Dat Destroyer Chem #41

[userah]

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hey guys, i'm trying to understand this one problem in general chem section of dat destroyer. i have the april 08 edition. I just don't understand the solution at all. It deals with temperature and Kw. Not sure if i can post the problem here, so i'm hoping someone will look and explain to me. please and thank you.

 

[doc3232]

You can post it. Many people don't have destroyer that want to help.
Me<---

 

[userah]

haha ok thanks. here's the problem

H2O(l) + H2O(l) ----> OH-(aq) + H3O+(aq)

When the temperature is decreased from 25* C, it is found that the Kw has decreased. Which of the following is true?

A. The reaction must be exothermic
B. The reaction must be endothermic
C. The reaction is thermoneutral
D. Not possible. Kw = 1 X 10^-14 at all temperatures
E. None of these

Not sure if you want to take a crack at it before I post the solution by Dr. Romano. So i'll leave the solution for now. Please and thank you =)

 

[IdahoDoc]

The pH will stay the same at all temps. this is because both of your reactants are water. If you had reactants other than water then heat would change the pH. This means the Kw will be the same but only for this reaction

 

[eatkabab]

haha ok thanks. here's the problem

H2O(l) + H2O(l) ----> OH-(aq) + H3O+(aq)

When the temperature is decreased from 25* C, it is found that the Kw has decreased. Which of the following is true?

A. The reaction must be exothermic
B. The reaction must be endothermic
C. The reaction is thermoneutral
D. Not possible. Kw = 1 X 10^-14 at all temperatures
E. None of these

Not sure if you want to take a crack at it before I post the solution by Dr. Romano. So i'll leave the solution for now. Please and thank you =)

I'm not so sure about the pH of the system, but I know that D is not correct because the problem it self is directly telling you that the "Kw has decreased." therefore, it must be possible...

a reduction in K indicates a shift towards the reactant side (left in this case). if you imagine "temperature/heat" as just another reactant or product this problem is very easy to solve. using le'chevalier's principal (or however you spell that weird guys name), for a reaction to shift towards the reactants, the concentration of something must have been reduced on the reactant side. that "something" is heat... so the reaction must be endothermic.

le'chevalier also deals with gases and a whole mess of other junk, but none of that other stuff applies to this problem. (no gases)

 

[bigstix808]

for a reaction to shift towards the reactants, the concentration of something must have been reduced on the reactant side. that "something" is heat... so the reaction must be endothermic.

^^^this is false!!

first off, to the OP...Le Chatelier's principle deals with any sort of equalibrium rxn as it was derived as a concept on how equalibrium rxn's shift in order to relieve certain "pressures" placed upon it.

now for the reasoning behind why the Kw would have lowered with a decrease in temp. In order for a system to lower in temp, heat must be taken away via some endothermic rxn (evaporation for example like in your freezer). the process of the heat leaving the system is a result of an endothermic rxn, so thus the water in the OP's question will undergo an EXOTHERMIC RXN (shift to the left) in order to counter those effects (explained by Le C's principle). you can also deduce this another way...when bonds are formed, energy is ALWAYS released. this is what's happening in this problem thus making the answer to this question exothermic.

H2O + heat ----> H+ + OH- (remove heat, and guess what happens

)


i hope that made sense, i wrote it pretty quick. gotta go enjoy the sunshine!!😎

 

[DDSelin2mori]

eatkabab is right. The reaction is endothermic.

 

[eatkabab]

^^^this is false!!

first off, to the OP...Le Chatelier's principle deals with any sort of equalibrium rxn as it was derived as a concept on how equalibrium rxn's shift in order to relieve certain "pressures" placed upon it.

now for the reasoning behind why the Kw would have lowered with a decrease in temp. In order for a system to lower in temp, heat must be taken away via some endothermic rxn (evaporation for example like in your freezer). the process of the heat leaving the system is a result of an endothermic rxn, so thus the water in the OP's question will undergo an EXOTHERMIC RXN (shift to the left) in order to counter those effects (explained by Le C's principle). you can also deduce this another way...when bonds are formed, energy is ALWAYS released. this is what's happening in this problem thus making the answer to this question exothermic.

H2O + heat ----> H+ + OH- (remove heat, and guess what happens

)


i hope that made sense, i wrote it pretty quick. gotta go enjoy the sunshine!!😎

Dude, the book says its endothermic...

 

[eatkabab]

perhaps some clarification is in order.

what I originally wrote:

a reduction in K indicates a shift towards the reactant side (left in this case). if you imagine "temperature/heat" as just another reactant or product this problem is very easy to solve. using le'chevalier's principal (or however you spell that weird guys name), for a reaction to shift towards the reactants, the concentration of something must have been reduced on the reactant side. that "something" is heat... so the reaction must be endothermic.


alright so, the way the reaction is written
H2O(l) + H2O(l) ----> OH-(aq) + H3O+(aq)
I'm placing the term "reactants" on the left side, and "products" on the right side. true that these terms are arbitrary, but traditionally, reactants are on the left and products are on the right, and I believe about 99.9% of problems won't stray from that orientation.

if heat, which is usually written as "q", is placed on the left side (reactants), the reaction is termed endothermic. if q is placed on the right side, its termed exothermic.

for the reaction to move towards the reactant side, either the concentration of something on the reactant side must be reduced, or the concentration of something on the product side must be increased. the move towards the reactant side is compensatory in order to regain the original balance the system once had.

NOTE!!!: a system will NEVER EVER be able to regain the SAME equilibrium point once a stress as been placed upon it. I had to get a question wrong to learn that one...

in this case we are reducing the "concentration" of something, taking away heat, and the reaction is shifting to the reactant side (left). therefore, the only logical conclusion that can be drawn is that the q term, or, heat, is on the reactant(left) side of the eqtn, making the reaction one of endothermic origins.




I personally think le'chevalier's principal is one of the greatest tools I've ever learned. it makes sense in ALLLL aspects of EVERYTHING. from general life, feelings, finances, biology, and of course but not limited to, chemistry. worst comes to worst, I always try to apply this weird guys crazy idea to a problem

 

[jp370]

haha ok thanks. here's the problem

H2O(l) + H2O(l) ----> OH-(aq) + H3O+(aq)

When the temperature is decreased from 25* C, it is found that the Kw has decreased. Which of the following is true?

A. The reaction must be exothermic
B. The reaction must be endothermic
C. The reaction is thermoneutral
D. Not possible. Kw = 1 X 10^-14 at all temperatures
E. None of these

Not sure if you want to take a crack at it before I post the solution by Dr. Romano. So i'll leave the solution for now. Please and thank you =)

[roblem with this difficulty wont be on dat....at least i hope not.....dat is straight forward from what i hear...am i not right?

 

[doc3232]

I think endothermic, WHATS THE ANSWER, we all excited, If I am right I will explain what I did. 😛

 

[doc3232]

[roblem with this difficulty wont be on dat....at least i hope not.....dat is straight forward from what i hear...am i not right?

I think it is a tricky question but not hard, but yes as you said, most questions from all these gen chem and ochem study guides are much harder than the DAT.

 

[eatkabab]

my exam date is on the 30th, but from my study experience, it seems as if this is just about one of the more difficult questions maybe on the DAT. I say difficult because you either know it or you're left guessing. if you know it, its mindless and takes minimal thought, but if you don't, you're stuck.

well maybe many questions are like that, but this one might be a little more than others...

 

[Zerconia2921]

haha ok thanks. here's the problem

H2O(l) + H2O(l) ----> OH-(aq) + H3O+(aq)

When the temperature is decreased from 25* C, it is found that the Kw has decreased. Which of the following is true?

A. The reaction must be exothermic
B. The reaction must be endothermic
C. The reaction is thermoneutral
D. Not possible. Kw = 1 X 10^-14 at all temperatures
E. None of these

Not sure if you want to take a crack at it before I post the solution by Dr. Romano. So i'll leave the solution for now. Please and thank you =)

Hmmm
Okay for a reaction to be exothermic it must be giving off heat being hotter and be on the product side. For it to be endo its getting colder and would be on the reactant side got that.

Alright now kw is products / reactants got it.

Alright now kw is small meaning more reactants than products would give us a an over all smaller kw got that.

Alright if temperture increases the reactions is exothermic all exos reactions are hot if you rmember from lab.
Alright now endos are all cold got that.

If the temp decreased the reaction must be endothermic.

 

[vvvv]

endothermic

 

[doc toothache]

http://www.chemguide.co.uk/physical/acidbaseeqia/kw.html

 

[bigstix808]

http://www.chemguide.co.uk/physical/acidbaseeqia/kw.html

THANKS DOC!! again another beautiful post. i'm not an idiot i swear! if Kw is decreased, as that is a result of lowering the temperature, the rxn will be EXOTHERMIC!!!

AGAIN ... K = [products] / [reactants] (agreed?); if Kw is getting smaller then there are more reactants forming (agreed?).

now lets look at that equation again

H2O + heat ----> OH- + H+

now take the heat away and the reation MUST go to the left to make more reactants (and heat) to account for the heat being lost to the SYSTEM. remember, the system is losing heat because of an outside endothermic rxn, NOT because of the rxn taking place within the water. the water is undergoing an exothermic rxn because it is trying to counter that of the endothermic rxn. if you guys dont want to believe me that is your choice but bond formation (ie: 2H+ + OH- ---> H20) is ALWAYS exothermic.


edit: i don't care if the book says its endothermic, the literature states otherwise. must be a poorly written question.

 

[bigstix808]

The pH will stay the same at all temps. this is because both of your reactants are water. If you had reactants other than water then heat would change the pH. This means the Kw will be the same but only for this reaction

^^^ yea, thats not right either.

 

[bigstix808]

eatkabab, you did such a good job with this explanation, until you contradicted your self at the end...

alright so, the way the reaction is written
H2O(l) + H2O(l) ----> OH-(aq) + H3O+(aq)
I'm placing the term "reactants" on the left side, and "products" on the right side. true that these terms are arbitrary, but traditionally, reactants are on the left and products are on the right, and I believe about 99.9% of problems won't stray from that orientation.

if heat, which is usually written as "q", is placed on the left side (reactants), the reaction is termed endothermic. if q is placed on the right side, its termed exothermic.

forward rxn: H2O + q ---> H+ + OH- ==> endothermic

for the reaction to move towards the reactant side, either the concentration of something on the reactant side must be reduced, or the concentration of something on the product side must be increased. the move towards the reactant side is compensatory in order to regain the original balance the system once had.

true!

in this case we are reducing the "concentration" of something, taking away heat, and the reaction is shifting to the reactant side (left). therefore, the only logical conclusion that can be drawn is that the q term, or, heat, is on the reactant(left) side of the eqtn, making the reaction one of endothermic origins.

(really??)

reverse rxn: H+ + OH- ---> H20 + q ====> EXOTHERMIC!!!!

 

[eatkabab]

eatkabab, you did such a good job with this explanation, until you contradicted your self at the end...


forward rxn: H2O + q ---> H+ + OH- ==> endothermic


true!


(really??)

reverse rxn: H+ + OH- ---> H20 + q ====> EXOTHERMIC!!!!

are you like bent on trying to prove people wrong or somethin? I had already set the reaction orientation in the forward direction. I don't think a reaction is gonna flip on you in the middle of doing the problem.

as you've written in bold, the part that seems to have triggered you was when I said "the reaction one of endothermic origins." If thats the case, well its obvious that I meant 'the reaction is endothermic as written.' but maybe I should have just said that, sorry I guess I got a little excited at the end of my explanation...

 

[bigstix808]

are you like bent on trying to prove people wrong or somethin? I had already set the reaction orientation in the forward direction. I don't think a reaction is gonna flip on you in the middle of doing the problem.

as you've written in bold, the part that seems to have triggered you was when I said "the reaction one of endothermic origins." If thats the case, well its obvious that I meant 'the reaction is endothermic as written.' but maybe I should have just said that, sorry I guess I got a little excited at the end of my explanation...

i'd like to appologize to anyone reading my rants above. ive been sounding out a bit much today. i've had a horrible day and i guess i took it out on here. my b - seriously! studying for the DAT and finals at the same time is beginning to take a toll on me...

bygones???

sweet!😎

 

[userah]

yes the answer is endothermic as others have stated. just a tad confusing haha. i'm gonna go over everyone's posts again and hopefully it will be a little clearer. thanks again everyone =)

btw bigstix, you're not the only one dealing with finals and studyin for the DAT'S haha. i understand how you feel. good luck man =)
my date's set for the 2nd of June.

 

[doc3232]

no no, don't stop, keep arguing over who is more right 🙄

 

[eatkabab]

no no, don't stop, keep arguing over who is more right 🙄

nobody is wrong, some people are just more stressed than others... totally understandable. this forum is FULL of stressed out, sick, and tired people just trying to prove that they are good enough for some ******ed school that thinks a test can define someone...

 

[doc3232]

nobody is wrong, some people are just more stressed than others... totally understandable. this forum is FULL of stressed out, sick, and tired people just trying to prove that they are good enough for some ******ed school that thinks a test can define someone...

Don't use the word "******ed" please, some may take offense to it.
Thanks

 

[joh020340]

nobody is wrong, some people are just more stressed than others... totally understandable. this forum is FULL of stressed out, sick, and tired people just trying to prove that they are good enough for some ******ed school that thinks a test can define someone...

And how do you propose to standardize applicants from hundreds of different undergrad schools? Please enlighten us on a more effective method.

 

[Zerconia2921]

nobody is wrong, some people are just more stressed than others... totally understandable. this forum is FULL of stressed out, sick, and tired people just trying to prove that they are good enough for some ******ed school that thinks a test can define someone...

Hey no school is ******ed they are the ones that realize our dreams of becoming dentists/doctors. Life is worth challenges and you must bend yourself if you want to reach success. If one is seeking an easy route you do not know know the meaning of a ture profession.

I still thinks it endothermic!!

 

[eatkabab]

Hey no school is ******ed they are the ones that realize our dreams of becoming dentists/doctors. Life is worth challenges and you must bend yourself if you want to reach success. If one is seeking an easy route you do not know know the meaning of a ture profession.

I still thinks it endothermic!!

no no, you're both correct, I'm wrong... I realize it, I was just excessively upset at that moment probably. sorry. but it does bother me that so much is riding on one exam. I do realize that they've gotta do something to sort through applicants though. I know, I know, but its just frustrating, you know?

whatever, sorry to offend, didn't mean to. back to studying.

and it is endothermic! 🙂

 

[userah]

sorry this is sort of an unrelated question but i didn't want to open another thread cuz it might be a dumb question haha

I'm figuring out enthalpy and i noticed that there are two equations
bonds broken - bonds formed
and
products - reactants

and i'm trying to figure out the difference between the two? like i keep getting confused when there's question about calculating enthalpy, of which one to use, if someone could clarify for me which is specifically used for which, that would be super. sorry it might be simple but mind's just overloaded and I might not be seeing it...😕

 

[doc3232]

sorry this is sort of an unrelated question but i didn't want to open another thread cuz it might be a dumb question haha

I'm figuring out enthalpy and i noticed that there are two equations
bonds broken - bonds formed
and
products - reactants

and i'm trying to figure out the difference between the two? like i keep getting confused when there's question about calculating enthalpy, of which one to use, if someone could clarify for me which is specifically used for which, that would be super. sorry it might be simple but mind's just overloaded and I might not be seeing it...😕

ok, so I think that each are two separate concepts. Some questions can be answered from bonds broken/bonds formed and some can be answered from reactants/products.
Those ideas cannot be intertwined.🙂

 

[eatkabab]

ok, so I think that each are two separate concepts. Some questions can be answered from bonds broken/bonds formed and some can be answered from reactants/products.
Those ideas cannot be intertwined.🙂

I agree with this, and I also agree with the original poster. I often get confused my self. I kinda just throw both methods at it and find the one that works. thats the benefit of a multiple choice exam...

 

[userah]

haha the problem with that is that it's also an exam that tests how well you can manage time. I'd rather understand when to use which so I don't waste time using both, plugging numbers and seeing if the answer matches one of the answer choices. so no one has any ideas on specifics on when to use which equation?

 

[userah]

noooobody knows? 🙁

 

[eatkabab]

noooobody knows? 🙁

I swear 90 min is an eternity on this exam... there was only 2 calculations on my entire science section. and one of those calculations only required you to choose the correct setup, not actually compute...

I finished the science section with 38 min left over.