General Chemistry Question Thread 2

[gridiron]

This is a continuation of the original General Chemistry Thread.

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[tennisboy85]

I don't think this was carried over. But anyhoo:

Q: Sample of neon is at temperature 285K and pressure of 58 atm. If pressure is lowered to 48 atm, there would be a phase change from:

I said solid to liquid, but the answer is liquid to gas. Can anyone explain? =S

Another Q:

As the temperature at which a reaction takes place is increased:
The reaction rate and the rate constant will increase.

Can someone explain those two terms?

Thanks!

 

[Nevadanteater]

Q: Sample of neon is at temperature 285K and pressure of 58 atm. If pressure is lowered to 48 atm, there would be a phase change from:

I said solid to liquid, but the answer is liquid to gas. Can anyone explain? =S

I'm assuming you need to assume that because neon is a noble gas it isn't going to form a solid unless you're MUCH MUCH colder than 285 and maybe even higher pressures than 58.

Were you given a phase diagram or other information?

As the temperature at which a reaction takes place is increased:
The reaction rate and the rate constant will increase.

Can someone explain those two terms?

Thanks!

the rate constant (k) is a quantification of the speed or a reaction.

as k goes up, the speed (reaction rate) of the reaction goes up.

this is related to temperature via the Arrhenius Equation

k = Ae^-(Ea/RT)

A (frequency constant), Ea (activation energy) and R (gas constant) are all constants.

Only T is variable and as T goes up k also increases as does the reaction rate....

 

[tennisboy85]

I'm assuming you need to assume that because neon is a noble gas it isn't going to form a solid unless you're MUCH MUCH colder than 285 and maybe even higher pressures than 58.

Were you given a phase diagram or other information?

No, I wasn't. That's why I was confused.

Thanks for the 2nd explanation!

 

[mpatricksweeney]

Are there periodic trends to help predict relative acidity among a list of compounds? We'd predict stronger acidity for compounds whose conjugate bases are stable-- what are the factors influencing stability of A-? (Are these the exact same traits which we attribute to good leaving groups in SN1 orgo?) If you don't recognize a compound, and are asked if it's acidic-- what questions do you consider in order to decide?

Specifically, there was an earlier thread which said acidity of HX increases as X moves down a group (for elements in same group). For example, acidity of H2O < H2S < H2Se <H2Te. Is this correct? If so, is it because the larger atoms can better distribute the negative charge on X-? Or some other reason?

2) As for basicity, where are hydrides (generally) compared to hydroxides, oxyanions, etc.? As a rule, can we consider them strong bases?

Thanks, all.

 

[axp107]

Just a general Q about acidity in general..

1) lets say you just have H2SO4.. no water nothing. Is a bottle of H2SO4 without water still considered acidic? I mean, if you measured the pH, what would it read? Considering H+ ions wouldnt form


2) Also, what would the litmus test read... would it turn pink b/c H2SO4 is acidic? Or if H2SO4 doesnt dissociate into H+ ions, will the litmus test do nothing?

 

[jjoe00]

hi, i am doing some practice questions and iwould like to get some clarification if anyone can help

antifreeze is separated from h2o and replaced with an equal mass of substance X. If substance X is soluble and has a low molecular weight than antifreeze then :

the ans is the resulting solution will show increased freezing depression. why?

i am thinking introduction of nonvolatile solutes alway decreases freezing point right so why will the resulting solution have an increased freezing point this time .

pls help, thanks so much

 

[theman120]

joe- you know the basics of the question, and are aware of colligative props based on your knowledge, and you are right that the freezing point gets depressed w/ the addition of a solute. The only point that you are missing is that solute X has a lower MW than antifreeze. Having a lower MW implies increased solubility, therefore solute X is more soluble than antifreeze in water, and therefore if something is more soluble in water it will decrease the freezing point even more.

You have to think of it as: if it's more soluble in water, then it will interrupt the crystal lattice freezing structure of water even more, therefore depressing the freezing point further. This is a theoretical perspective.


If you are more mathematically oriented: Delta Tf=Kf*molality. If the same mass of X is added as the antifreeze and X has a lower MW....than more moles of X have been added. and molality=moles/kg solvent....so an increase in moles would increase molality....therefore increase delta Tf even more....so the freezing pt would be depressed even more.....

hope this helps man. Good luck studying.

 

[HristosKaran]

I have a fairly easy question, but I just don't see it. Can someone explain to me how NaCH3 is a lewis base?? How does it donate electrons?

 

[elayem]

i have a quick question on beta decay (conversion of neutron into proton, electron, and antineutrino). when the electron is ejected, wouldn't that cause the daughter isotope to have a + charge?

e.g., (A/Z X) 14/6 C -> 14/7 N + 0/-1 e- this balances out. but then the 0/-1 e- is ejected, so i would assume that 14/7 N would become 14/7 N+

would this happen? why or why not? thanks in advance...

 

[pezzang]

Let's say that CaCl2 (s) -> Ca2+ (aq) + 2Cl- (aq)
If Ksp were 4 x 10^4, we would set Ksp = (x)(2x)^2

However, if we have an equilibrium equation (non-aq dissolved) like 2A + B -> 2C and we know Keq and [C], we set Keq = (x^2)(x)/[C]^2 to find the conc of A and B.

My question is why don't we use (2x)^2 for the conc of A in the Keq equation while we use (2x)^2 in the dissolution (Ksp) equation? I realize dissolution of one CaCl2 results in two Cl- so for every mole of CaCl2, we get two Cl-. However, shouldn't that be the same for Keq equation where 2A and one B are used to make 2C? I am confused.. Please help..

 

[axp107]

Let's say that CaCl2 (s) -> Ca2+ (aq) + 2Cl- (aq)
If Ksp were 4 x 10^4, we would set Ksp = (x)(2x)^2

However, if we have an equilibrium equation (non-aq dissolved) like 2A + B -> 2C and we know Keq and [C], we set Keq = (x^2)(x)/[C]^2 to find the conc of A and B.

My question is why don't we use (2x)^2 for the conc of A in the Keq equation while we use (2x)^2 in the dissolution (Ksp) equation? I realize dissolution of one CaCl2 results in two Cl- so for every mole of CaCl2, we get two Cl-. However, shouldn't that be the same for Keq equation where 2A and one B are used to make 2C? I am confused.. Please help..

No, you b/c think about what x is...

CaCl2 (s) -> Ca2+ (aq) + 2Cl-

x = Concentration of Ca2+

You could just write it as [Ca2+] [Cl-]^2

These two things are equivalent:

[Ca2+][Cl-]^2 = [x][2x]^2 = Ksp

where x is the concentration of Ca2+

Look at it.. keep looking at it. it should make sense.

 

[pezzang]

No, you b/c think about what x is...

CaCl2 (s) -> Ca2+ (aq) + 2Cl-

x = Concentration of Ca2+

You could just write it as [Ca2+] [Cl-]^2

These two things are equivalent:

[Ca2+][Cl-]^2 = [x][2x]^2 = Ksp

where x is the concentration of Ca2+

Look at it.. keep looking at it. it should make sense.

Thanks but it still doesn't make sense to me. I am not talking about how the concentration is expressed. I am talking about why one (Ksp) has [Cl-] expressed as (2x)^2 while Keq has [A] in (x^2).

 

[axp107]

2A + B -> 2C

define x.. what do you want x to be?

I'm going to call x k?

so it'll be

k eq = (C)^2 / (2x^2)(x) or [C]^2 /[A]^2

 

[3dee]

Question: H4P2O7 and H3AsO3 have acid dissociation constants of 3.0*10^-2 and 6.6*10^-10, respectively at room temp. Find Gibbs Free energy of each dissociation rxn and determine if it is spontaneous. What does this mean for H and S for these reactions.

I am confused as to how the acid dissociation constant relates to solving for Gibbs free energy??

Since the acid dissociation constant is less than 0, that means there are more reactants than products and that H is -, right?

thanks

 

[cheezer]

Let's say that CaCl2 (s) -> Ca2+ (aq) + 2Cl- (aq)
If Ksp were 4 x 10^4, we would set Ksp = (x)(2x)^2

However, if we have an equilibrium equation (non-aq dissolved) like 2A + B -> 2C and we know Keq and [C], we set Keq = (x^2)(x)/[C]^2 to find the conc of A and B.

My question is why don't we use (2x)^2 for the conc of A in the Keq equation while we use (2x)^2 in the dissolution (Ksp) equation? I realize dissolution of one CaCl2 results in two Cl- so for every mole of CaCl2, we get two Cl-. However, shouldn't that be the same for Keq equation where 2A and one B are used to make 2C? I am confused.. Please help..

you are confusing a specific example with a general one. for reaction 2A + B -> 2C

Keq = [C]^2 /([A]^2)() (products over reactants, yea or nay?)

"A" could very well equal 2x if x is defined correctly

anyways the way you define Keq = [C]^2 /(x^2)(x) does not take stoichometry of A and B into account. if B is x, then A is most certainly 2x. if A is x, then B is going to be (1/2)x

 

[cheezer]

I have a fairly easy question, but I just don't see it. Can someone explain to me how NaCH3 is a lewis base?? How does it donate electrons?

Hmmm, I don't know? I drew the structure with C as the central atom and four single bonds around it, H, H, H, Na. I didn't see any lone pairs. Someone help!

 

[cheezer]

Just a general Q about acidity in general..

1) lets say you just have H2SO4.. no water nothing. Is a bottle of H2SO4 without water still considered acidic? I mean, if you measured the pH, what would it read? Considering H+ ions wouldnt form

I'm going to take a shot at this. Yes, H2SO4 should still be considered acidic. The question you should ask yourself is whether H2S04 can still accept electrons as a solid and I think it can. As for pH, no idea, anyways the def of pH I learned -log[h30+] is supposed to be pretty restrictive...I think my professor said it's actually measured with activitiy series. blah.


2) Also, what would the litmus test read... would it turn pink b/c H2SO4 is acidic? Or if H2SO4 doesnt dissociate into H+ ions, will the litmus test do nothing?




I know. Im a huge help.

 

[Foghorn]

Hmmm, I don't know? I drew the structure with C as the central atom and four single bonds around it, H, H, H, Na. I didn't see any lone pairs. Someone help!

NaCH3 is an ionic compound. Anytime you see a metal with something else, besides another metal, think ionic compound.
The components to forming this compound can be expressed
Na+(g) + CH3-(g) ----> NaCH3(s); the negative charge on CH3- is localized on the carbon atom because it' more electronegative that H or that H is more electropositive element

When NaCH3 dissociates in water the dissociation equation is
NaCH3(s) -----> Na+(aq) + CH3-(aq)

 

[cheezer]

NaCH3 is an ionic compound. Anytime you see a metal with something else, besides another metal, think ionic compound.
The components to forming this compound can be expressed
Na+(g) + CH3-(g) ----> NaCH3(s); the negative charge on CH3- is localized on the carbon atom because it' more electronegative that H or that H is more electropositive element

When NaCH3 dissociates in water the dissociation equation is
NaCH3(s) -----> Na+(aq) + CH3-(aq)

okay i'm ******ed. totally forgot about that

well there you go HristosKaran. Na is neutral, and CH3- is a bronsted base, thus a lewis base (or another way to look at it is its single unbonded electron so it can donate)

foghorn, am i getting this right?

 

[Foghorn]

Just a general Q about acidity in general..

1) lets say you just have H2SO4.. no water nothing. Is a bottle of H2SO4 without water still considered acidic? I mean, if you measured the pH, what would it read? Considering H+ ions wouldnt form

I'm going to take a shot at this. Yes, H2SO4 should still be considered acidic. The question you should ask yourself is whether H2S04 can still accept electrons as a solid and I think it can. As for pH, no idea, anyways the def of pH I learned -log[h30+] is supposed to be pretty restrictive...I think my professor said it's actually measured with activitiy series. blah.

It'd be considered an organic acid in it's solid form if water isn't present. It's actually a hygroscopic compound too i.e. the solid, assuming it's in an enclosed jar, when exposed to the environment will absorb water from the air. The pH can't be measured because there would be no H+ ions. No H+ ions would be around since compounds in their solid form have to be electrically neutral.

2) Also, what would the litmus test read... would it turn pink b/c H2SO4 is acidic? Or if H2SO4 doesnt dissociate into H+ ions, will the litmus test do nothing?

Yes it will turn pink as you said even if you took the solid from a jar with a spatula and put it on the paper. Remember it's hygroscopic. A few solid particles will go into solution with the water it absorbs from air.

 

[Foghorn]

okay i'm ******ed. totally forgot about that

well there you go HristosKaran. Na is neutral, and CH3- is a bronsted base, thus a lewis base (or another way to look at it is its single unbonded electron so it can donate)

foghorn, am i getting this right?

Not quite. The Na will still have a + charge because charge conservation must be maintained. Since it's going to be Na+, ionic sodium will act as a lewis acid.

 

[cheezer]

It'd be considered an organic acid in it's solid form if water isn't present. It's actually a hygroscopic compound too i.e. the solid, assuming it's in an enclosed jar, when exposed to the environment will absorb water from the air. The pH can't be measured because there would be no H+ ions. No H+ ions would be around since compounds in their solid form have to be electrically neutral.


Yes it will turn pink as you said even if you took the solid from a jar with a spatula and put it on the paper. Remember it's hygroscopic. A few solid particles will go into solution with the water it absorbs from air.

you are a gentleman and a scholar

 

[Foghorn]

you are a gentleman and a scholar

No problemo. So are those ebooks of any help?

 

[cheezer]

i'll look over them on the weekend. working on lab at the moment.

 

[BananaMonkey]

I had a passage in a PS section that involved the following:

There's an experiment with instructor and students. The instructor preps a .2M NaOH solution. The students then standardized the solution by titration against a pure sample of KHP
KHP(aq) + NaOH(aq) &#8594; KNaP(aq) + H2O(&#8467

After this the students are trying to identify some unknown compounds by titrating them with their standardized NaOH solution.
____
The question I got wrong was:

If a student did NOT remove all the moisture from the KHP before the titration with NaOH(aq), then the molarity determined for the NaOH(aq) would be:

A) too high because the actual number of moles of KHP titrated would be less than the number used in the calculations.
B) too low because the actual number of moles of KHP titrated would be more than the number used in the calculations.
C) too low because the actual number of moles of KHP titrated would be less than the number used in the calculations.
D) unaffected because the weighed KHP was dissolved in water, making any moisture in the sample unimportant.

I said C because I figured if the KHP had water, then a supposed amount X of KHP would be smaller than X because water would account for the rest. So, I figured you have less actual KHP, so when you titrate you use less NaOH.
And then my brain gets confused from there.

The solution said A is correct because: In the titration of KHP with NaOH (Equation 2), the number of moles of KHP titrated equals the number of moles of NaOH needed to reach the equivalence point. If a student weighs a sample of KHP that contains moisture, the measurement will include the mass of both KHP and H2O and will be larger than the mass of the KHP alone. The conversion of the mass of KHP into moles of KHP will carry the error along, making the calculated number of moles of KHP too high. Since the number of moles of NaOH equals the number of moles of KHP at the equivalence point, the number of moles of NaOH will also be too high. The molarity calculated for the NaOH will also be too high, because molarity equals moles/liter.

I don't understand

 

[axp107]

I had a passage in a PS section that involved the following:

There's an experiment with instructor and students. The instructor preps a .2M NaOH solution. The students then standardized the solution by titration against a pure sample of KHP
KHP(aq) + NaOH(aq) &#8594; KNaP(aq) + H2O(&#8467

After this the students are trying to identify some unknown compounds by titrating them with their standardized NaOH solution.
____
The question I got wrong was:

If a student did NOT remove all the moisture from the KHP before the titration with NaOH(aq), then the molarity determined for the NaOH(aq) would be:

A) too high because the actual number of moles of KHP titrated would be less than the number used in the calculations.
B) too low because the actual number of moles of KHP titrated would be more than the number used in the calculations.
C) too low because the actual number of moles of KHP titrated would be less than the number used in the calculations.
D) unaffected because the weighed KHP was dissolved in water, making any moisture in the sample unimportant.

I said C because I figured if the KHP had water, then a supposed amount X of KHP would be smaller than X because water would account for the rest. So, I figured you have less actual KHP, so when you titrate you use less NaOH.
And then my brain gets confused from there.

The solution said A is correct because: In the titration of KHP with NaOH (Equation 2), the number of moles of KHP titrated equals the number of moles of NaOH needed to reach the equivalence point. If a student weighs a sample of KHP that contains moisture, the measurement will include the mass of both KHP and H2O and will be larger than the mass of the KHP alone. The conversion of the mass of KHP into moles of KHP will carry the error along, making the calculated number of moles of KHP too high. Since the number of moles of NaOH equals the number of moles of KHP at the equivalence point, the number of moles of NaOH will also be too high. The molarity calculated for the NaOH will also be too high, because molarity equals moles/liter.

I don't understand

You understand that water will add weight to the KHP right? Let's say your actual molarity of KHP is 2M. With extra water weight in it, it would "appear" to be 3M b/c there would appear to be more KHP. (REMEMBER: You don't know its water! So the MOLARITY would appear to be higher, not lower)

So when you do your calculations, you would think that more NaOH would need to be added.

But in reality, more NaOH would not be needed b/c the KHP amount is still the same (or reduced b/c the extra water takes space). The extra water gives YOU the illusion that there's more KHP.. so when you do your calculations, you'll think, "hey, I need more NaOH to titrate it!"

So your calculations would predict a HIGHER amount of NaOH that would be needed.

Does that make sense? Hope that explanation helped.

PS: I see what you're saying with your logic.. except, you don't know there's water! Thats the thing! You assume its all KHP! Yeah, these questions take time... you gotta sit down and think a lot. I suggest you write stuff down for a problem like this.. it can only help.. of skip it and come back to it later!

 

[BananaMonkey]

You understand that water will add weight to the KHP right? Let's say your actual molarity of KHP is 2M. With extra water weight in it, it would "appear" to be 3M b/c there would appear to be more KHP. (REMEMBER: You don't know its water! So the MOLARITY would appear to be higher, not lower)

So when you do your calculations, you would think that more NaOH would need to be added.

But in reality, more NaOH would not be needed b/c the KHP amount is still the same (or reduced b/c the extra water takes space). The extra water gives YOU the illusion that there's more KHP.. so when you do your calculations, you'll think, "hey, I need more NaOH to titrate it!"

So your calculations would predict a HIGHER amount of NaOH that would be needed.

Does that make sense? Hope that explanation helped.

PS: I see what you're saying with your logic.. except, you don't know there's water! Thats the thing! You assume its all KHP! Yeah, these questions take time... you gotta sit down and think a lot. I suggest you write stuff down for a problem like this.. it can only help.. of skip it and come back to it later!

So THAT was my brain fart. Hehe thanks for the helpful explanation. But, I'm still a little confused. OK so you *think* you have 3M b/c of the extra water, but you really only have 2M. Wouldn't you still need the "real" amount of NaOH during titration? the OH- would only be neutralizing the KHP stuff, so I figured the numbers would still be "real". Perhaps standardization is confusing me...

 

[gymgirl]

Decreasing the pressure would increase the water vapor production. Looking at the equilibrium below, it is clear that reduction of pressure would favor the formation of gas by Le Chatelier​

’s principle.
H2O(l) + heat <-> H2O(g)​

I understand the above statement, except that, I always thought that Le Chateliers Principle does not deal with any phase except for the gaseous state of reactants or products. So, how would we even be able to apply the Le Chateliers principle to this reaction if the reactant is in the liquid state?

Thanks!

 

[axp107]

Decreasing the pressure would increase the water vapor production. Looking at the equilibrium below, it is clear that reduction of pressure would favor the formation of gas by Le Chatelier​

’s principle.
H2O(l) + heat <-> H2O(g)​

I understand the above statement, except that, I always thought that Le Chateliers Principle does not deal with any phase except for the gaseous state of reactants or products. So, how would we even be able to apply the Le Chateliers principle to this reaction if the reactant is in the liquid state?

Thanks!

You can always apply Le Chatlier's principle as long as there is atleast ONE gas somewhere in the reaction (unless its a noble gas of course). That's basically the rule. 2 solids, 2 liquids.. it wouldn't work. It only works b/c Le Chatlier's principle favors gases naturally.

 

[gymgirl]

Thanks axp!

 

[tennisboy85]

This question is from AAMC CBT 8, which I'm not quite sure I understand. Can someone clarify the answer? Anyway, the question asks:

For a given magnitude of B1 (a magnetic field), the nucleus with nonzero procession frequency will be which of the following?

19
9F

In case that doesn't make sense, it basically is saying: Mass number = 19, Atomic number =9

Reason: because protons and neutrons have spin just as electrons do, to guarantee a nonzero net spin, an odd number of nucleons is needed.

 

[RPedigo]

A stranger is being shown around a village that he has just become part of. He is shown a well and his guide says, "on any day except Wednesday, you can shout any question down that well and you'll be told the answer."

The man seems pretty impressed, and so he shouts down, "why not on Wednesday?"
The voice from in the well shouts back, "because on Wednesday, it's your day in the well."

The SDN forums have been an invaluable source of information for me ever since I registered, so I figure that I should give back to the same community that has helped me. This is my time in the well.

I will answer your questions related to MCAT general chemistry.

Qualifications:

• Have taught general chemistry at UC Riverside as a tutor for over two years
• Received a 14 on physical sciences on my MCAT
• Chemistry major, 3.9+ major GPA

Rules:

• MCAT-related questions only; do not post your advanced statistical mechanics class questions in here
• You may ask only one question at a time; do not post a laundry list of fifty questions
• I am not a substitute for your studying; please only post things that you have studied yet do not understand
• If you post a MCAT question, please list the question and all the possible answers [and the correct answer, if available]

Sometimes I might not check back that frequently, but rest assured that I'll answer all your questions. Good luck on your MCAT!

 

[Hindiana_Jones]

Dude, this is awesome. thank you.

 

[SigurRos]

I can also offer my services. I was a general chemistry TA for a year and will be an organic chemistry TA next year. I scored a 13 in PS, but I'm unfortunately no chemistry major (psych major). I don't want to steal RPedigo's thunder, so I will just add second opinions where needed.

 

[axp107]

Hey .. awesome scores there. Thanks a lot for the help

I guess I'll start haha... with a simple question

The rate of the reaction is dependant on temperature, catalyst blah blah.. and concentration?

Rate = k [A] I understand that that's the equation.. except I've seen many posts on SDN saying that the concentration of the reactants does not affect the rate... it seems as though the concentration would affect it, atleast from the equation...

Also, what is k affected by?

Thanks a lot

 

[RPedigo]

Hey .. awesome scores there. Thanks a lot for the help

I guess I'll start haha... with a simple question

The rate of the reaction is dependant on temperature, catalyst blah blah.. and concentration? Is this true?

Rate = k [A] I understand that that's the equation.. except I've seen many posts on SDN saying that the concentration of the reactants does not affect the rate... it seems as though the concentration would affect it, atleast from the equation...

Thanks a lot

Concentration will affect the overall rate, but it will not affect the rate constant k. The reaction that you depicted is first order in respect to both A and B, and second order overall. This basically means that a molecule of "A" has to hit a molecule of "B" with a certain orientation. As you increase the concentration of A and B, the amount of times that a molecule of A and B find each other increases, and therefore you are going to make more molecules of product per second.

In the image I attached, we have A depicted by red molecules, and B depicted by green molecules.

Look at the top part of the image. There is only one molecule of each, so the only possible combination is: A(1) + B(1) --> C

Now, on the bottom part of the image, we tripled the concentration of A and tripled the concentration of B.

Looking at our formula rate = k [A], we would expect the rate to be 9 times as fast. Let's see if it is.

We now have A(1) A(2) and A(3)... and B(1) B(2) and B(3)

Our possible collisions are:

A(1) B(1)
A(1) B(2)
A(1) B(3)
A(2) B(1)
A(2) B(2)
A(2) B(3)
A(3) B(1)
A(3) B(2)
A(3) B(3)

As we can see... 9 combinations, so now it will happen nine times as fast. This is easiest to see, since it's only first order in respect to both A and B.

Now remember this has nothing to do with the rate constant k. The rate constant k is a magical number given to you that was found using the Arrhenius equation.

You will see that the Arrhenius equation is dependent on activation energy and temperature, and therefore the rate constant k will change depending on activation energy and temperature. You might say, "wait, but activation energy should be constant for a reaction!" But then remember... what does a catalyst do to the activation energy? It lowers it. This will make the reaction occur faster, by making k bigger.

Hope this helps!

 

[SigurRos]

Hey .. awesome scores there. Thanks a lot for the help

I guess I'll start haha... with a simple question

The rate of the reaction is dependant on temperature, catalyst blah blah.. and concentration? Is this true?

Rate = k [A] I understand that that's the equation.. except I've seen many posts on SDN saying that the concentration of the reactants does not affect the rate... it seems as though the concentration would affect it, atleast from the equation...

Thanks a lot

Here's my take on this. Please feel free to amend rpedigo

When you say that rate = k[a], you are showing the equation for what is called a 2nd order reaction (but you have to assume that this is the most elementary form of a reaction, otherwise you cannot judge the order of the reaction based on the equation)

There are instances when changing the conc. of a reactant doesn't change the rate.... this is in what is called a zero order reaction where rate = k

remember that if there is an exponent next to a certain reactant, that reactant is the order of the exponent.

for example....

rate = k[a]^2
the overall reaction rate is the third order, and is second order in respect to a and first order in respect to b. So if you double the concentration of a, you quadruple the rate of the reaction. Likewise, if you double the reaction of b, you double the reaction rate.

I know this is a very scatterbrained post, but I hope it helps. I'll try not to drink so much coffee next time haha

 

[axp107]

Concentration will affect the overall rate, but it will not affect the rate constant k. The reaction that you depicted is first order in respect to both A and B, and second order overall. This basically means that a molecule of "A" has to hit a molecule of "B" with a certain orientation. As you increase the concentration of A and B, the amount of times that a molecule of A and B find each other increases, and therefore you are going to make more molecules of product per second.

In the image I attached, we have A depicted by red molecules, and B depicted by green molecules.

Look at the top part of the image. There is only one molecule of each, so the only possible combination is: A(1) + B(1) --> C

Now, on the bottom part of the image, we tripled the concentration of A and tripled the concentration of B.

Looking at our formula rate = k [A], we would expect the rate to be 9 times as fast. Let's see if it is.

We now have A(1) A(2) and A(3)... and B(1) B(2) and B(3)

Our possible collisions are:

A(1) B(1)
A(1) B(2)
A(1) B(3)
A(2) B(1)
A(2) B(2)
A(2) B(3)
A(3) B(1)
A(3) B(2)
A(3) B(3)

As we can see... 9 combinations, so now it will happen nine times as fast. This is easiest to see, since it's only first order in respect to both A and B.

Now remember this has nothing to do with the rate constant k. The rate constant k is a magical number given to you that was found using the Arrhenius equation.

You will see that the Arrhenius equation is dependent on activation energy and temperature, and therefore the rate constant k will change depending on activation energy and temperature. You might say, "wait, but activation energy should be constant for a reaction!" But then remember... what does a catalyst do to the activation energy? It lowers it. This will make the reaction occur faster, by making k bigger.

Hope this helps!

Thanks for a reply.. a long reply really wasn't necessary.. but it helped!

 

[RandomBlackManX]

The SDN forums have been an invaluable source of information for me ever since I registered, so I figure that I should give back to the same community that has helped me. This is my time in the well.

I will answer your questions related to MCAT general chemistry.

Qualifications:

• Have taught general chemistry at UC Riverside as a tutor for over two years
• Received a 14 on physical sciences on my MCAT
• Chemistry major, 3.9+ major GPA

Rules:

• MCAT-related questions only; do not post your advanced statistical mechanics class questions in here
• You may ask only one question at a time; do not post a laundry list of fifty questions
• I am not a substitute for your studying; please only post things that you have studied yet do not understand
• If you post a MCAT question, please list the question and all the possible answers [and the correct answer, if available]

Sometimes I might not check back that frequently, but rest assured that I'll answer all your questions. Good luck on your MCAT!

RPedigo is my hero.

 

[ToledoPike]

Hey RPedigo,

My main question is on Electrochemistry. I can remember almost everything I learn except this area! I always get confused on galvanic and electrolytic cells, anodes, cathodes, and more than anything, reduction potentials!

Are there any easy ways to remember these things or can you give a quick summary of these.

Thanks!

 

[mozdef]

Hey I have a question about PV=nRT. Now I was reading a passage from Ek chemistry and it said "we see that volume decreases with either increasing pressure or decreasing temperature". Now I can clearly see these relationships from the equation. But how come when you increase pressure the temperature doesn't increase instead of just volume decreasing. Or how come when you decrease temperature the pressure doesn't decrease instead of volume decreasing.

This seems so simple but sometimes the simple things confuse me the most lol

 

[gymgirl]

hey guys..

Anyone know why it is that when the temperature of water is raised, its pH becomes higher?
Thanks!

 

[gymgirl]

Holla RPedigo!

GOt a question for ya..
Why is it, that when the temperature of water is raised, its pH increases?
Is it because more OH- ions are formed due to more dissociation?
thanks

 

[heymanooh1]

hey guys..

Anyone know why it is that when the temperature of water is raised, its pH becomes higher?
Thanks!

higher temp increases the dissociation rate of pure water and in effect increases both the concentrations of H+ and OH-, where [H+] = [OH-]. since both ion concentrations increase Kw will also increase.

at temp > 25 celsius: [H+] is larger --> lower pH; [OH-] is larger --> lower pOH; pH + pOH < 14

what do you think happens to [H+], [OH-], pH, pOH, Kw when the temperature is less than 25 celsius?

 

[RPedigo]

Hey RPedigo,

My main question is on Electrochemistry. I can remember almost everything I learn except this area! I always get confused on galvanic and electrolytic cells, anodes, cathodes, and more than anything, reduction potentials!

Are there any easy ways to remember these things or can you give a quick summary of these.

Thanks!

I'd use the RED CAT trick... just remember a red cat. Reduction is always at the cathode. That means that oxidation has to occur at the anode (you can think about AN OX for that).

A galvanic cell goes on its own... it doesn't need any outside work put in. An example is the battery that runs your cellphone or car. If it happens without work put in, that means it is spontaneous, so that means the sign of the Gibb's free energy is negative.

delta G = - n F E(cell)

This is a key formula for electrochemical cells. I'll refer to it often.

n is the number of electrons transferred (can never be negative), F is Faraday's constant (positive). The only way the sign of delta G can change is by the E(cell). We know a galvanic cell is spontaneous, therefore the delta G is negative. But we see on the right that there's already a negative sign. That means that the E(cell) must be positive to keep the delta G negative.

The inverse is true for an electrolytic cell. In an electrolytic cell we use work to drive an electrochemical reaction a way it doesn't want to go. That means that the reaction is nonspontaneous, since we have to add work to get it moving (plug it into a wall, add a battery, etc). Since it's nonspontaneous, delta G has to be positive. That means that the E(cell) must be negative in this case.

As far as reduction potentials, they're always given as the E(cell) for the half reaction. Let's look at two of them:

Hg(2+) + 2e- ---> Hg E=+0.85V
Zn(2+) + 2e- ---> Zn E=-0.76V

Ok, one's negative, and one's positive. What does that mean?

Well, lets refer back to the delta G formula. The mercury (Hg) reaction has a positive E, so that means the delta G for that reaction would be negative. That means that this is spontaneous, so it would be reduced without us having to do anything special.

As for the zinc, as it's written the E is negative, so that means the delta G is positive. That means that this reaction does NOT want to happen. So what we can say is:

For a galvanic cell, we know it's spontaneous. That means that we need both reactions to occur spontaneously.

That means that the mercury reaction can happen as written (so that reaction is good). Since mercury is reduced, that means it must be the cathode (remember RED CAT). Now, we need to balance the electrons, so we have to FLIP the other reaction. Zinc metal will be oxidized instead of reduced, so since we're running the reaction the other way, it is now spontaneous. (recall that if you flip a reaction, you flip the sign of the delta G).

Hopefully that kinda-sorta helps. It's a broad topic, so let me know where you are having specific trouble and I can help!

 

[RPedigo]

Hey I have a question about PV=nRT. Now I was reading a passage from Ek chemistry and it said "we see that volume decreases with either increasing pressure or decreasing temperature". Now I can clearly see these relationships from the equation. But how come when you increase pressure the temperature doesn't increase instead of just volume decreasing. Or how come when you decrease temperature the pressure doesn't decrease instead of volume decreasing.

This seems so simple but sometimes the simple things confuse me the most lol

That all depends on the conditions of the reaction you're doing.

"how come when you decrease temperature the pressure doesn't decrease instead of volume decreasing"

Well, let's say we have a balloon. We decrease the temperature. The air inside the balloon is pushing the walls out so it's inflated (recall that a balloon with no air in it is really flaccid). If we decrease the temperature, the elastic walls can relax inwards and therefore you'll have a decrease in volume.

[YOUTUBE]http://youtube.com/watch?v=rPbE2KSPxuU[/YOUTUBE]

However, let's take the same case with gas trapped inside a metal box with 1' thick walls. If you decrease the temperature of this, the volume certainly won't be able to decrease since the container is what we call rigid. Now, the pressure will just decrease.

"how come when you increase pressure the temperature doesn't increase instead of just volume decreasing"

Again, the question is how you're going to increase the pressure. Let's take a piston and push on it. To increase the pressure, we're decreasing the volume

However, if you heat up our rigid container as we talked about before, the pressure will increase, since there cannot be a volume change.

Hopefully this helps... the MCAT would have to describe the container in question to remove the ambiguity.

 

[RPedigo]

Holla RPedigo!

GOt a question for ya..
Why is it, that when the temperature of water is raised, its pH increases?
Is it because more OH- ions are formed due to more dissociation?
thanks

I think you meant that the pH decreases when you heat water; the pH does decrease with increasing water temperature. This is a kinda tricky question, and hopefully would not be on your MCAT.

As you heat up the water, the Kw (self-ionization constant of water) gets larger, which means we have more H+ and more OH- coming off. The important thing to remember is that for each H+ we have exactly one OH-. Water is H2O, and so the only way we can get another H+ is to have another partner OH-, because that's what's left when we lose an H+ from water.

Remember that a pH meter measures just the [H+], and that's true. So we'll see more H+ in the water, and your pH meter will say "hey! More H+! Let's lower the pH!"... although really, you just have more H+ and more OH-.

Water is a tricky case, and I hope someone can elaborate on this if need be.

So yes, the water has more H+ at higher temperatures because the Kw changes, but it does not become more "acidic", since the H+ and OH- ratio is exactly 1:1 still, and always will be.

This website might help explain it better than I did. Hope this clarifies things, but this is a very weird case.

 

[Ludacris]

Can you please explain the effective nuclear charge a bit? Thanks

 

[RPedigo]

Why not become a contributor to the Study Q&A thread? There's a thread there on Chemistry, I'm sure many will appreciate your feedback there. It'd help keep the forums more organized, otherwise, we end up with 2 threads on the same basic topic.

lol... I didn't even see that subforum; I'd never checked it out. If you think it's redundant, you can do whatever you think is best with this thread. Otherwise, as long as it's open, I'll do explanations. If this one ends up getting shut down, I could contribute to the other forum of course.