/

heather

yes

can only mods see that?

Kenshin

maybe i'll have to test

heather

hmm okay

Kenshin

why do you think it should be hidden tho

heather

if they find the user searching, that's fine, but it's like SE - we don't want to parade them as shameful users. and, fwiw, we should probably delete all the accounts except xxxx's.

(all the completely spamming accounts)

Kenshin

12:34 PM

maybe

it's better to block than delete in my opinion

because this stops them coming back with the same email

that said, I don't think these particular users will come back due to the anti-spam measures in place

heather

true

but I mean, SE deletes accounts like that, and probably blocks their emails to keep them from coming back.

Kenshin

yeah that would be easy to implement too

heather

i guess what i'm suggesting is - delete accounts that have made no legitimate contribution to the site and block their emails.

I <3 hats

Swapnil Das

1:57 PM

Hey @Doraemonドラえもん

Do you have objective mathematics by RD Sharma? How's the book? Also @Kaumudi.H

user228700

No, I don't have it, sorry.

Swapnil Das

Oh, that's fine :)

I heard that it is a very good book, so thought to start JEE Math.

user228700

Oh, I see. That's great :-)

Swapnil Das

Thanks :-)

user228700

@heather More like faces, really :-P

ACuriousMind

2:14 PM

@MAFIA36790 Your edit here was purely cosmetic and wholly unnecessary to do on a post more than a year old. Try to make edits substantial, especially on old posts that haven't been active for a while.

user116211

Sure, ACM; noted.

user116211

The SO guys have already deciphered the secret hats; damn.

Secret

0

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