General Chemistry Question Thread 2

[gridiron]

This is a continuation of the original General Chemistry Thread.

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[Kaustikos]

Here's something that's been annoying me:
You have a stock solution of, say, Acetic Acid (we'll say 20 mil of 1 M) and want to get to a certain pH before equivalence. Let's say you want to reach a pH of 4. What would you do? I'm basically asking how you solve these types of questions. I found this question from another poster and i'll just quote it:

Answer the questions below concerning the following recipe for 10x TBE (Tris-Borate-EDTA) electrophoresis buffer:


750 mL dd H2O
108.0 g Tris base
55.0 g Boric Acid
40 mL 0.5 M EDTA, pH 8.0 Adjust pH to 8.3 with NaOH (sodium hydroxide).

Anyone?

 

[armybound]

Here's something that's been annoying me:
You have a stock solution of, say, Acetic Acid (we'll say 20 mil of 1 M) and want to get to a certain pH before equivalence. Let's say you want to reach a pH of 4. What would you do? I'm basically asking how you solve these types of questions. I found this question from another poster and i'll just quote it:



Anyone?

I'm not great in chemistry, but I'll give you what came to mind to me..

since you want pH = 4 and pH = -log [H+], you're looking for 1 x 10^-4 = [H+]. your acid is 1M so you need 1 x 10^-4 L or 100 microliters of it if iti dissociates completely? is that right? otherwise you need the Ka and can do the math from there.

so uhhh you need .. that much acid. or something. that concentration. yeah, I suck at chemistry.

 

[Heyeon]

Percentage weight per volume [ % (w/v)]: grams of solute in a total volume of 100 mL.

Example: a solution 25% (w/v) is made by using 25 grams of solute and adding solvent until the total volume is 100 mL.

When the solvent is not specified, it is assumed to be water.

Now just do your algebra

I know, I don't like dilution problems either. Try this one on for size (these aren't MCAT questions by the way):

Answer the questions below concerning the following recipe for 10x TBE (Tris-Borate-EDTA) electrophoresis buffer:

750 mL dd H2O
108.0 g Tris base
55.0 g Boric Acid
40 mL 0.5 M EDTA, pH 8.0

Adjust pH to 8.3 with NaOH (sodium hydroxide). Adjust volume to 1 L.


Notice that Tris, Tris base, Trizma, and Trizma base are all the same reagent, namely Tris (Hydroxymethyl) Aminomethane (C4H11NO3).

Questions:

1) The MW of Tris is 121.2. What is the millimolar concentration of Tris base in a 10x TBE buffer?

2) The MW of Boric Acid is 61.83. What is the millimolar concentration of boric acid in a 1x TBE buffer?

3) What is the molar concentration of EDTA in 10x TBE buffer?

4) How much 10 mM EDTA do you need to make 10x TBE? Discuss your results.

5) How much 1x TBE can you make if you have only 23g of Tris base?

6) What recipe would you use to make 1 L of 1X TBE starting with a solution of 10X TBE?

Answers
1.900mM
2.889mM
3.0.02M
4
5.about 2L?
6.100 ml of 10X+ 900ml H2O (Cant remember!!)

 

[werd]

end point is technically when the indicator changes color. the point where the titrate equals the amount of analyte is known as the equivalence point. for most (good) indicators, the end point is very close to the equivalence point.

 

[werd]

Here's something that's been annoying me:
You have a stock solution of, say, Acetic Acid (we'll say 20 mil of 1 M) and want to get to a certain pH before equivalence. Let's say you want to reach a pH of 4. What would you do? I'm basically asking how you solve these types of questions. I found this question from another poster and i'll just quote it:



alright... so any weak acid "AH" is in equillibrium... AH + H20 <--> H3O+ + A-.

The equillibrium equation is K = Ka = [H] [A-] / [AH]. rearranging, Ka/[H+] = [A-]/[AH]. Ka is known for a given acid, and use the [H+] that corresponds to the desired pH. let's say Ka = 2x10^-5 and desired pH of 4 (H+ = 0.0001)... that gives ***Ka/H = 0.2 = [A-]/[AH].

the total amount of "A" in solution is (from above) is 1 M * 0.002L = 0.002 mol. we can say that the sum of the "A" species is ***[A-] + [AH] = 0.002.

the two ***'d equations give us 2 equations with 2 unknowns, so we can solve for the amount of A-. solving using simlpe algebra, A- = 0.000333 moles and AH = 0.001666 moles.

Now, in the titration, we start with .002 moles of AH (and essentially no A-) and quantitatively make A- as we add strong base ( AH + OH- --> A- +H20). 1 mole of A- is made for every mole of NaOH added. The amount of A- in the pH = 4 solution is the amount of NaOH that needs to be added.... in this example you'd expect adding 0.000333 moles NaOH to create a solution with pH = 4.

 

[nycfella]

If the temperature of a reaction mixture increases after a solution is mixed and undergoes a chemical reaction, does this make the reaction exothermic or endothermic?

 

[smeagol]

Exothermic - You know that heat is released because the temperature has increased.

 

[BloodySurgeon]

[edit] misread.

 

[BerkReviewTeach]

...heating carbon dioxide and solid carbon to give carbon monoxide (CO2(g) + C(s) = 2CO(g) ) and we're doing this in a sealed furnace in order to keep constant pressure.
The question asks if we were to stabilize the system at 1200K and inject a sample of helium into the furnace, what would happen to the amount of carbon dioxide in the system?
A) it should increase
B) it should decrease
C) it should be completely converted to carbon monoxide
D) it will remain the same

Le Chatelier's principle predicts no shift. The reason is this: upon adding an inert gas (one not involved in the reaction), the partial pressures of the reactant and product gases did not change. There was no stress applied to the reaction, so it remains at the same equilibrium before and after the helium is added.

If you were to change the volume of the container, then you would change the partial pressures of the reactants and products, which would in fact displace the system from equilibrium. In that case, you would use Le Chatelier's principle to predict the shift.

 

[BlackSails]

Exothermic - You know that heat is released because the temperature has increased.

I would be careful.

You know that the reaction is, overall, exothermic, but there may be endothermic steps between some intermediates.

 

[MLT2MT2DO]

Percentage weight per volume [ % (w/v)]: grams of solute in a total volume of 100 mL.

Example: a solution 25% (w/v) is made by using 25 grams of solute and adding solvent until the total volume is 100 mL.

When the solvent is not specified, it is assumed to be water.

Now just do your algebra

I know, I don't like dilution problems either. Try this one on for size (these aren't MCAT questions by the way):

Answer the questions below concerning the following recipe for 10x TBE (Tris-Borate-EDTA) electrophoresis buffer:

750 mL dd H2O
108.0 g Tris base
55.0 g Boric Acid
40 mL 0.5 M EDTA, pH 8.0

Adjust pH to 8.3 with NaOH (sodium hydroxide). Adjust volume to 1 L.


Notice that Tris, Tris base, Trizma, and Trizma base are all the same reagent, namely Tris (Hydroxymethyl) Aminomethane (C4H11NO3).

Questions:

1) The MW of Tris is 121.2. What is the millimolar concentration of Tris base in a 10x TBE buffer?

2) The MW of Boric Acid is 61.83. What is the millimolar concentration of boric acid in a 1x TBE buffer?

3) What is the molar concentration of EDTA in 10x TBE buffer?

4) How much 10 mM EDTA do you need to make 10x TBE? Discuss your results.

5) How much 1x TBE can you make if you have only 23g of Tris base?

6) What recipe would you use to make 1 L of 1X TBE starting with a solution of 10X TBE?

4. K2 or K3 EDTA?

 

[SketchLazy]

Can you guys help me solve these dilution problems. These problems are pretty different from the ones I've done in G-Chem. I guess the percentage is throwing me off as I am confused on how to approach it. Any help is appreciated. Thanks.

1) Make 500 milliliters of a 20% solution of glucose (weight/volume). The molecular weight of glucose is 180.2 g/mole.

2) Make 60 milliliters of 0.2% NaCl (weight/volume) from 1% NaCl.

The first one already has been answered, for the second one use this equation for dilutions.

[C]1V1=[C]2V2

[C]1= initial concentration (or in this case weight/volume percent)
V1 = initial volume
[C]2 = final concentration (or in this case weight/volume percent)
V2 = final volume

you have [C]1, [C]2, and V2, now all you have to do is solve for V1.

The answer you get is the volume of 1% NaCl you need to make 60mL of .2% NaCl.

 

[Kaustikos]

Here's something that's been annoying me:
You have a stock solution of, say, Acetic Acid (we'll say 20 mil of 1 M) and want to get to a certain pH before equivalence. Let's say you want to reach a pH of 4. What would you do? I'm basically asking how you solve these types of questions. I found this question from another poster and i'll just quote it:



alright... so any weak acid "AH" is in equillibrium... AH + H20 <--> H3O+ + A-.

The equillibrium equation is K = Ka = [H] [A-] / [AH]. rearranging, Ka/[H+] = [A-]/[AH]. Ka is known for a given acid, and use the [H+] that corresponds to the desired pH. let's say Ka = 2x10^-5 and desired pH of 4 (H+ = 0.0001)... that gives ***Ka/H = 0.2 = [A-]/[AH].

the total amount of "A" in solution is (from above) is 1 M * 0.002L = 0.002 mol. we can say that the sum of the "A" species is ***[A-] + [AH] = 0.002.

the two ***'d equations give us 2 equations with 2 unknowns, so we can solve for the amount of A-. solving using simlpe algebra, A- = 0.000333 moles and AH = 0.001666 moles.

Now, in the titration, we start with .002 moles of AH (and essentially no A-) and quantitatively make A- as we add strong base ( AH + OH- --> A- +H20). 1 mole of A- is made for every mole of NaOH added. The amount of A- in the pH = 4 solution is the amount of NaOH that needs to be added.... in this example you'd expect adding 0.000333 moles NaOH to create a solution with pH = 4.

Thank you so much. I had forgotten how to calculate that. It didn't help that my brain didn't use the ICE box I was taught either.
But thank you for that refresher.

 

[tf2medic]

doing some gen. chem review...this comes from the huge Kaplan 2007 edition book. Under applications of the kinetic molecular theory of gases...it gives the equation for average molecular speed: c=(3RT)squared / MM. Next page it gives r1/r2 = (MM2/MM1)squared for Graham's law of diffusion and effusion. do these actually needed to be memorized for the MCAT? I've never even seen these equations before! seem pretty useless to me....anyone know?

 

[Vihsadas]

doing some gen. chem review...this comes from the huge Kaplan 2007 edition book. Under applications of the kinetic molecular theory of gases...it gives the equation for average molecular speed: c=(3RT)squared / MM. Next page it gives r1/r2 = (MM2/MM1)squared for Graham's law of diffusion and effusion. do these actually needed to be memorized for the MCAT? I've never even seen these equations before! seem pretty useless to me....anyone know?

Know the concepts behind the eqns. Know that the rates of diffusion for two substances are inversely proportional to their MM. The MCAT has definitely tested both of these equations from a conceptual standpoint:
"Which of the following will diffuse faster?"

Besides, it's just two more equations. The eqn relating speed and temperature you should know for sure. The other one is easy enough to figure out if you understand the concepts. GL.

 

[bluemonkey]

Yes, you need to know all of the equations, but as Vihsadas said, knowing how to use them is even more important. The second equation is an easy derivation: since at the same temperature two gases have the same kinetic energy, mava^2=mbvb^2...

 

[physics junkie]

You should understand them conceptually...and yeah, you might need to use them. Learn them.

 

[armymt]

My buddies Jon and Jordan says it is not.

 

[Leathaface2g]

This can't be a serious question...or maybe there's something I'm not getting?

 

[Nasem]

I totally forgot how to do this:

Dissolve 0.10 mol of HCN to make 1.0 L of solution.... what is the pH? (no other info is provided)

Here is what I have:
HCN + H2O <--> H3O(+) + CN(-)

well I know that
K = [H3O+]^2 / [HCN]
and I know that the concentration of HCN = 0.10/1 = 0.10 M,
but what do I do after this??

 

[shippudenfan]

I totally forgot how to do this:

Dissolve 0.10 mol of HCN to make 1.0 L of solution.... what is the pH? (no other info is provided)

Here is what I have:
HCN + H2O <--> H3O(+) + CN(-)

well I know that
K = [H3O+]^2 / [HCN]
and I know that the concentration of HCN = 0.10/1 = 0.10 M,
but what do I do after this??

Okay, either they gave you the Keq (is there a given table somewhere?) or it is a multiple choice question.

If they gave you a Keq:
Then use the eqn:
Keq = x^2/0.1

If it is a multiple choice question:
then you know that pH is going to be bigger than 1 (i.e., if all of the HCN dissolved so that [H+] would be 0.1. Then the -log0.1 = 1. I can't see any other way to narrow it down...can anyone else see?)

I can't see this being done any other way...good thing I opted out of the Jan.MCAT...whew.

 

[Nasem]

its not multiple choice,.... its a problem from my chemistry prof, its weird, I feel like he should have told us what the value of K is....

 

[shippudenfan]

its not multiple choice,.... its a problem from my chemistry prof, its weird, I feel like he should have told us what the value of K is....

I bet you anything there is some table you're not looking at that he assumes you've looked at. Otherwise, the problem is impossible. It is like asking what is x if x+y =3. It can be anything.

 

[Phlame217]

In most chem books, K's are in the back. you should know the HCN is a very weak acid meaning the pH should be in the upper area.

 

[Phlame217]

ehh it depends... some people say molecules are multiple atoms such as water is a molecule (H2O) and other people define molecules as small particles. Since table salt is typically NaCl, it could be defined as a molecule itself or as made up of molecule.

 

[armymt]

ehh it depends... some people say molecules are multiple atoms such as water is a molecule (H2O) and other people define molecules as small particles. Since table salt is typically NaCl, it could be defined as a molecule itself or as made up of molecule.

Thanks . I bought the Audio osmosis CD's and Jon and Jordan said the salt is not made up of molecules. NaCl is a crystalline substance. Perhaps there's some truth to that, but still there is some stuff they say that pretty contradictory to my beliefs. lol

 

[werd]

not heard of these jon and jordan folks, but NaCl is a molecule. while solid table salt exists as a crystal lattice, it is still considered as made of NaCl molecules. lattice structures are common in solids; metals atoms form "electron seas" with nearby metal atoms; even most liquids have some semblance of a structure (eg water each O hydrogen bonds with 2 H's from other molecules, like a water lattice)...despite the secondary level of organization, these things are still considered to be made up of individual molecules.

 

[dondada1]

I think what jon and jordan are saying is that ionic compounds are not technically called molecules. only covalently bonding atoms are given the term - molecular compounds

 

[tf2medic]

I feel like an idiot. I haven't done any aqueous chem since high school and I just started studying. Anyway, here's the question:

If you add 500 ml of 0.2M sodium hydroxide to 500 ml of 0.4M of an unknown weak acid, the final pH is 6.2. What is the pK of the weak acid?

So, you have .1 moles of NaOH and .2 moles of weak acid. The NaOH would subtract .1 mol from the acid and form .1 mol of conjugate base, right? So, you end up with .1 mol of weak acid and .1 mol of conjugate base, so the pK = pH = 6.2?

is that right?

thanks

what? it's pH=pKa + log(unprot/prot). that'd be 6.2 = pKa + log(0.1/0.2) and then you'd solve for the pKa. how did you get 0.1/0.1. the bottom protonated acid refers to the reactant, not the "you end up with .1 mol of weak acid." the unprotonated top part does refer to the conjugate base like you said, so it is 0.1...a few of you agreed with him so either i'm wrong or three of you are wrong haha. assuming i'm wrong, do you have an explanation as to why 0.1/0.2 is wrong? I'm guessing because I did that in terms of moles and those refer to concentrations so they need to be in mol/L. is that my problem? or are you guys wrong? in any case, someone care to explain it in terms of writing the equation out? thanks

 

[tf2medic]

these guys are right but they're somewhat overcomplicating it. you don't really need to know the pH of the stomach for this. on an empty stomach that means there is no MgOH in there. MgOH is a strong base, so if you have that in there with the aspirin, which is decently acidic, of course there is going to be more deprotonated aspirin since it is losing H+'s to that strong base. on an emtpy stomach, there's no strong base, you only have the aspirin. so there's no strong base that is deprotonating it, so on an empty stomach of course there's more protonated aspirin. make sense? maybe a little confusing...but the stomach pH is really unnecessary. that's an unnecessary explanation on the book's part too...too much detail having to worry about that the "pH will raise no more than 4-5 blah blah blah". the point is having that strong base will deprotonate the aspirin to whatever extent and thus will be less protonated than without the MgOH. that's the only info you need. you don't need to think about what direction the equilibrium equation is going or anything, the stomach's <2 pH, or any of that.

 

[BerkReviewTeach]

what? it's pH=pKa + log(unprot/prot). that'd be 6.2 = pKa + log(0.1/0.2) and then you'd solve for the pKa. how did you get 0.1/0.1. the bottom protonated acid refers to the reactant, not the "you end up with .1 mol of weak acid." the unprotonated top part does refer to the conjugate base like you said, so it is 0.1...a few of you agreed with him so either i'm wrong or three of you are wrong haha. assuming i'm wrong, do you have an explanation as to why 0.1/0.2 is wrong? I'm guessing because I did that in terms of moles and those refer to concentrations so they need to be in mol/L. is that my problem? or are you guys wrong? in any case, someone care to explain it in terms of writing the equation out? thanks

Sorry, but the three beat the one in this case.

The HH equation refers to a buffer system, which forms when the 0.1 OH- deprotonates HA, resulting in a solution with 0.1 A- (conjugate base) and 0.1 HA (unreacted HA). The HH equation is for a solution that contains a conjugate pair. So, in this case we have pH = pKa, just like we see with any weak acid titration curve at the half-deprotonated point.

 

[tf2medic]

oooh, so the log(deprot/prot) does not imply products/reactants. it simply is as is stated: deprotonated acid (conjugate base)/protonated acid (unreacted). that makes sense...k thanks! i was getting it confused with the Keq equation in which Q=products/reactants. thanks for the clarification. i think i went through gen. chem, organic, and biochem with the mixup of deprot/prot also meaning product/reactants haha.

hmm take CH3COOH--->CH3COO- + H+ for example, though. the prot. is the reactant and the unprot. is at least one of the products...i suppose if you flip the reaction around, though, then it'll all be messed up. i know in biochem i actually did have it straight...i just never had both equations fresh in my mind at the sametime before until this recent mcat studying. BUT WAIT, the prot. as the denominator then refers to what acid is left over after the reaction?? is that where I was mistaken then? I thought prot. always referred to the original amount of acid...but yea you're right, putting into titration terms, it cant' be the original acid concentration. it has to refer to what's in the solution at the time because as the titration curve goes down/up that y-scale IS the acid concentration. since th HH equation obviously pertains to the particular point you're looking at on the curve, the protonated denominator thus refers to that y-axis that gives what acid concentration it's at. or the x-axis depending on how the graph is labeled...hmm yea okay i knew all this in biochem last semester but forgot. thanks for the clarification Berkley man

 

[LittleRocker]

nope, they're not molecules because they're ionic. I believe they're actually called "formula units". It has to do with the lattice structure of the crystal. There aren't individual molecules of one Na atom and one Cl atom, its more of a ratio. In the crystal you have alternating Na and Cl atoms in a 3 dimensional structure, its not made up of a bunch of "molecules".

 

[manimworried]

nope, they're not molecules because they're ionic. I believe they're actually called "formula units". It has to do with the lattice structure of the crystal. There aren't individual molecules of one Na atom and one Cl atom, its more of a ratio. In the crystal you have alternating Na and Cl atoms in a 3 dimensional structure, its not made up of a bunch of "molecules".

BASICALLY

Table Salt = NaCl
Na+ = ionized atom = cation
Cl- = ionoized aton = anion
NaCl = Table Salt = Ionic Compound.

Molecule = Covalent Compound.

therefore, NaCl is NOT a molecule, but rather, an Ionic Compound.
Table Salt the crystallized form of NaCl.

 

[xrevision]

I think it forms a face centered cubic crystal lattice structure. NaCl is the ratio of each atom.

 

[BerkReviewTeach]

I totally forgot how to do this:

Dissolve 0.10 mol of HCN to make 1.0 L of solution.... what is the pH? (no other info is provided)

Here is what I have:
HCN + H2O <--> H3O(+) + CN(-)

well I know that
K = [H3O+]^2 / [HCN]
and I know that the concentration of HCN = 0.10/1 = 0.10 M,
but what do I do after this??

The acid is weak, so you need to use the Ka expression, and thus need the Ka. Luckily, because you don't need to show your work on the MCAT, we can start at a simplified point of the Bennett Equation:

• pH = 0.5 (pKa + pH if the acid were strong).

Given that pKa for HCN is about 9.2, pH = 0.5 (9.2 + 1.0) = 0.5 x 10.2 = 5.1.

 

[BrokenGlass]

I totally forgot how to do this:

Dissolve 0.10 mol of HCN to make 1.0 L of solution.... what is the pH? (no other info is provided)

Here is what I have:
HCN + H2O <--> H3O(+) + CN(-)

well I know that
K = [H3O+]^2 / [HCN]
and I know that the concentration of HCN = 0.10/1 = 0.10 M,
but what do I do after this??

If you think general chemistry is stupid, you ain't seen nothin' yet. Wait until you apply and you will see stupid in all its mighty glory. Yes, I am talking about the application process.

 

[RySerr21]

okay, either i'm not understanding the concept or this book is wrong but its been bugging me for the last half hour and i'd like to move on. so i thought i'd put it open for discussion.

i'm the in the thermochemistry section of the Kaplan 08 review book.

When talking about gibbs free energy, it gives an example:

When water vapor condenses:

1. h bonds form and energy is released

2. the reaction is exothermic (change of H is negative) and entropy decreases, as a liquid is forming from a gas (T change of S is negative)

3. condenstation will be spontaneous only if T(change)S > than (change)H



i am pretty sure the 3rd statement should be < than and not >. My reasoning is because both enthalpy and entropy are negative, so spontaneous reactious would only occur at low temps....which would mean that we'd have to have T small so that T(change)S is < (change) H and the overall free energy is negative

if i am wrong, someone please help me understand.

 

[BerkReviewTeach]

okay, either i'm not understanding the concept or this book is wrong but its been bugging me for the last half hour and i'd like to move on. so i thought i'd put it open for discussion.

i'm the in the thermochemistry section of the Kaplan 08 review book.

When talking about gibbs free energy, it gives an example:

When water vapor condenses:

1. h bonds form and energy is released

2. the reaction is exothermic (change of H is negative) and entropy decreases, as a liquid is forming from a gas (T change of S is negative)

3. condenstation will be spontaneous only if T(change)S > than (change)H



i am pretty sure the 3rd statement should be < than and not >. My reasoning is because both enthalpy and entropy are negative, so spontaneous reactious would only occur at low temps....which would mean that we'd have to have T small so that T(change)S is < (change) H and the overall free energy is negative

if i am wrong, someone please help me understand.

You're thinking terms of magnitude (absolute value). On a numerical scale, if deltaH is -300 and TdeltaS is -200, then delta G = -300 - (-200) = -100 and thus is spontaneous. Sign conventions are a pin sometimes.

 

[blastula]

You are correct. Condensation will be spontaneous only if delta H < T delta S. I think they may be comparing delta H with -T delta S

 

[blastula]

Condensation is the case where both delta H and T delta S are both negative so temperature would decide spontaneity. delta H has to be < T delta S to get negative delta G. negative minus smaller negative gives negative but negative minus large negative gives positive. I just covered this stuff so happens to be fresh. Hope that helps!

 

[RySerr21]

You're thinking terms of magnitude (absolute value). On a numerical scale, if deltaH is -300 and TdeltaS is -200, then delta G = -300 - (-200) = -100 and thus is spontaneous. Sign conventions are a pin sometimes.

ohhhh. so in your example....-200 is the low temperature i mentioned in the oriignal post, but it is greater than deltaH because it is less negative, which is what the kaplan book said. correct? if so, that makes sense.

thanks!

 

[BerkReviewTeach]

ohhhh. so in your example....-200 is the low temperature i mentioned in the oriignal post, but it is greater than deltaH because it is less negative, which is what the kaplan book said. correct? if so, that makes sense.

thanks!

Statement 3 is correct. Consider the following:

deltaG = deltaH - TdeltaS

deltaG is negative if you have a (smaller #) - (larger #)

For condensation, deltaH is a negative # (exothermic = bond forming) and deltaS is a negative # (liquid is more ordered than gas).

This means that the equation is:

-# = (-#) - T(-#)

This can only be true is the magnitude of deltaH is greater than the magnitude of TdeltaS. So, it is easy to think that |deltaH| > |TdeltaS|. However, their Statement 3 does not state magnitude, it states deltaH and TdeltaS. Given that they are both negative numbers, it works out that:

TdeltaS > deltaH

The example I randomly chose was TdeltaS = -200 and deltaH = -300

-200 > -300 and deltaG = (-300) - (-200) = -300 + 200 = -100, so deltaG is a negative number, making it spontaneous.

Hopefully this makes sense.

In a nut shell, this statement looks weird because we aren't used to comparing the relative sizes of negative numbers.

 

[RySerr21]

Statement 3 is correct. Consider the following:

deltaG = deltaH - TdeltaS

deltaG is negative if you have a (smaller #) - (larger #)

For condensation, deltaH is a negative # (exothermic = bond forming) and deltaS is a negative # (liquid is more ordered than gas).

This means that the equation is:

-# = (-#) - T(-#)

This can only be true is the magnitude of deltaH is greater than the magnitude of TdeltaS. So, it is easy to think that |deltaH| > |TdeltaS|. However, their Statement 3 does not state magnitude, it states deltaH and TdeltaS. Given that they are both negative numbers, it works out that:

TdeltaS > deltaH

The example I randomly chose was TdeltaS = -200 and deltaH = -300

-200 > -300 and deltaG = (-300) - (-200) = -300 + 200 = -100, so deltaG is a negative number, making it spontaneous.

Hopefully this makes sense.

In a nut shell, this statement looks weird because we aren't used to comparing the relative sizes of negative numbers.

right. thats what i thought you were saying from your first post. maybe i just worded my response wrong, but in my head thats what i was telling myself . thanks for the help.

 

[BerkReviewTeach]

right. thats what i thought you were saying from your first post. maybe i just worded my response wrong, but in my head thats what i was telling myself . thanks for the help.

Not a problem... glad it helped. I need to move this thread to the study questions section though.

 

[Nma]

#61 (p.58)

The heats of combustion for graphite and diamond are as follows:

Cgraphite(s) + O2(g) ---> CO2(g) (change of)H = -394kJ
Cdiamond(s) + O2(g) ---> CO2(g) (change of)H = -396kJ

Diamond spontaneously changes to graphite. What is the change in enthalpy accompanying the conversion of two moles of diamond to graphite?

A. -790kJ
B. -4kJ
C. 2kJ
D. 4kJ

The book chose B. as the answer.

Meanwhile, they're asking for Cdiamond(s) ---> Cgraphite. Which means the graphite reaction has to be flipped inorder for graphite to be on the product side, and also, the enthalpy value of -394 must be flipped to +394

Shouldn't the answer then be:
(change of)H= (change of)Hproducts - (change of)Hreactants
which is: (change of)H= (+394) - (-396)= +790 X 2 (wich isnt even in the options.

Am i wrong? Could someone pls. explain?

 

[blastula]

1 mole of Diamond to graphite with have delta H of -2
2 moles with have 2 (-2) = 4

 

[Nma]

Hey Blastula, dont mean to be a pest but could u pls explain? i showed how i understood the calculation but seems i got it wrong. Looking at my explanation in the previous post, could u point out where i went wrong?

It seems u added...y? when the equation for (change of)Hreaction= Hproducts - Hreactants

 

[Nma]

Question#2 for n e other kind hearted citizen

So #62 on the same pg is equally buggn me

How is it possible that "formation of liquid water" is more exothermic than "formation of water vapor"?

 

[blastula]

Flip the first equation to get CO2 -> C + 02 delat H=394. This gives you the conversion of diamond to graphite. Since heats of reactions are additive in exactly the same way that the reactions they belong to are additive -396 + 394 = -2 per mole x 2 mole = -4

Hope that helps!