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[QofQuimica]

All users may post questions about MCAT, DAT, OAT, or PCAT general chemistry here. We will answer the questions as soon as we reasonably can. If you would like to know what general chemistry topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm)

Acceptable topics:
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-particular MCAT-level gen chem problems, whether your own or from study material
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Unacceptable topics:
-actual MCAT questions or passages, or close paraphrasings thereof
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If you really know your gen chem, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current members of the General Chemistry Team:

-QofQuimica (thread moderator): I have my M.S. in organic chemistry and I'm currently finishing my Ph.D., also in organic chemistry. I have several years of university general chemistry TA teaching experience. In addition, I teach general chemistry classes through Kaplan for their MCAT, DAT, OAT, and PCAT courses. On the MCAT, I scored 14 on PS, 43 overall.

-Learfan: Learfan has his Ph.D. in organic chemistry and several years worth of industrial chemistry experience. He scored 13 on the PS section of the MCAT, and 36 overall.

-Sparky Man: Sparky Man has his Ph.D. in physical chemistry. He scored 14 on the PS section of the MCAT, and 36 overall.

-GCT: GCT scored in the 99th percentile on the PCAT. He has also taught introductory physics and general chemistry.

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[QofQuimica]

Here's a heat transfer question for ya. This is from memory but here goes:
you've got a 50g piece of metal and heat it up to 100 degrees C. The metal is put in 50g of water starting at 25 degrees C. What's the final temperature of the water? You're given the c of the metal: 25 - also it's in units such that c for water is 1.

so I set it up like this: (50)(.25)(Tfinal-100) = (50)(1)(Tfinal-25)

the answer has it set up such that the delta Ts are reversed ie. (100-Tfinal) and (Tfinal - 25).

I guess I just don't understand why the delta T of the metal initial minus final. If someone could clarify this for me that'd be killer - thanks.

Because the final T is intermediate between the T of the metal and the T of the water. In other words, you have three temperatures. The lowest is that of the water, which is at 25 degrees. The middle temperature is Tf, which is what you are solving for. And the highest temperature is that of the metal, which is 100 degrees. If you subtract it backwards like you did, then you will end up with a negative delta T for the metal. You can do it this way if you want, but then you must remember to add your two terms instead of subtracting them, the way you have them set up now. (If you put everything on the same side of the equation, you would be subtracting one term from the other.) You could do that by putting a negative sign in front of the metal side, and then you will come up with the correct answer.

 

[Mr. Three-Wiggle]

Thanks for the reply Q. So essentially I'll go right if I always set it up with positive values for temperature change? Or is that only true for this specific case? I guess it's not completely clicking.

 

[QofQuimica]

Thanks for the reply Q. So essentially I'll go right if I always set it up with positive values for temperature change? Or is that only true for this specific case? I guess it's not completely clicking.

Right, make your delta T positive on both sides. This is because of the way that you have set up the equation. If you want to make one delta T negative, you must add the two terms instead of subtracting them. In other words, this is how you have the equation set up:

X = Y

This is the same as saying:

X-Y = 0

So if you want to make the delta T negative on one side, then you must add the two terms instead of subtracting. In other words:

X+(-Y) = 0

Which is still the same as saying:

X = Y

So you'll come up with the same answer either way. Your problem is that you are trying to subtract in two places. In other words, you are trying to essentially do this:

X-(-Y)=0

This comes out to be

X+Y = 0

and therefore

X = (-Y)

which is not correct.

 

[Lindyhopper]

Hi I seem to me confusing the universal gas law & thermo. For an ideal gas PV = nRT. Therefore, if we halve the Volume the pressure is said to double so long as T is constant. But if we halve the volume while keeping the number of moles constant do not we have to do work on the system increasing the temperature? This seems suggest that to that if we halve the volume, T also increase requiring P to more than double. Help Please!

 

[QofQuimica]

Hi I seem to me confusing the universal gas law & thermo. For an ideal gas PV = nRT. Therefore, if we halve the Volume the pressure is said to double so long as T is constant. But if we halve the volume while keeping the number of moles constant do not we have to do work on the system increasing the temperature? This seems suggest that to that if we halve the volume, T also increase requiring P to more than double. Help Please!

If your process is isothermal and you do work on it, you will lose the extra energy as heat. Remember the first law of thermodynamics: U = Q + W. So if T stays constant, U basically stays constant. (U is the internal energy and a measure of T.) Thus, if you increase W, you must have to decrease Q to maintain a constant U (and T).

 

[QofQuimica]

Lindyhopper: I moved your free energy question to the Beyond the Scope of the MCAT Questions thread.

 

[jessica_says]

Hey, what a great website. I have a question that's been nagging me, and I can't seem to find the answer anywhere.

How is phosphate acidic enough to give DNA its acidic property? I learned that an acid is a proton donor or an electrons acceptor, and phosphate has two negative charges on it, along with a lot of lone pairs of electrons. It seems to me it should be classified as a BASE, based on its lack of protons on the phosphate, but I am really confused.

Please help! Thanks

 

[QofQuimica]

Hey, what a great website. I have a question that's been nagging me, and I can't seem to find the answer anywhere.

How is phosphate acidic enough to give DNA its acidic property? I learned that an acid is a proton donor or an electrons acceptor, and phosphate has two negative charges on it, along with a lot of lone pairs of electrons. It seems to me it should be classified as a BASE, based on its lack of protons on the phosphate, but I am really confused.

Please help! Thanks

Hi Jessica,

Please don't post in any of the stickied threads; those are meant for reference posts by the advisors only. I've moved your question here to the general chemistry question thread.

To answer your question: remember that according to the Bronsted-Lowry definitions, an acid and its conjugate base are related by the transfer of a proton. The negatively charged phosphates that you are wondering about are the conjugate bases of the phosphoric acids that are a part of the DNA. Since phosphoric acid is deprotonated at physiological pH (pKa of the acid is about 2, versus a body pH of slightly over 7), the acid will be in its deprotonated form and hence negatively charged. It is true that the negatively charged conjugate base could accept a proton if the environment around it were acidic enough, but that won't happen at a pH of 7. Therefore, phosphate does not act as a base at physiological pH.

Conversely, most organic bases (read amines here) will be POSITIVELY charged at physiological pH. This happens because their pKas are around 10, and they are therefore not able to lose their protons to act as acids in an environment of pH 7.

 

[Lindyhopper]

Are there many types of energy but only two way to transfer energy into or out of a system. Work & heat. Is this accurate or does equation;
delta E = q + w assume that other posible Energy transfers are = 0?

Energy is an idea I have trouble getting a handle on. It seems like there are several types of energy including Radiant,Thermal, Chemical, kinetic, & potential.
But we often talk about changes in potential energy always being completely converted to KE or transferred to the surrounding as heat. The heat thus increasing the thermal energy of the surroundings by increaseing the internal KE of the surroundings by 3/2kT. Does this imply that all the other types of energy are subsets of PE & KE with the transfer energy as work resulting in a change in PE & the transfer of energy as heat resulting in a change in internal KE?
In this little formulation changes in H might be the conversion of internal PE into q transferred to the surrounding. The q then changes the internal KE of the surroundings.

Finally, is there a reversal of sign conventions in physics & chem? In chemistry if the volume of the system expands w= -(P) delta(V). Work is negative as energy is leaving the system. In physics is a system (say a piston expands) w= (P) delta (V) = Fd. Work here would be positive.

I realize there are a few related (& not particularly clear) questions in here but any insights are appreciated.

 

[DrChandy]

Are there many types of energy but only two way to transfer energy into or out of a system. Work & heat. Is this accurate or does equation;
delta E = q + w assume that other posible Energy transfers are = 0?

Energy is an idea I have trouble getting a handle on. It seems like there are several types of energy including Radiant,Thermal, Chemical, kinetic, & potential.
But we often talk about changes in potential energy always being completely converted to KE or transferred to the surrounding as heat. The heat thus increasing the thermal energy of the surroundings by increaseing the internal KE of the surroundings by 3/2kT. Does this imply that all the other types of energy are subsets of PE & KE with the transfer energy as work resulting in a change in PE & the transfer of energy as heat resulting in a change in internal KE?
In this little formulation changes in H might be the conversion of internal PE into q transferred to the surrounding. The q then changes the internal KE of the surroundings.

Finally, is there a reversal of sign conventions in physics & chem? In chemistry if the volume of the system expands w= -(P) delta(V). Work is negative as energy is leaving the system. In physics is a system (say a piston expands) w= (P) delta (V) = Fd. Work here would be positive.

I realize there are a few related (& not particularly clear) questions in here but any insights are appreciated.

Just a quick note to state that delta E=Q-W, not Q+W

I understand your question about the relationship between PE, KE, and internal thermal energy. It is better to think of the different forms of energy as being equally related, rather than some forms being subsets of another.

Finally, in terms of sign conventions, they are the same for both chemistry and for physics. In physics, work is usually asked for in terms of the absolute value (you will know whether it is positive or negative because the nature of the work [ie if it is done on the system or by the system] is usually established in the problem's wording itself)

 

[QofQuimica]

Just a quick note to state that delta E=Q-W, not Q+W

U (or E) = Q + W is the chemistry convention for writing the First Law of Thermodynamics. The signs for work are then reversed versus the physics convention. In other words, if you use the plus version of the equation, you must make work done BY the system to expand ("exowork") negative, and work done ON the system to compress it ("endowork") positive. In contrast, if you use the U = Q - W version, you must make work done BY the system positive, and work done ON the system negative. I think that is what Lindyhopper is asking about.

Lindyhopper, I think that the answer to your first question is that KE vs. PE and Q vs. W are two different ways of looking at a system's energetics. One is not a subset of another, as DrChandy said. But like you said, all types of energy are interrelated. This is not my area of expertise, and I'm sorry that I can't give you a more detailed answer.

 

[thadarknyte]

can someone explain why a catalyst can change the rate constant and rate law, but not the equilibrium constant of a reaction?

 

[QofQuimica]

can someone explain why a catalyst can change the rate constant and rate law, but not the equilibrium constant of a reaction?

Catalysts affect the kinetics of a reaction, not its thermodynamics. In other words, the catalyst makes the reaction happen faster. It does not affect the relative stability of the reactants versus the products. The catalyst gets the reaction to wherever it "wants" to go faster, but it does not change the final destination (product ratio) that the reaction would like to have. Since Keq is the ratio of product concentrations to reactant concentrations at EQUILIBRIUM, Keq stays constant for a given set of environmental conditions, even in the presence of a catalyst. On the other hand, Q, the reaction quotient when the reaction is not at equilibrium, DOES change if you add a catalyst. Perhaps that is part of what is confusing you.

One example to help explain this is that of a diamond ring. Thermodynamically, the carbon in the diamond would be more stable as graphite here on the earth's surface. However, this reaction occurs so slowly that you can count on being able to pass your diamonds on to your heirs, and they in turn to theirs. On the other hand, say I had a catalyst for this diamond and graphite conversion reaction. In that case, I would make the diamond turn into graphite much faster. But I could not use the same catalyst to turn the graphite back into diamond (assuming that I'm at STP), because thermodynamically the carbon "prefers" to be in its graphite form. The final ratio at equilibrium of diamond and graphite will always be the same, but the catalyst will get me to the preferred graphite product faster than I'd get there without it.

Somehow, I don't think that most people would want to let their diamond rings anywhere near my catalyst.

 

[thadarknyte]

thank you. that really helped!

 

[kevin86]

1. So back to the catalyst question. Can you add unlimited amounts of catalyst, and not alter the reaction as oppose to the amount needed to reach Vmax. Also, so if a question asks something about the function of the catalysts, it will never have any effects on the thermodynamics right. But what about the effects on stability of the intermediates, would that be considered thermodynamics and never kinetics?

2. does low density automatically imply higher volume in one of those glass tubes that measure atomsphere pressure where you have one with mercury and another compound with higher density. I forgot which equation combines all that.

 

[QofQuimica]

1. So back to the catalyst question. Can you add unlimited amounts of catalyst, and not alter the reaction as oppose to the amount needed to reach Vmax. Also, so if a question asks something about the function of the catalysts, it will never have any effects on the thermodynamics right. But what about the effects on stability of the intermediates, would that be considered thermodynamics and never kinetics?

Right. The catalyst is regenerated after each use, and it works fast, so you don't generally need to add more. As long as there is excess substrate, the catalyst will work as fast as it can. Adding more catalyst will speed things up to a point, but eventually you will slow down because of running out of substrate. Catalysts should also not affect the stability of the intermediates; that's thermodynamics, as you said.

2. does low density automatically imply higher volume in one of those glass tubes that measure atomsphere pressure where you have one with mercury and another compound with higher density. I forgot which equation combines all that.

Density is mass over volume, and assuming that mass stays constant, the only thing that can be changed is the volume. So yes, a larger volume will correspond to a smaller density, assuming that no change in mass can occur.

 

[kevin86]

Also is there a connection between density and viscosity. Is the reason mercury rise higher than water given the same volume and atm pressure due to density only or is viscosity involved too.

 

[QofQuimica]

Also is there a connection between density and viscosity. Is the reason mercury rise higher than water given the same volume and atm pressure due to density only or is viscosity involved too.

Hmm, I don't think so. Mercury isn't particularly viscous; it's only slightly more viscous than water. Viscosity is a measure of resistance to flow. When you think of a viscous liquid, think of something like maple syrup, which flows slowly, in comparison to something that flows rapidly like water or mercury.

 

[thadarknyte]

The question is:

What is the solubility of a Ca(OH)2 in a 0.1 M solution of KOH? You are given that the Ksp of Ca(OH)2 is 1.3x10^-6

How do you start this?

 

[QofQuimica]

The question is:

What is the solubility of a Ca(OH)2 in a 0.1 M solution of KOH? You are given that the Ksp of Ca(OH)2 is 1.3x10^-6

How do you start this?

Start by writing down the equation for the dissociation of calcium hydroxide. It will be difficult for you to come up with a Ksp equation without it.

 

[thadarknyte]

Start by writing down the equation for the dissociation of calcium hydroxide. It will be difficult for you to come up with a Ksp equation without it.

haha sorry, I didnt fully complete my question.

So the dissociation is:

Ca(OH)2 --> [Ca^2+][2OH]^2

So you have [x][2x+.1]^2

Now I dont know where to go from there? Usually in problems I would make something zero, but for some reason I am not too sure how to complete this...

 

[QofQuimica]

haha sorry, I didnt fully complete my question.

So the dissociation is:

Ca(OH)2 --> [Ca^2+][2OH]^2

So you have [x][2x+.1]^2

Now I dont know where to go from there? Usually in problems I would make something zero, but for some reason I am not too sure how to complete this...

So far you've been doing great; your set up is correct. Now, you don't want to have to use the quadratic formula if you can avoid it, esp. b/c you don't have a calculator on the MCAT. Generally, x will be a very small number, several orders of magnitude smaller than 0.1. So you can neglect the x and just say that the hydroxide concentration is approximately 0.1 M still. That simplifies the math considerably. On the other hand, you have to leave the calcium concentration as x, b/c the only source of calcium ion is the calcium hydroxide.

 

[Orthodoc40]

So far you've been doing great; your set up is correct. Now, you don't want to have to use the quadratic formula if you can avoid it, esp. b/c you don't have a calculator on the MCAT. Generally, x will be a very small number, several orders of magnitude smaller than 0.1. So you can neglect the x and just say that the hydroxide concentration is approximately 0.1 M still. That simplifies the math considerably. On the other hand, you have to leave the calcium concentration as x, b/c the only source of calcium ion is the calcium hydroxide.

Sorry to jump in but these questions gave me hell last year so I'm hoping to see if I get it...Don't you just go:

Ksp (value given above as 1.6 x 10^-6) = 4x^3?

And by the way, in KOH, is it going to dissociate very much ANYway?? Or is that what you were trying to say?

 

[QofQuimica]

Sorry to jump in but these questions gave me hell last year so I'm hoping to see if I get it...Don't you just go:

Ksp (value given above as 1.6 x 10^-6) = 4x^3?

No, because this is a common ion problem. The KOH is a strong electrolyte, meaning that it dissociates completely into K and OH ions. Therefore, a 0.1 M solution of KOH will give you 0.1 M of K ions and 0.1 M of OH ions. We don't really care about the K ions; they're just a spectator. But we do care about the OH ions, because they are going to affect the dissociation of the Ca(OH)2. So your final concentration of OH will be 0.1 M, from the KOH, + 2x, from the Ca(OH)2. You can round 0.1 + 2x to just 0.1, since 0.1 is much larger than x.

And by the way, in KOH, is it going to dissociate very much ANYway?? Or is that what you were trying to say?

No, as I explained above, KOH is a strong electrolyte and dissociates completely.

 

[Orthodoc40]

No, because this is a common ion problem. The KOH is a strong electrolyte, meaning that it dissociates completely into K and OH ions. Therefore, a 0.1 M solution of KOH will give you 0.1 M of K ions and 0.1 M of OH ions. We don't really care about the K ions; they're just a spectator. But we do care about the OH ions, because they are going to affect the dissociation of the Ca(OH)2. So your final concentration of OH will be 0.1 M, from the KOH, + 2x, from the Ca(OH)2. You can round 0.1 + 2x to just 0.1, since 0.1 is much larger than x.

No, as I explained above, KOH is a strong electrolyte and dissociates completely.

Okay, I think I get your answer to the first part, but I thought the question was about the Ca(OH)2's solubility in KOH? So if there's a lot of OH ions from the KOH hanging around already, the Ca(OH)2 wouldn't completely dissociate?? Maybe I don't get it.

 

[QofQuimica]

Okay, I think I get your answer to the first part, but I thought the question was about the Ca(OH)2's solubility in KOH? So if there's a lot of OH ions from the KOH hanging around already, the Ca(OH)2 wouldn't completely dissociate?? Maybe I don't get it.

Right. The calcium hydroxide doesn't completely dissociate anyway, b/c it's a weak electrolyte. But in the presence of KOH, it dissociates even less than it would in pure water, as predicted by Le Chatelier's Principle.

 

[93106]

Hi!

I am frequently confused by ionization energy. I ran across a passage regarding the haber process, and a question asks why h and n react slugglishly while o and h react explosively. The answer I chose (incorrectly) says that the ionization energy for o is less than that of n.

The trend is up and to the right for ionization energy, right? and in those directions the ionization energy is more negative, right? so o's ionization energy should be more negative, and thus "less" than n's, right??

harrumph!!

 

[93106]

Hello!

When is water left out of the equilibrium expression?

 

[93106]

nevermind about the ionization energy question! i was confusing it with electron affinity!

 

[QofQuimica]

Hi!

I am frequently confused by ionization energy. I ran across a passage regarding the haber process, and a question asks why h and n react slugglishly while o and h react explosively. The answer I chose (incorrectly) says that the ionization energy for o is less than that of n.

The trend is up and to the right for ionization energy, right? and in those directions the ionization energy is more negative, right? so o's ionization energy should be more negative, and thus "less" than n's, right??

harrumph!!

I wrote a post about the periodic trends in the General Chemistry Explanations Thread.

Water is a pure liquid and should never be included in a Keq expression.

 

[QofQuimica]

nevermind about the ionization energy question! i was confusing it with electron affinity!

Just to be clear: electron affinity is reported as an absolute value, meaning that you drop the negative sign. So as you go from left to right, the electron affinity is said to increase, even though technically yes, it is releasing more energy as you move from left to right.

 

[kevin86]

Hey can someone please give me a good definition of catalysts and reaction rates. Like does a catalyst stablilize the transtion state, and how? One of the tpr says it stabilizes the transition state by makeing it thermodynamically favorable. It was the wrong answer before of it mentioned thermo thats what i thought. But then how does it stabilizes the transtion state.

2. reaction rates, increasing temperature always speed it up right, but not necessarily push the reaction forward or reverse right?

3. is negative gibbs free energy exothermic, or does that just corresponds to enthapy.

4. What does vapor pressure depend on, is it just how strong the molecules are held together, like stronger molecules have low vapor pressures. And how does temperature and pressure increase, decrease or not do anything to vp.

 

[QofQuimica]

Hey can someone please give me a good definition of catalysts and reaction rates. Like does a catalyst stablilize the transtion state, and how? One of the tpr says it stabilizes the transition state by makeing it thermodynamically favorable. It was the wrong answer before of it mentioned thermo thats what i thought. But then how does it stabilizes the transtion state.

Catalysts lower the activation energy, right? So that is what your book is trying to tell you. I can't draw one here, but take a look at a reaction coordinate diagram. You will see that the transition state is at the top of an energy "hill." If you lower the hill, then you make the activation energy lower and you stabilize the transition state; that is essentially what the catalyst is doing. This is different than thermodynamic stability, where you're talking about stabilizing the PRODUCT of the reaction.

2. reaction rates, increasing temperature always speed it up right, but not necessarily push the reaction forward or reverse right?

Most reactions speed up if you heat them because most reactions that you are familiar with are endothermic. Heat can be considered a reactant in an endothermic reaction, so adding more of it pushes the reaction toward the products according to LeChatelier's Principle. If the reaction is exothermic, where heat is like a product, you will push it in the reverse direction by heating it up.

3. is negative gibbs free energy exothermic, or does that just corresponds to enthapy.

The term "exothermic" refers to a negative enthalpy only. A reaction that is spontaneous (negative delta G) is called "exergonic." The opposite (for a positive delta G) is "endergonic."

4. What does vapor pressure depend on, is it just how strong the molecules are held together, like stronger molecules have low vapor pressures. And how does temperature and pressure increase, decrease or not do anything to vp.

It is not the strength of the bonds comprising the molecules that matters, but rather the strength of the intermolecular forces holding one molecule to the next. Liquids that hydrogen bond have lower vapor pressures than liquids with dipole-dipole interactions, which in turn have lower vapor pressures than liquids with van der Waals interactions. I wrote a post about intramolecular versus intermolecular forces in the gen chem explanations thread; you can check it out if you need to review your intermolecular forces versus intramolecular bonds.

Increasing temperature will increase vapor pressure. Remember that temperature is a measure of the kinetic energy of a substance. So as you warm the liquid, more and more of its molecules on average will have enough KE to escape and become gaseous. Eventually, you reach a constant temperature above which you cannot heat the liquid (the boiling temperature). This occurs when the vapor pressure is equal to the atmospheric pressure.

Increasing atmospheric pressure makes it more difficult to boil your liquid. This is because you have the same amount of vapor pressure, but you have a higher atmospheric pressure. Again, the vapor pressure must equal the atmospheric pressure in order for boiling to occur.

 

[Lests55]

Hey Q! I really appreciate all your help! Ever decided where you are going next year...I think it is the big soap opera on SDN, you are certainly the most famous celebrity...Where will Q end up? I assume you are racking up the acceptances, and if not I might as well quit now...anyway...it would be awesome to meet you one of these days!

Will you answer this one first or the one in the organismal biology?

OK on to the question....

I know Temperature affects equil. K and rate constant k. And increasing T, increases rate and thus increases rate constant (little k) because they are proportional. rate=k{a}....

But can we predict if Temp will make K increases or decrease? My first instinct is no. And setting G=H-TS (sorry for the lack of deltas) equal to G=-RTln(K) really doesn't make anything intuitively obvious to me...

Thanks!

 

[KAM]

No magic Q here, but if I were to educate guess i would say simply yes. Yes, with everything else constant, an increase in temp would increase the rate and a decrease in temp would decrease the rate. no if's and's or but's about it

 

[QofQuimica]

Hey Q! I really appreciate all your help! Ever decided where you are going next year...I think it is the big soap opera on SDN, you are certainly the most famous celebrity...Where will Q end up? I assume you are racking up the acceptances, and if not I might as well quit now...anyway...it would be awesome to meet you one of these days!

lol, I don't think that most people care quite that much. But no, I haven't decided yet.

Will you answer this one first or the one in the organismal biology?

Generally I start at the bottom and work my way up. So since you posted in the other thread first, I read that one first. Unfortunately, I don't know the answer, but hopefully MollyMalone will.

OK on to the question....

I know Temperature affects equil. K and rate constant k. And increasing T, increases rate and thus increases rate constant (little k) because they are proportional. rate=k{a}....

But can we predict if Temp will make K increases or decrease? My first instinct is no. And setting G=H-TS (sorry for the lack of deltas) equal to G=-RTln(K) really doesn't make anything intuitively obvious to me...

Thanks!

This one I can answer. Yes, we can predict what K does as long as we know whether the reaction is endothermic or exothermic. Remember that K is equal to the product of the concentrations of the products divided by the product of the concentrations of the reactants. For an endothermic reaction, where heat is like a reactant, warming up the reaction will drive it to the right according to Le Chatelier's principle. Thus, you will form more product, and K will increase because you've increased the numerator (product concentrations) and decreased the denominator (reactant concentrations). For an exothermic reaction, where heat is like a product, warming up the reaction will drive it to the left, and K will decrease because the numerator decreases and the denominator increases. Good question.

 

[QofQuimica]

No magic Q here, but if I were to educate guess i would say simply yes. Yes, with everything else constant, an increase in temp would increase the rate and a decrease in temp would decrease the rate. no if's and's or but's about it

Good guess. Most reactions you are familiar with are endothermic, and so K will increase with increased temperature, as I explained above.

 

[jsong812]

I have a quick question about enthalpy. So, enthalpy is equal to the change in internal energy plus the product of pressure and the change in volume. I understand why at constant volume, enthalpy is equal to the change in internal energy; because the change in volume is zero, that cancels out the PV work. However, I do not understand why at constant pressure, enthalpy is equal to heat, q. Examkrackers tells me to just memorize it but I think it'd much more beneficial if I was able to understand why this is so. Any input would be much appreciated. Thanks!

 

[QofQuimica]

I have a quick question about enthalpy. So, enthalpy is equal to the change in internal energy plus the product of pressure and the change in volume. I understand why at constant volume, enthalpy is equal to the change in internal energy; because the change in volume is zero, that cancels out the PV work. However, I do not understand why at constant pressure, enthalpy is equal to heat, q. Examkrackers tells me to just memorize it but I think it'd much more beneficial if I was able to understand why this is so. Any input would be much appreciated. Thanks!

This is beyond the scope of the MCAT, but it relates to the fact that the equation you gave should technically be written in differential form, substituted first, and then integrated:

dH = dQ + V(dP) -> H = Q + VP

Since an isobaric (constant pressure) process has dP (change in P) = 0, then the second term will drop out, and you will wind up with dH = dQ, which integrates to H = Q.

 

[jsong812]

aaah.. that makes sense. Thanks so much.

 

[kevin86]

Again about the heat and the reaction rate thing. I didn't mean which direction it goes to, but the rate. When they say increase the heat increases the kinetic energy and then the rate, does it mean toward the equilibrium? they are talking about activiation energy is being overcome by kinetic energy right? And whether it stays in equilibrium would be thermodynamic with the exo and endo stuff.

I have a question about specific heat/heat capacity. So if an object can hold on to more energy when droped into room temperature water where the change in the temperature for the object is smaller than another object, then it has a high specific heat right?

So since the final temperature is the same as the water, then if the 100Celcius metal only changed to 90 Celsius while the other object changed from 100 to 60, it would mean the the 90 celsius one held on to more energy. But if ther water is 90celsius as oppose to 60 celsious, wouldnt that mean the object released more energy? So confused on that one.

could you show it to me in equation form

By the way whats the difference between standard condition and stp

Also what happens to a buffer''s pH if you dilute it.

 

[Lests55]

Hello again Q, two unrelated topics.

(1). Kw/pH changing with temperature
I had a practice passage about Kw changing with temperature and how pH also changes. That's all well and good that the hydrogen ion concentration increases because the Kw also increases, and thus by definition the pH must increase. I understand this by "definition" but not by intuition. Since pH is defined by the conc. of protons it makes since. But intuitively it seems like pH would stay the same in that it is not any more acidic because the hydroxide ion concentration also increases. Probably poor explanation, but...

(2) Reactivity
I am aware of the existence of a metal displacement chart and that it is not required knowledge for el MCAT. If I am remembering correctly it simply says that one metal can replace certain metals but not others. Does this relate to the periodic trends...for instance if I were asked which of the following metals was most reactive would that mean that I have to choose the one that has the smallest electron affinity/electroneg? Can I predict reactivity with these trends?

THANKS!

 

[Orthodoc40]

Okay, in this TPR science workbook, they have this question:

Of the following, which is the strongest base?

A) HO-
B) NH3
C) NH2-
D) CH3-

Answer is D. How's that? I thought OH- was about as strong a base as you can get?

 

[Lests55]

Okay, in this TPR science workbook, they have this question:

Of the following, which is the strongest base?

A) HO-
B) NH3
C) NH2-
D) CH3-

Answer is D. How's that? I thought OH- was about as strong a base as you can get?

Which is a stronger acid, water or methane?
It's obviously water so stronger acid=weaker conjugate base (stable). Weaker acid=stronger conjugate base=unstable.

 

[Orthodoc40]

Which is a stronger acid, water or methane?
It's obviously water so stronger acid=weaker conjugate base (stable). Weaker acid=stronger conjugate base=unstable.

Okay that makes sense...why didn't I think of that??!

 

[QofQuimica]

Again about the heat and the reaction rate thing. I didn't mean which direction it goes to, but the rate. When they say increase the heat increases the kinetic energy and then the rate, does it mean toward the equilibrium? they are talking about activiation energy is being overcome by kinetic energy right? And whether it stays in equilibrium would be thermodynamic with the exo and endo stuff.

I have a question about specific heat/heat capacity. So if an object can hold on to more energy when droped into room temperature water where the change in the temperature for the object is smaller than another object, then it has a high specific heat right?

So since the final temperature is the same as the water, then if the 100Celcius metal only changed to 90 Celsius while the other object changed from 100 to 60, it would mean the the 90 celsius one held on to more energy. But if ther water is 90celsius as oppose to 60 celsious, wouldnt that mean the object released more energy? So confused on that one.

could you show it to me in equation form

By the way whats the difference between standard condition and stp

Also what happens to a buffer''s pH if you dilute it.

1) I think you are confused because there are two things changing here: both the value of k (the kinetic constant) and K (the equilibrium constant) change with temperature.

2) Yes, the one with the smaller delta T will have the higher value of c, assuming that the masses are the same.

3) Yes. Specific heat is not the same thing as energy. Remember, Q is equivalent to energy, not c. I'm not sure off the top of my head how to show this in equation form, but I'll think about it some more this weekend.

4) Both STP and standard conditions occur at 1 atm of pressure. STP has a temperature of 273K, while standard conditions is 298 K. Use STP for ideal gas law problems, and standard conditions for thermodynamics problems. Remember that your temperature must always be in K, not C.

 

[QofQuimica]

Hello again Q, two unrelated topics.

(1). Kw/pH changing with temperature
I had a practice passage about Kw changing with temperature and how pH also changes. That's all well and good that the hydrogen ion concentration increases because the Kw also increases, and thus by definition the pH must increase. I understand this by "definition" but not by intuition. Since pH is defined by the conc. of protons it makes since. But intuitively it seems like pH would stay the same in that it is not any more acidic because the hydroxide ion concentration also increases. Probably poor explanation, but...

(2) Reactivity
I am aware of the existence of a metal displacement chart and that it is not required knowledge for el MCAT. If I am remembering correctly it simply says that one metal can replace certain metals but not others. Does this relate to the periodic trends...for instance if I were asked which of the following metals was most reactive would that mean that I have to choose the one that has the smallest electron affinity/electroneg? Can I predict reactivity with these trends?

THANKS!

1) pH should actually DECREASE with increased temperature. It's still neutral if you don't have any outside acid or base, but it will be lower than 7. The entire pH scale is actually different at a higher temperature. The 1-14 scale you are familiar with only holds at 25C b/c that's where Kw = 1 x 10^-14. If you raise the temp, Kw also increases, and you wind up with a larger K. At 65C, for example, Kw is around 1 x 10^-13. That makes the maximum pH now 13, not 14, and a neutral solution will have pH<7.

2) I think that would work, because the metals at the bottom left of the periodic table (say Cs) will be the most amenable to giving up their valence electrons, and will have the most metallic character. It would certainly be exciting if you dropped some Cs metal in water.

 

[QofQuimica]

Which is a stronger acid, water or methane?
It's obviously water so stronger acid=weaker conjugate base (stable). Weaker acid=stronger conjugate base=unstable.

Right.

 

[Caboose]

Hey, hey! How come AsH3 can't hydrogen bond and NH3 can?
We should change that.

Caboose.

 

[QofQuimica]

Hey, hey! How come AsH3 can't hydrogen bond and NH3 can?
We should change that.

Caboose.

Because of electronegativity differences. Only elements that are much more electronegative than hydrogen can form hydrogen bonds. Basically the only ones that fit the bill are F, O and N. Other X(H)n compounds where X is an element with a smaller but still significant electronegativity difference to H will still be polar and have dipole-dipole moments, but will not be protic (no H-bonds).