Good morning,

I had a question about Galvanic cells. Since oxidation happens at the anode, that means electrons are leaving the anode and therefore anions should migrate to it to make up for the lost charge, right? I recently ran across a practice problem that contradicted this (possibly their error), so I wanted to ask if my logic was correct. Thanks!

Good morning,

I had a question about Galvanic cells. Since oxidation happens at the anode, that means electrons are leaving the anode and therefore anions should migrate to it to make up for the lost charge, right? I recently ran across a practice problem that contradicted this (possibly their error), so I wanted to ask if my logic was correct. Thanks!
Yes, anions should flow from the salt bridge into the oxidation half-cell.

(1)In my textbook's explanation of deviation from PV=nRT bliss, they state that "it is expected PV/RT=1 for one mole of ideal gas at any temperature and pressure". They go on to show V's + deviation and P's negative deviation. Why is the ratio equal to 1?

(2) When a reaction is shown as broken down into 2 or 3 steps, can I assume that the steps are elementary steps (and thus can use the coefficients of the reaction in the slow step to calculate the rate constant and rate)? Or do they have to be explicity labeled as an "elemantary step".

If so, I never made this distinction while taking gen chem. It was always "rate law must be determined by experiment" and the "stupid answer" on the tests were always with reaction orders equal to the coefficients. So for instance, if asked for the rate law I would mark "can't be determined" unless they explicity told me elementary step.

(3) Did you find the PS of the AAMC practice exams to be relevant difficulty-wise to the real deal? I am most concerned about physics of any of the sciences, but PS has been my highest score on AAMC practices. Also, did you find yourself "educatedly guessing" on a lot of the questions rather than knowing for sure (obviously more likely on VR and the why? science questions as to the calculation questions, but...). Thanks so much.

Q, I am seriously going to send you a check for as much work as you have done for me. Drop me a PM with your address.

(1)In my textbook's explanation of deviation from PV=nRT bliss, they state that "it is expected PV/RT=1 for one mole of ideal gas at any temperature and pressure". They go on to show V's + deviation and P's negative deviation. Why is the ratio equal to 1?

(2) When a reaction is shown as broken down into 2 or 3 steps, can I assume that the steps are elementary steps (and thus can use the coefficients of the reaction in the slow step to calculate the rate constant and rate)? Or do they have to be explicity labeled as an "elemantary step".

If so, I never made this distinction while taking gen chem. It was always "rate law must be determined by experiment" and the "stupid answer" on the tests were always with reaction orders equal to the coefficients. So for instance, if asked for the rate law I would mark "can't be determined" unless they explicity told me elementary step.

(3) Did you find the PS of the AAMC practice exams to be relevant difficulty-wise to the real deal? I am most concerned about physics of any of the sciences, but PS has been my highest score on AAMC practices. Also, did you find yourself "educatedly guessing" on a lot of the questions rather than knowing for sure (obviously more likely on VR and the why? science questions as to the calculation questions, but...). Thanks so much.

Q, I am seriously going to send you a check for as much work as you have done for me. Drop me a PM with your address.
1) You may not want to pay me after I answer this one for you....that equation equals one because you have one mole of gas. In other words, PV/RT = n, and you have one mole of gas, so.... ;)

2) By elementary step, do you mean RDS? If so, you need to have some way to know which one is rate-determining. You can't figure that out from the individual steps alone, but if you know the overall reaction equation, you can figure out which one is the RDS. Rate law exponents have to be determined experimentally as you said.

3) I actually didn't take any of the AAMC tests, just the five Kaplan tests, so I can't answer that question. I would say that I did not have to make an educated guess on most questions, but of course there were some that I didn't know, and those I did guess on. If you didn't know this already, don't leave any questions blank on the MCAT; there is no penalty for guessing like there is on the SAT.

hi this is just a simple question. But I recently ran into a question involving determines the color of a bunch ionic solutions. I never really learned how the D oribital affected color. Can you explain it to me. Is the D orbital the only thing that determines the color in a transitional metal + some nonmetal solution. So does the nonmetal molecule have no effect on color then?

If I am calculating equilibrium for gases, I can just use pressures in Kp, right? What if the reaction contains different phases?

hi this is just a simple question. But I recently ran into a question involving determines the color of a bunch ionic solutions. I never really learned how the D oribital affected color. Can you explain it to me. Is the D orbital the only thing that determines the color in a transitional metal + some nonmetal solution. So does the nonmetal molecule have no effect on color then?
This question is definitely beyond the scope of the MCAT; if you take inorganic chemistry, you will learn about ligand-metal interactions. But the simple answer to your question is that yes, the ligands present DO affect the color of the complex. This happens because the ligands affect the energy of the metal's d-orbitals. In a lone atom, the five d-orbitals are normally all degenerate (have the same energy), but the presence of ligands will "split" the energies of the d-orbitals. The amount of difference in energy between the d-orbitals will affect what color the complex becomes. Again, this is beyond the scope of freshman gen chem. If you got asked about something like this on the MCAT, they would have to give you a passage explaining it all.

If I am calculating equilibrium for gases, I can just use pressures in Kp, right? What if the reaction contains different phases?
In that case, use Kc. You should only use Kp if all species present are gases.

Hello Q,

Can you shed some light on this equation that I found on AAMC's topic list under ionic bonding:

"Force of att=R(n+e)(n-e)/d^squared"

It kind of looks like a constant (R), two charges, and the inverse square conservative force form. I don't think I have ever seen anything like that in my life....

Any last words of advice one week out?

Hello Q,

Can you shed some light on this equation that I found on AAMC's topic list under ionic bonding:

"Force of att=R(n+e)(n-e)/d^squared"

It kind of looks like a constant (R), two charges, and the inverse square conservative force form. I don't think I have ever seen anything like that in my life....

Any last words of advice one week out?
I promise you that this is an equation you are (or should be!) intimately familiar with from electrostatics in physics. They changed the letters they used to denote the variables and constants, but the relationship is identical. If you still can't recognize it, take a look at the physics explanations thread; I think I wrote about it in the very last post. ;) Electrostatics applies to chemistry, too. Don't forget about your physics just because you're doing a chemistry problem.

Hi

When asked to find molar solubility do they want the X variable in the Ksp eqn. If you have Ba(OH)2 for example will the molar solubility of Ba be X and OH be 2x or (2x)^2?

Also, I have a question about flow speed. Let's say you have a test tube and you put a funnel in. The flow speed is fastest in the neck of the funnel since it has the smallest area, but according to Bernoulli's equation it would be fastest in the test tube since the elevation head (H) is lowest Pressure is constant since everything is at atmospheric pressure and so the speed v must increase right?

oh hey can you explain the difference between ferromagnetic, paramagnetic and dimagnetic. And would a simple definition be sufficient, should we know the maths too

oh hey can you explain the difference between ferromagnetic, paramagnetic and dimagnetic. And would a simple definition be sufficient, should we know the maths too

Diamagnetic: the individual atoms have no net magnetic field
-repelled from the pole of a strong bar magnet
-sometimes called "weakly antimagnetic"

Paramagnetic: can be a degree of alignment to create a net magnetic field, attracted towards the pole of a strong bar magnet- sometimes called "weakly magnetic"

Ferromagnetic: below a certain temperature (dependent on the material) there is a high degree of alignment Ex: iron, nickel

I think you should know the basic math of magnetic fields, if that's what you're referring to... Otherwise, Q will know....

Hi

When asked to find molar solubility do they want the X variable in the Ksp eqn. If you have Ba(OH)2 for example will the molar solubility of Ba be X and OH be 2x or (2x)^2?

Also, I have a question about flow speed. Let's say you have a test tube and you put a funnel in. The flow speed is fastest in the neck of the funnel since it has the smallest area, but according to Bernoulli's equation it would be fastest in the test tube since the elevation head (H) is lowest Pressure is constant since everything is at atmospheric pressure and so the speed v must increase right?
Yes, molar solubility is how much of that stuff you have dissolved, which is usually represented as "X."

Ok, so the continuity equation would predict that the tube with the smaller cross-sectional area (the funnel stem) should have a faster velocity, like you said. Bernoulli's equation says that when Pi is the same everywhere (atmospheric pressure), you get an inverse relationship between v and P. So since the velocity is slower in the test tube, the pressure there should be higher.

Hey Q,

I had a real simple question; how does the reaction between HCl and a carbonate ion produce carbon dioxide gas? Thanks!

Hey Q,

I had a real simple question; how does the reaction between HCl and a carbonate ion produce carbon dioxide gas? Thanks!
You would form carbonic acid, which decomposes to water and carbon dioxide. It isn't easy for me to write the equations using text, but I'll do my best. This assumes you are using sodium carbonate:

2HCl + Na2CO3 -> H2CO3 <-> H2O + CO2

Ok this is probably crucial so I'd like to clear it up.


Regarding pka's and Kas in relation to titration curves. When Ka is greater than 1 this means it is strong, less than 1 weak. when taking the -log of ka to get pka this means that a positive pka indicates a weak acid and a negative pka a strong acid. But on the titration curve of a strong acid, 1/2 way to the endpt. pH=pka and this means pka is positive for a strong acid! Please explain based on the titration curve. How can I figure out based on initial pH on the curve whether an acid/base is strong or weak.

Ok this is probably crucial so I'd like to clear it up.


Regarding pka's and Kas in relation to titration curves. When Ka is greater than 1 this means it is strong, less than 1 weak. when taking the -log of ka to get pka this means that a positive pka indicates a weak acid and a negative pka a strong acid. But on the titration curve of a strong acid, 1/2 way to the endpt. pH=pka and this means pka is positive for a strong acid! Please explain based on the titration curve. How can I figure out based on initial pH on the curve whether an acid/base is strong or weak.
Ok, I think what is confusing you is that you're looking for a buffering region. But strong acids do not ever function as buffers, because their conjugate bases are too weak. You can tell very easily from the titration graph whether your acid is strong or weak just by the shape of it. If it is weak, there will be an initial buffering region (flat part of the curve) before you get much change in pH. It is in this buffering region that pH = pKa. In addition, the equivalence point will occur at a basic pH. On the other hand, if the acid is strong, there will be no buffering region (flat area), and the pH will not ever be equal to the pKa. (The pH range only goes from 0-14 in water at 25 C, so it can't be equal to a negative pKa!) The equivalence point will occur at a neutral pH (7 if you're at 25 C) for strong acids.

Hi
I have a question about superheated liquids, where the pressure is lowered while keeping the temperature constant so as to avoid a phase transition, resulting in a liquid with a temperature higher than the boiling point for the new low pressure. How is this possible on a phase diagram? With constant temp and lowered pressure, doesn't it just look like a straight line down on the diagram, so how is it possible to avoid a transition into gas?

Thanks so much!

Hey Q,

Just watned to clear this up...

Covalent bonds are stronger than ionic bonds, but ionic bonded compounds have stronger intermolecular forces (salts are solids). True?

Hi
I have a question about superheated liquids, where the pressure is lowered while keeping the temperature constant so as to avoid a phase transition, resulting in a liquid with a temperature higher than the boiling point for the new low pressure. How is this possible on a phase diagram? With constant temp and lowered pressure, doesn't it just look like a straight line down on the diagram, so how is it possible to avoid a transition into gas?

Thanks so much!
If you look at the phase diagrams, you will see that the lines separating the phases are not perfectly vertical. In fact, they usually slope in the positive direction. (The MP line for water is a notable exception to this rule.) Since the line between the liquid and gas phases represents the boiling point, that sloping takes into account how the boiling point changes with pressure. (BP will decrease as pressure decreases, as you said.) Here's a phase diagram of water so you can see what I mean. Note that the line from O to P (representing BP) slopes in the positive direction. Thus, a decrease in pressure will lead to a corresponding decrease in the temperature where the BP line is reached as well:

http://www.nzifst.org.nz/unitoperations/unopsassets/fig7-1.gif

Hey Q,

Just watned to clear this up...

Covalent bonds are stronger than ionic bonds, but ionic bonded compounds have stronger intermolecular forces (salts are solids). True?
Covalent bonds are not necessarily stronger than ionic bonds. The bond energies of both types of bonds actually overlap, and so you will have to evaluate each particular case individually instead of relying on a rule like this. Ionic compounds are held together by electrostatic forces, and these ARE stronger than any of the inTERmolecular interactions that hold covalent molecules together. This is in contrast to the inTRAmolecular covalent bonds, which, as I said, are of comparable strength to the ionic bonds.

What is the difference between saying that a solution is saturated with 1.0 g of a salt left undissolved and saying that the solution is supersaturated? thanks :)

Hi! Thanks in advance for the reply. I was wondering whether the ionic configuration for the Cobalt-3+ ion is

a. [Ar]3d6 OR
b. [Ar]3d5 4s1

Kaplan test says it is a, but my hunch following Hund's rule says it is b.

Also, which electron comes off first, is it the 4s2 electron or the 3d7.

Thanks a lot!

What is the difference between saying that a solution is saturated with 1.0 g of a salt left undissolved and saying that the solution is supersaturated? thanks :)
A supersaturated solution has more than the normal amount of solute dissolved in it (i.e., beyond the saturation point). If you raise the temperature of the solution, you can force more salt to dissolve. If you then cool it back to the original temperature, the extra salt may not precipitate out immediately, giving rise to a supersaturated solution. Eventually the extra salt will precipitate out until the solution reaches the saturation point.

Hi! Thanks in advance for the reply. I was wondering whether the ionic configuration for the Cobalt-3+ ion is

a. [Ar]3d6 OR
b. [Ar]3d5 4s1

Kaplan test says it is a, but my hunch following Hund's rule says it is b.

Also, which electron comes off first, is it the 4s2 electron or the 3d7.

Thanks a lot!
Kaplan's answer is actually correct. The reason why is that when the 4s and 3d orbitals are empty, the 4s orbital is lower in energy, and therefore it fills first. Once all of these orbitals have electrons in them, the 4s orbital is now higher in energy than the 3d orbitals, and it therefore empties first. In general, you should fill the 4s orbital before the 3d, and you should also empty the 4s orbital before the 3d. Thus, the two 4s electrons would be removed first, followed by the 3d7 electron.

Thank you Q of Q! I also received a reply from Kaplan saying that their answer was correct. I appreciate your response as well. It clarified the issue nicely.

Best of luck to everyone on the real one!

On AAMC 6R I noticed that they asked a question about geometry. However, in the solutions, they use the words geometry and shape interchangably saying just to look at the bonds to atoms. (essentially, the shape). For example, XeF4 would be square planar (as opposed to octahedral). I've learned from g-chem classes that geometry and shape were two separate things and would have called the geometry of XeF4 as octahedral. Would it be safe to assume they are asking for shape only in these types of questions? Another example was SO2, was 'bent' as opposed to trigonal planar. Hmmm...

On AAMC 6R I noticed that they asked a question about geometry. However, in the solutions, they use the words geometry and shape interchangably saying just to look at the bonds to atoms. (essentially, the shape). For example, XeF4 would be square planar (as opposed to octahedral). I've learned from g-chem classes that geometry and shape were two separate things and would have called the geometry of XeF4 as octahedral. Would it be safe to assume they are asking for shape only in these types of questions? Another example was SO2, was 'bent' as opposed to trigonal planar. Hmmm...
It sounds like you are thinking of electronic geometry and they are describing molecular geometry. In general, I would recommend always picking molecular geometry unless you are specifically asked for the electronic geometry. In other words, do not include the lone pairs when you pick the "shape" of the molecule.

It sounds like you are thinking of electronic geometry and they are describing molecular geometry. In general, I would recommend always picking molecular geometry unless you are specifically asked for the electronic geometry. In other words, do not include the lone pairs when you pick the "shape" of the molecule.

Got it. Thank you. I appreciate your help in the midst of all my flustered knowledge I'm trying to organize.

Hey,

would someone please give me a nice, comprehensive explaination as to how to calculate bond orders?
I know the formula, I just can't grasp it all.

Thanks a lot.

Hey,

would someone please give me a nice, comprehensive explaination as to how to calculate bond orders?
I know the formula, I just can't grasp it all.

Thanks a lot.
The easiest thing to do is to count each sigma and pi bond as contributing one to the bond order, and dispense with the formula altogether. So a single bond has a bond order of 1, a double bond has a bond order of 2, and a triple bond has a bond order of 3. That's pretty simple, right? So what do you do when you have multiple resonance structures? You have to average the bond order over each resonance contributor, giving you a fractional bond order. For example, benzene has a bond order of 1.5, because there are two equal resonance contributors. Carbonate ion has a bond order of 1.33 because there are three equal resonance contributors. And so on.

Hey Q,

Is the melting point or decomposition rate related to the strength of a compound's IMF? If so, which one?

Hey Q,

Is the melting point or decomposition rate related to the strength of a compound's IMF? If so, which one?
I'm assuming IMF = intermolecular forces? If so, mp does depend on them in part, but it is more complex than that. This is b/c mp also depends heavily on packing ability (how well the molecules can fit together into a lattice), so the pattern isn't as consistently based on intermolecular forces as it is for, say, bp. I would say that decomposition is probably most closely correlated with temperature, and it would depend on mp too. Organic compounds with very high melting points are going to tend to want to decompose rather than melt when you heat them up to high temperatures.

Does that answer your question?

I'm assuming IMF = intermolecular forces? If so, mp does depend on them in part, but it is more complex than that. This is b/c mp also depends heavily on packing ability (how well the molecules can fit together into a lattice), so the pattern isn't as consistently based on intermolecular forces as it is for, say, bp. I would say that decomposition is probably most closely correlated with temperature, and it would depend on mp too. Organic compounds with very high melting points are going to tend to want to decompose rather than melt when you heat them up to high temperatures.

Does that answer your question?

Kinda...

bump

Don't blindly memorize formulas, because that is bound to get you into trouble if you are asked a question that is even the slightest bit different than one you've seen before. Instead, the best way to solve Ksp problems IMHO is to make up one of those little charts, with the starting concentrations, change in concentrations, and final concentrations for each species. (Technically, you can ignore the salt, since it's a pure solid and it won't appear in the equilibrium expression.) You might be asked to calculate Ksp using the molar solubility, or you might be given Ksp and asked to calculate Qsp to determine whether your salt will precipate or not. If Ksp<Qsp, the salt will precipitate (supersaturated solution), but if Ksp>Qsp (unsaturated solution), it will not.

Again, I would like to emphasize to everyone: to score well on these tests (MCAT, PCAT, DAT, or OAT), you must understand what you are doing; don't blindly memorize formulas.

Is the reason that the sal will precipiate when Ksp<Qsp is because Qsp =(product^coefficient) /(reactant^coefficient) and that there must be a lot more product than reactant as compared to Ksp. And higher concentration of product in Qsp than that in Ksp indicates superaturated solution? Am I on the right track? Thank you moderators. You are awesome!!!

pKa is the equilibrium constant for an acid dissociating into its conjugate base and H+. Thus, you could get asked to calculate pKa given the molar solubility of an acid, or you could get asked to calculate the molar solubility given the pKa. pKa values are also useful to gauge relative strengths of acids and bases: stronger acids have lower pKas, while stronger bases have higher pKas.

Isn't it Ka that is teh equilibrium constant for an acid dissciating into its conj base and H+, not pKa? I thought pKa was -log[Ka] like pH or pOH? Am I wrong?

can anyone explain this for me?

how come when you 'fill up' electrons (i.e. the Aufbau principle) you fill up orbitals/subshells in the order determined by the (n+l) values

ie. 4s fills before 3d because the (n+l) values are 4 and 5 respectively

yet when you take electrons away (for example, forming cations) the order is determined by only the principle quantum number

ie. 4s electron gets removed first, then the 3d, then 3p, then 3s etc etc.

and also, when forming Anions, which rule is followed? the former or the latter?

Thanks.

I think there is a typo in your book; check the EK website to see if anyone has reported an error for that question. The cathode is positive, not negative, for a galvanic cell, and the cathode is negative for electrolytic cells. In all types of cells, the electrons always move to the cathode, which is where reduction occurs. If you have a spontaneous cell (galvanic), then your cathode will be positive. This makes intuitive sense; the negatively charged electrons "want" to go to a positively charged electrode if they can. Cations in the salt bridge must move to balance the electrons so that you do not have a buildup of charge in the cathode. Thus, they will also move to the cathode, following the electrons. The salt bridge anions will go toward the anode to balance out the electrode cations left behind after the electrons leave.

I am not getting the concept of cations in the salt bridge. I am aware that electrons always move from anode to cathode through salt bridge. This occurs because of the oxidation allows electrons to be removed from anode (in galvanic cell) to cathode. Do cations actually move too (just like electrons)? In physics, I was taught that it's electrons that generate electric field and etc..always electrons move and positively charged ions (=protons?) are stationary. So, in galvanic cell, cations (protons?) are generated by cathode and move to anode? But I thought due to the movement of electrons in anode, oxidation occurs, and if cations move to anode, further oxication should take place, right? Please correct me.

Is the reason that the sal will precipiate when Ksp<Qsp is because Qsp =(product^coefficient) /(reactant^coefficient) and that there must be a lot more product than reactant as compared to Ksp. And higher concentration of product in Qsp than that in Ksp indicates superaturated solution? Am I on the right track? Thank you moderators. You are awesome!!!
Sounds like you've got it.

Isn't it Ka that is teh equilibrium constant for an acid dissciating into its conj base and H+, not pKa? I thought pKa was -log[Ka] like pH or pOH? Am I wrong?
No, you're right. pKa is the -log of Ka, and Ka is the dissociation constant for an acid.

can anyone explain this for me?

how come when you 'fill up' electrons (i.e. the Aufbau principle) you fill up orbitals/subshells in the order determined by the (n+l) values

ie. 4s fills before 3d because the (n+l) values are 4 and 5 respectively

yet when you take electrons away (for example, forming cations) the order is determined by only the principle quantum number

ie. 4s electron gets removed first, then the 3d, then 3p, then 3s etc etc.

and also, when forming Anions, which rule is followed? the former or the latter?

Thanks.
If you look up at the top of this page, I answered that exact same question for someone else. :)

I am not getting the concept of cations in the salt bridge. I am aware that electrons always move from anode to cathode through salt bridge. This occurs because of the oxidation allows electrons to be removed from anode (in galvanic cell) to cathode. Do cations actually move too (just like electrons)? In physics, I was taught that it's electrons that generate electric field and etc..always electrons move and positively charged ions (=protons?) are stationary. So, in galvanic cell, cations (protons?) are generated by cathode and move to anode? But I thought due to the movement of electrons in anode, oxidation occurs, and if cations move to anode, further oxication should take place, right? Please correct me.
The electrons don't pass through the salt bridge. They go through the wire. The salt bridge is there to complete the circuit. It contains an inert salt like KCl that dissociates to counter the charges being gained and lost by motion of the electrons. The Cl anions go to the anode cell to counter the cations released from the anode after the electrons leave, and the K cations go to the cathode cell to replace the lost cations from solution that deposit on the cathode.

I read in my gen chem text book that
"Provided that the reaction is sufficiently exothermic, K will decrease so rapidly with increasing temporature."
Can you briefly explain how this can occur? I thought le Chantiler principle said that reaction will proceed in a direction that will alleviate the heat. So I thought the reaction will forward direction since heat is released and thus heat appears in the product. I think my reasoning is incorrect. Correct me if I am wrong. Thanks.

I read in my gen chem text book that
"Provided that the reaction is sufficiently exothermic, K will decrease so rapidly with increasing temporature."
Can you briefly explain how this can occur? I thought le Chantiler principle said that reaction will proceed in a direction that will alleviate the heat. So I thought the reaction will forward direction since heat is released and thus heat appears in the product. I think my reasoning is incorrect. Correct me if I am wrong. Thanks.
It's not simple to apply Le Chatelier's Principle with regard to heat, because the value of K actually changes with temperature. So you have two things going on at one time: the reaction does shift according to Le Chatelier's Principle when you change temperature, and the equilibrium position also changes with temperature (i.e., you have a different value of K when you heat or cool the reaction). That is what they are trying to tell you there.

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.

According to your model of treating endo/exothermic as addition of heat in reactants and products, respectively, increasing heat (adding heat) moves the reaction to the right for endothermic and to the left for the exothermic. It is contrary to my last question about Le Chantelier's principle that alleviates the heat added. In other words, my last posting said that endothermic will shift the reaction to the left to move toward where heat is added.
Secondly, since K changes with temperature, should we assume that the temperature drop/rise for exo/endothermic will cause the change in value of K? In that case, how do we balance the two factors you mentioned above?
"So you have two things going on at one time: the reaction does shift according to Le Chatelier's Principle when you change temperature, and the equilibrium position also changes with temperature (i.e., you have a different value of K when you heat or cool the reaction). "
It is a long post and I really appreciate your help. I may not have been to clear. Please let me know if anything is too ambiguous. Thanks!

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. ;)

I came across one of the questions/answers in this thread and was curious about this:
"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?"
You said that it's the one with the most metallic character, which I think means high conductivity of electrons (and thus heat+electricity). Why would it be the metals with most metallic character but not something like Al (13)? Does it have to do with the number of valence electrons? Because Al has three while Cs has one valence electron? Also, because Cs has higher n than Al, meaning that Cs has higherPE and more unstable (=more likely to lose electrons)? Increasing n (principal number) increase PE and consequently decrease stability?
Lastly, if i were to choose which one is more reactive bw Li and Al (both of the same n but different orbit), i should choose Li because it's in s orbit, which is more stable? How is s more stable if s is in lower energy state than p, where electrons fill in after s orbit is filled?
i am confused... :eek:

According to your model of treating endo/exothermic as addition of heat in reactants and products, respectively, increasing heat (adding heat) moves the reaction to the right for endothermic and to the left for the exothermic. It is contrary to my last question about Le Chantelier's principle that alleviates the heat added. In other words, my last posting said that endothermic will shift the reaction to the left to move toward where heat is added.
Secondly, since K changes with temperature, should we assume that the temperature drop/rise for exo/endothermic will cause the change in value of K? In that case, how do we balance the two factors you mentioned above?
"So you have two things going on at one time: the reaction does shift according to Le Chatelier's Principle when you change temperature, and the equilibrium position also changes with temperature (i.e., you have a different value of K when you heat or cool the reaction). "
It is a long post and I really appreciate your help. I may not have been to clear. Please let me know if anything is too ambiguous. Thanks!
Ok, I think I misunderstood your question. If you have an exothermic reaction, heat is considered to be a "product." Therefore, we would predict according to Le Chatelier's Principle that heating an exothermic reaction should cause it to go in the reverse direction (back toward reactants.) The reverse would be true for endothermic reactions, where heat is a "reactant."

Knowing how to balance the change in K with temperature with the effect of Le Chatelier's Principle is definitely beyond the scope of the MCAT. And it's also beyond the scope of my knowledge; sorry. :p

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? :confused:

How are we to know that CH3- is weaker acid than the rest of the choices? I can't see why. I tried to make sense with the defitions of acid (h donor and e acceptor) and inverse relationship between the strength of conj. acid and its base but I can't tell how each choice will be weaker/stronger conj. acid in relation to other choices. Thanks for your help!

I came across one of the questions/answers in this thread and was curious about this:
"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?"
You said that it's the one with the most metallic character, which I think means high conductivity of electrons (and thus heat+electricity). Why would it be the metals with most metallic character but not something like Al (13)? Does it have to do with the number of valence electrons? Because Al has three while Cs has one valence electron? Also, because Cs has higher n than Al, meaning that Cs has higherPE and more unstable (=more likely to lose electrons)? Increasing n (principal number) increase PE and consequently decrease stability?
Lastly, if i were to choose which one is more reactive bw Li and Al (both of the same n but different orbit), i should choose Li because it's in s orbit, which is more stable? How is s more stable if s is in lower energy state than p, where electrons fill in after s orbit is filled?
i am confused... :eek:
Sorry, I should have been more clear about what I meant by "metallic character." I was actually referring to the periodic trends. Aluminum is a fairly electronegative metal, and thus would not have as much metallic character compared to highly electropositive metals like the alkalis. The periodic trends would predict that the most metallic elements (low electronegativity, low electron affinity, large radius, and low ionization energy) would be found in the bottom left of the table. Hence why I specifically mentioned Cs, which is the most metallic naturally occurring element.

How are we to know that CH3- is weaker acid than the rest of the choices? I can't see why. I tried to make sense with the defitions of acid (h donor and e acceptor) and inverse relationship between the strength of conj. acid and its base but I can't tell how each choice will be weaker/stronger conj. acid in relation to other choices. Thanks for your help!
LMAO, you're really going through this entire thread, aren't you???

Ok, here you are mixing up conjugate bases and acids. Remember that if CH3- is the base, then its conjugate acid is CH4. The others are NH3, H2O, and NH4+. Of those four conjugate acids, CH4 definitely has the highest pKa, estimated to be around 50. Ammonia is around 35. Water is about 16. I forget exactly what ammonium is, but I think it's about 9 or so. So ammonium would be the strongest acid, and methane would be the weakest, exactly as you'd expect based on the relative conjugate base strengths.

LMAO, you're really going through this entire thread, aren't you???

Ok, here you are mixing up conjugate bases and acids. Remember that if CH3- is the base, then its conjugate acid is CH4. The others are NH3, H2O, and NH4+. Of those four conjugate acids, CH4 definitely has the highest pKa, estimated to be around 50. Ammonia is around 35. Water is about 16. I forget exactly what ammonium is, but I think it's about 9 or so. So ammonium would be the strongest acid, and methane would be the weakest, exactly as you'd expect based on the relative conjugate base strengths.
So are we supposed to have memorized the relative pKa's (strength) of different molecules in order to tell the order of strength? If that is the case, generally, are there any general molecules that appear frequenly in MCAT? Are there any easy way to know the relative pKa's of different complex ions? Thanks.

This is a question that I asked before but I am still not clear about reactivity.

If i were to choose which one is more reactive between Li and Al (both of the same n but different orbit), is Li less reactive because it's in s orbit (lower PE orbit than p), which is more stable?

So are we supposed to have memorized the relative pKa's (strength) of different molecules in order to tell the order of strength? If that is the case, generally, are there any general molecules that appear frequenly in MCAT? Are there any easy way to know the relative pKa's of different complex ions? Thanks.
No, you don't need to memorize pKa values. I just cited them so that you can see how far apart they are. (Remember that since pKa is a log scale, every unit is 10x different than the one above or below it, so these differences in acidity are huge.) You should learn to have a feel though for where things lie relative to one another on the acidity or basicity scale. More electronegative atoms like O make better (weaker) conjugate bases versus less electronegative atoms like C. So CH4 will make a weaker acid and a stronger base versus NH3, which is a weaker acid than H2O, and that's weaker than HF.

This is a question that I asked before but I am still not clear about reactivity.

If i were to choose which one is more reactive between Li and Al (both of the same n but different orbit), is Li less reactive because it's in s orbit (lower PE orbit than p), which is more stable?
I don't understand this question. They don't have the same n-value. Li's valence shell is the 2s, while Al's is the 3p. My interpretation of the previous poster's question was that the question was asking which elements behave most like metals in terms of the periodic trends (i.e., give up electrons most readily). Those elements are found at the bottom left of the periodic table. All of the alkali metals are very reactive in this sense. You generally don't find them as the neutral metals in nature; in fact, we have to keep our metallic sodium in oil, because it's so humid here in FL that the sodium will react with the water in the air and release hydrogen gas, which can actually catch fire. Don't ever stick a chunk of sodium in your pocket. ;)

For this question Calculate the percent ionization in 0.250 M HCrO4-.

I don't understand how to get the HCrO4- disassociate into. The Ka value for it is 3.0x10^-7.

Anyway, I tried and this is what I got.

Ax^2+Bx+C=0
1x^2 + 3.0E-7x - 7.5E-8 = 0
[H+] = 2.737113198E-4 M
Percent Ionization = [H+]/[Acid concentration]*100
2.7371E-4/0.250 * 100 = .109%

For this question Calculate the percent ionization in 0.250 M HCrO4-.

I don't understand how to get the HCrO4- disassociate into. The Ka value for it is 3.0x10^-7.

Anyway, I tried and this is what I got.

Ax^2+Bx+C=0
1x^2 + 3.0E-7x - 7.5E-8 = 0
[H+] = 2.737113198E-4 M
Percent Ionization = [H+]/[Acid concentration]*100
2.7371E-4/0.250 * 100 = .109%
If that's the acid you're supposed to use, it must be dissociating into CrO4^2- and H+. In that case, you wind up having Ka = [H+][A-]/[HA] ~ x^2/0.25, so x is about 2.74 ^-4 as you said. Then divide by .25 and multiply by 100 gives me 0.11%, same as you. Looks like you did it fine to me. Does your book say the answer is different?

MCAT prep books say that increase in temp in exothermic reaction causes the reaction to shift left. is the reverse true as well? for example, decrease in exo. reaction shifts the reaction to the right. similarly, increase/decrease in temp in endo. reaction causes teh reaction to shift right/left, respectively?

As for the volume of the container in which gases reside, does increase in volume shift the reaction toward the side that contains more molecules? (just opposite of decreasing vol.(threby increasing pressure) causes the reaction to shift from the side with more molecules to the side with less molecules)?

I understand the logic of how decreasing vol. causes the reaction to shift toward the side with less molecules but not the reverse. Similary, with the increase/decrease in temp. in endothermic reaction and decrease in temp in exothermic reaction. Thanks for your help in advance.

MCAT prep books say that increase in temp in exothermic reaction causes the reaction to shift left. is the reverse true as well? for example, decrease in exo. reaction shifts the reaction to the right. similarly, increase/decrease in temp in endo. reaction causes teh reaction to shift right/left, respectively?

As for the volume of the container in which gases reside, does increase in volume shift the reaction toward the side that contains more molecules? (just opposite of decreasing vol.(threby increasing pressure) causes the reaction to shift from the side with more molecules to the side with less molecules)?

I understand the logic of how decreasing vol. causes the reaction to shift toward the side with less molecules but not the reverse. Similary, with the increase/decrease in temp. in endothermic reaction and decrease in temp in exothermic reaction. Thanks for your help in advance.

Ok, I'll give you the explaination my professor gave us about this. The answer uses Le Châtelier's principle. Basically you can think of heat as either a reagent or product. So in an exothermic reaction one of the products is heat. So write out the equation

A + B <->AB + heat

So remember the principle? If you increase the temperature you basically have more heat. Since the system tries to undo the change it shifts left. Reduce the temperature and it shifts right.(If I remember right this is why cars have cooling systems. You have to get rid of the heat otherwise it builds up and stops the reaction.)

Similarly an endothermic reaction is one that requires heat
A + B + heat <->AB

So increasing temperature drives this one right, reducing drives it left.(Actually once our professor told us to just consider heat to be a reagent or product it was so much easier to understand.)

Ok, I'll give you the explaination my professor gave us about this. The answer uses Le Châtelier's principle. Basically you can think of heat as either a reagent or product. So in an exothermic reaction one of the products is heat. So write out the equation

A + B <->AB + heat

So remember the principle? If you increase the temperature you basically have more heat. Since the system tries to undo the change it shifts left. Reduce the temperature and it shifts right.(If I remember right this is why cars have cooling systems. You have to get rid of the heat otherwise it builds up and stops the reaction.)

Similarly an endothermic reaction is one that requires heat
A + B + heat <->AB

So increasing temperature drives this one right, reducing drives it left.(Actually once our professor told us to just consider heat to be a reagent or product it was so much easier to understand.)

Thanks, Dave. How about volume change?

Thanks, Dave. How about volume change?
Volume is inversely proportional to pressure. If volume is increased, pressure is decreased, and the reaction moves to the side with more moles of gas. If volume is decreased, pressure is increased, and the reaction moves to the side with fewer moles of gas.

Volume is inversely proportional to pressure. If volume is increased, pressure is decreased, and the reaction moves to the side with more moles of gas. If volume is decreased, pressure is increased, and the reaction moves to the side with fewer moles of gas.

Yes you are right. But my intuition says that even when volume is increased (pressure decreased), the reaction should proceed from the side with more moles of gas to that with less moles of gas because of diffusion. What's wrong with this?

Yes you are right. But my intuition says that even when volume is increased (pressure decreased), the reaction should proceed from the side with more moles of gas to that with less moles of gas because of diffusion. What's wrong with this?
This isn't a case of diffusion. Diffusion occurs when you have a substance spreading out down its concentration gradient. You've probably had your chem teacher open a bottle of perfume and over time, the entire class can smell it, even the people in the back of the room. That's diffusion. You are dealing with an equilibrium here. If you perturb your system away from equilibrium by changing its pressure, you will cause the reaction to shift to restore equilibrium. But it isn't an issue of substances moving down their concentration gradients.

I realize that E (or U) = q + w according to the 1st law of thermo. Is w = p(delta v) or w = fd + p(delta v)?

At constant volume, E = q because delta v = and hence w = p(delta ) = 0. Why at constant pressure, E = q as well if only delta p is 0 and delta v is not necessarily 0 and thus w = p(deltav) does not have to be equal to 0? I think E = q + w where w = p(delta v) and p is constant but not zero.

Lastly, books claim that at only constant pressure q = delta H. But because H = E (or U) + P(delta v) and P is constant but not necessarily zero, i think P(delta V) doesn't necessarily have to be zero, which makes the eq. at const pressure H = E + P(delta v). On the other hand, books claim that at const volume, H is not equal to E. How is P(delta V) not equal to zero at const volume is not changing and therefore delta V is equal to zero?

I am so confused about these equations involving E, q, H, P and V. Please help me. Thank you,

Examkracker Gen chem book says that weak acids does not necessarily have to have strong conjugate base. For the sake of MCAT, should we neglect this fact and assume that weak acids generally have strong conjugate base?
And if we shouldn't neglect this fact and want to know the relative strength of conjugate base of a weak acid, what factors should we take account? Thank you for your help.

Lol another question...

in a reaction,

NaH + H2O -> Na+ + OH- + H2

After a careful review I said that H2O is an acid because it donates H+ and OH- is an obvious conjugate acid, accepting H+ (proton). However, NaH in the left side looks like acid because it donates H+ as well. But I thought because it actually dissociates into Na+ and H- instead of donating H+, it is not an acid. But how do we explain H2? How is this made and is my reasoning correct? What would you add to solidify my understanding?

Also, is the reason that NO3- is neutral is because HNO3 is a strong acid and hence NO3- is a weak conjugate base, which is more or less equal to neutral? Examkracker says NO3- is neutral.

Thanks.

I realize that E (or U) = q + w according to the 1st law of thermo. Is w = p(delta v) or w = fd + p(delta v)?

At constant volume, E = q because delta v = and hence w = p(delta ) = 0. Why at constant pressure, E = q as well if only delta p is 0 and delta v is not necessarily 0 and thus w = p(deltav) does not have to be equal to 0? I think E = q + w where w = p(delta v) and p is constant but not zero.

Lastly, books claim that at only constant pressure q = delta H. But because H = E (or U) + P(delta v) and P is constant but not necessarily zero, i think P(delta V) doesn't necessarily have to be zero, which makes the eq. at const pressure H = E + P(delta v). On the other hand, books claim that at const volume, H is not equal to E. How is P(delta V) not equal to zero at const volume is not changing and therefore delta V is equal to zero?

I am so confused about these equations involving E, q, H, P and V. Please help me. Thank you,
After reading this post, I'm confused myself. :p My advice is to not rely on memorizing equations. Work on actually understanding what is going on. A lot of your confusion is probably because the equations you're using are actually derivatives that must be integrated. If you try to solve them algebraically without taking that into account, you will get some strange results like you did.

Examkracker Gen chem book says that weak acids does not necessarily have to have strong conjugate base. For the sake of MCAT, should we neglect this fact and assume that weak acids generally have strong conjugate base?
And if we shouldn't neglect this fact and want to know the relative strength of conjugate base of a weak acid, what factors should we take account? Thank you for your help.
Strength of conjugate bases is relative and correlated with the strength of the corresponding weak acids. A very weak acid, like a hydrocarbon (pKa around 50) will have a very strong conjugate base. A medium weak acid, like an alcohol (pKa around 17) will have a strong conjugate base. A slightly weak acid, like acetic acid (pKa around 5) will have a weak conjugate base as well. Understand that there are different degrees of being weak or strong for bases and acids, and it's a continuum from extremely strong to extremely weak, with everything else in between.

Lol another question...

in a reaction,

NaH + H2O -> Na+ + OH- + H2

After a careful review I said that H2O is an acid because it donates H+ and OH- is an obvious conjugate acid, accepting H+ (proton). However, NaH in the left side looks like acid because it donates H+ as well. But I thought because it actually dissociates into Na+ and H- instead of donating H+, it is not an acid. But how do we explain H2? How is this made and is my reasoning correct? What would you add to solidify my understanding?

Also, is the reason that NO3- is neutral is because HNO3 is a strong acid and hence NO3- is a weak conjugate base, which is more or less equal to neutral? Examkracker says NO3- is neutral.

Thanks.
NaH is a strong base, not an acid. The hydrogen is a hydride, H-, not a proton.

Nitrate ion is too weak of a base to be able to dissociate water. An aqueous solution of nitrate ion would not be basic because of this.

After reading this post, I'm confused myself. :p My advice is to not rely on memorizing equations. Work on actually understanding what is going on. A lot of your confusion is probably because the equations you're using are actually derivatives that must be integrated. If you try to solve them algebraically without taking that into account, you will get some strange results like you did.
Okay, so why is it that when P is constant, q = E? Also, why is it that when P is constant, H = q?

Is it only when P and V are both constant that q = E? As for the equation for heat of reaction (H), do both P and V have to be kept constant so as to make H = q? or just P?

In the Nernst equation, E = E(standard) - (0.06/n)logQ. Here Q is product/reactant. For galvanic cell, which concentration should be the product and reactnat? Is reactant the conc. of anode and product = conc. of cathode because e- travel from anode to cathode? Similarly, in electrolytic cell, anode is the conc. of product and cathod the concentration of reactant? Thankx

for S: Ca, Ba, Sr.....just remember the tv network CBS
for H: Ca, Ba, Sr + Happy...whats happy? Hb Ag Pb ...mercury, silvr and lead...add a py to the end and all the first letters spell HAPPY

I read this part in the soluubility rules and i just want to clarify. So those that are insolube for S, are CaSO4, BaSO4 etc etc. And for H, it would be CaCl2, BaF2, etc etc. Am I correct? Sorry, I just have a test on friday for general chem and I happen to run into this and I figured it would be very useful :cool:. Thanks again.

I need to stop writing here today but yet I have one more question.
Electrolytic runs by an external battery so it does not (or can't) use its chemical energy to work. And that's why electrons run from cathode to anode unlike galvanic cell where electrons run from anode to cathode toward the positive end. I read the Gen chem thread about EMF and electro/galvanic cell and I encountered this problem that said that electrons flow to the cathode in an electrolytic cell, which seems just too contrary to what I thought about electrolytic cell. The explanation says that because things are reduced in cathode, electrons flow to the cathode in electrolytic cell. But cathode in electrolytic cell is negative so electrons would be repelled if they move to cathode.....
IS this true? And if so, what was it that was flowing from cathode to anode in electrolytic cell? Thanks

Is it ever possible that such intermolecular forces as covalent is used also as intramolecular interaction? If so, can other intermolecular forces as ionic and others used as intramolecular interaction forces as well? Thank you.

Is it ever possible that such intermolecular forces as covalent is used also as intramolecular interaction? If so, can other intermolecular forces as ionic and others used as intramolecular interaction forces as well? Thank you.
I think I *am* going to limit you to one question per day. ;) Ok, this is a lot of questions, and I don't have time to get to them all now, but I'll work on them when I get the chance. In answer to this one:

No, covalent bonds are inTRAmolecular (within a single molecule), not inTERmolecular (among several molecules). Ionic bonds are also inTRAmolecular. However, it is correct to say that the inTERmolecular forces (H bonds, dipole-dipole, and dispersion forces) are all held together by electrostatic forces, as are ionic bonds. I wrote two very extensive explanations posts about intramolecular bonds and intermolecular interactions in the gen chem explanations thread that you might want to read through if you haven't already.

I need to stop writing here today but yet I have one more question.
Electrolytic runs by an external battery so it does not (or can't) use its chemical energy to work. And that's why electrons run from cathode to anode unlike galvanic cell where electrons run from anode to cathode toward the positive end. I read the Gen chem thread about EMF and electro/galvanic cell and I encountered this problem that said that electrons flow to the cathode in an electrolytic cell, which seems just too contrary to what I thought about electrolytic cell. The explanation says that because things are reduced in cathode, electrons flow to the cathode in electrolytic cell. But cathode in electrolytic cell is negative so electrons would be repelled if they move to cathode.....
IS this true? And if so, what was it that was flowing from cathode to anode in electrolytic cell? Thanks
Ok, you're kind of mixed up here. Electrons always go from the anode to the cathode. ALWAYS, no matter what kind of cell you have. This is because the cathode is defined as the reducing electrode, while the anode is being oxidized. Again, this is always the case, regardless of what kind of cell it is. What changes for galvanic vs. electrolytic cells is the SIGN of the electrodes. In a spontaneous (galvanic) cell, the negative electrons go from a negatively charged anode down their electrical gradient to a positively charged cathode. Makes sense, right? Negative charges would like to go to a positively charged electrode, given their druthers. On the other hand, in an electrolytic cell, you are forcing the negative electrons to go backward, against their electrical gradient, to a negative electrode. Thus, in electrolytic cells, the anode is now positive while the cathode is negative.

As for what flows to the anode, that would be the current. Recall that current is defined by physicists as the direction that imaginary positive charges would travel (though of course it is the electrons we are concerned with in chemistry, not the current). It is also possible if you were reading about a galvanic cell that the problem might mention ions from the salt bridge moving to the cathode and anode. (Since the electrodes are in separate cells for a galvanic cell, you need the salt bridge to complete the circuit or you won't get any electron flow.) What happens? Well, when the electrons move from the anode, they leave behind positive charges. So negative charges from the salt bridge (anions) travel to the anode cell to counterbalance them. Conversely, the electrons going to the cathode side would build up a net negative charge if positive ions (cations) from the salt bridge didn't travel to the cathode side to balance them. So salt bridge anions go to the anode side, and cations go to the cathode side.

for S: Ca, Ba, Sr.....just remember the tv network CBS
for H: Ca, Ba, Sr + Happy...whats happy? Hb Ag Pb ...mercury, silvr and lead...add a py to the end and all the first letters spell HAPPY

I read this part in the soluubility rules and i just want to clarify. So those that are insolube for S, are CaSO4, BaSO4 etc etc. And for H, it would be CaCl2, BaF2, etc etc. Am I correct? Sorry, I just have a test on friday for general chem and I happen to run into this and I figured it would be very useful :cool:. Thanks again.
Right, you've got the idea. Hope your test went/goes well. :luck:

Are there ever a case where the triple point has lower (triple point) pressure and temperature than the temperature and the pressure of any melting point (the equilibrium line between gas and liquid)?
Is it a rule for the sake of MCAT that the triple point will always has higher P and T than melting point P and T?
Thanks, QofQuimica!!!!! :thumbup:

Are there ever a case where the triple point has lower (triple point) pressure and temperature than the temperature and the pressure of any melting point (the equilibrium line between gas and liquid)?
Is it a rule for the sake of MCAT that the triple point will always has higher P and T than melting point P and T?
Thanks, QofQuimica!!!!! :thumbup:
Ok, first of all, the melting point is the line dividing the liquid phase from the SOLID phase, not the gas phase. Second, the triple point is at a LOWER pressure than either the MP or the BP. Below the triple point, you don't have any liquid phase at all any more. Instead, you have a sublimation/deposition line where the solid and the gas phases interconvert directly to one another without going through a liquid. (Think of dry ice; the reason why it sublimates instead of melting is because the triple point of carbon dioxide is at about 5.2 atm.) So the triple point pressure will always be lower than any melting point pressure. If you mean the vaporization/condensation line (separating the gas and liquid phases), then again, the triple point pressure will be lower than the vaporization pressure, because you can't have vaporization below the triple point.

If you look at the image that I've pasted below, the triple point is point B. Line BD is the melting point line, and line BC is the vaporization line. Pressure is on the Y-axis. Note that both lines BD and BC are at HIGHER pressure than point B is. The only line that is at a lower pressure than B is line AB, which is the sublimation line. Thus, the triple point is, by definition, the lowest pressure at which a liquid can exist. Below the triple point pressure, you can still have solid or gas phases, but no more liquid.

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch14/graphics/14_14fig.gif

Ok, first of all, the melting point is the line dividing the liquid phase from the SOLID phase, not the gas phase. Second, the triple point is at a LOWER pressure than either the MP or the BP. Below the triple point, you don't have any liquid phase at all any more. Instead, you have a sublimation/deposition line where the solid and the gas phases interconvert directly to one another without going through a liquid. (Think of dry ice; the reason why it sublimates instead of melting is because the triple point of carbon dioxide is at about 5.2 atm.) So the triple point pressure will always be lower than any melting point pressure. If you mean the vaporization/condensation line (separating the gas and liquid phases), then again, the triple point pressure will be lower than the vaporization pressure, because you can't have vaporization below the triple point.

If you look at the image that I've pasted below, the triple point is point B. Line BD is the melting point line, and line BC is the vaporization line. Pressure is on the Y-axis. Note that both lines BD and BC are at HIGHER pressure than point B is. The only line that is at a lower pressure than B is line AB, which is the sublimation line. Thus, the triple point is, by definition, the lowest pressure at which a liquid can exist. Below the triple point pressure, you can still have solid or gas phases, but no more liquid.

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch14/graphics/14_14fig.gif

Oh, thank you, QofQuimica. Yes, I meant vap/condensation line. Looking at the phase diagram, the boliling temp will always be higher than that of the temp at the triple point. Can there be any exceptions in the relative location of the triple point in relation to boiling temp? In other words, are there any exceptions where vap/condensation temp will be lower of equal to that of the triple point temperature at different pressure? (like the uniqueness of water in phase change from solid to liquid or liquid to solid as compared to other substances) So it would mean that from the triple point, the BC is parallel to x-axis for some pressure greater than the pressure of the triple point. Do I make sense?... Thanks again! :D

Oh, thank you, QofQuimica. Yes, I meant vap/condensation line. Looking at the phase diagram, the boliling temp will always be higher than that of the temp at the triple point. Can there be any exceptions in the relative location of the triple point in relation to boiling temp? In other words, are there any exceptions where vap/condensation temp will be lower of equal to that of the triple point temperature at different pressure? (like the uniqueness of water in phase change from solid to liquid or liquid to solid as compared to other substances) So it would mean that from the triple point, the BC is parallel to x-axis for some pressure greater than the pressure of the triple point. Do I make sense?... Thanks again! :D
Ok, so I think what you are asking is whether the boiling point line ever slants to the left the way the melting point line does for water? I can't think of any example where it would. Remember that the reason why the melting point line slants to the left for water is because unlike for most substances, the solid phase of water (ice) is less dense than the liquid phase. But I can't think of any substance that has a denser gas phase in comparison to its liquid phase. If there is such a substance, you certainly wouldn't be expected to know that for the MCAT. :p

"The lewis acids include all simple cations except teh alkali and the heavier alkaline earth metal cations."

Are the alkali and alkaline earth metal cations not a part of the lewis acids because they are basically conjugate bases of some of the strong acids? If so, wouldn't that leave such cations as Rb and Be because they, when reacted with H2O, don't form a strong acid? Or, is it because, regardless of the strength, the alkali and alkaline earth metal cations are weak/really weak conjugate bases of weak/strong acids?

I have reviewed EK Gen chem and now I am going to start working on physics. Thanks for answering so many questions everyday.. :)

"The lewis acids include all simple cations except teh alkali and the heavier alkaline earth metal cations."

Are the alkali and alkaline earth metal cations not a part of the lewis acids because they are basically conjugate bases of some of the strong acids? If so, wouldn't that leave such cations as Rb and Be because they, when reacted with H2O, don't form a strong acid? Or, is it because, regardless of the strength, the alkali and alkaline earth metal cations are weak/really weak conjugate bases of weak/strong acids?

I have reviewed EK Gen chem and now I am going to start working on physics. Thanks for answering so many questions everyday.. :)
A Lewis acid is defined as a species that can accept an electron pair. Alkali metal cations have only lost one electron, so they won't want to accept an electron pair; if they did, they'd go from a +1 to a -1 oxidation number. Metallic character increases as you go down a group on the periodic table, following the periodic trends. The larger alkaline earth metals have more metallic character than the smaller ones do, and apparently that's enough to prevent them from being able to accept an electron pair.

Here's a question I've been struggling with and I'd appreciate it if someone could talk me through it like I'm a small child because no matter how I try to solve it, I just don't understand.

I'm probably just overlooking something basic, but I had a friend go over it with me and we figured out a way to get to the right answer, but we don't know why, if that makes sense. In other words, we don't know if we can apply the concept on a test. It's been bugging me for days.

It's a two-parter. OK, here it is:

The gage pressure in a tire is 35.0 lb/inches-squared in January at 25 degrees F. In July at 96 degrees F, the gage pressure is at 26.0 lb/inches-squared. The atm pressure is at 14.7 lb/inches-squared for both.

(a) What precentage of air leaked out of the tire?

(b) If half the air leaked out, what would the gage pressure then be?

Here's a question I've been struggling with and I'd appreciate it if someone could talk me through it like I'm a small child because no matter how I try to solve it, I just don't understand.

I'm probably just overlooking something basic, but I had a friend go over it with me and we figured out a way to get to the right answer, but we don't know why, if that makes sense. In other words, we don't know if we can apply the concept on a test. It's been bugging me for days.

It's a two-parter. OK, here it is:

The gage pressure in a tire is 35.0 lb/inches-squared in January at 25 degrees F. In July at 96 degrees F, the gage pressure is at 26.0 lb/inches-squared. The atm pressure is at 14.7 lb/inches-squared for both.

(a) What precentage of air leaked out of the tire?

(b) If half the air leaked out, what would the gage pressure then be?

I have an idea on how to solve this bad boy....but I am not 100% sure...here it is....I would use the ideal gas law (P1V1/T1) = (P2V2/T2). First, make sure to convert F to K---as is the case when using the ideal gas law. Since P1, P2, T1, and T2 are known and since this occurs at constant atm. pressure set up a flow chart. What goes in is a cool gas and what comes out is a hot gas. And remember one more thing, total pressure is equal to gage pressure plus atmospheric pressure. So calculate the total pressure for each case first. Then, Divide both sides by V1 and then multiply each side by (T2 over P2). You should then get an answer that represents the ratio of V2/V1. This is the ratio of the final volume over the final volume. Convert the ratio to a percentage by multiplying the ratio by a 100 and subtract this from 100%. That should be how much air leaked out. For the second question, are you what the final gage pressure will be. Use the same formula (P1V1/T2)=(P2V2/T1). If the all the values are same as part (a), this time set V2=(1/2)V1. You should be able to cancel V1 from both sides of the equation because of this and than solve for P2. Hope this helps....if I am wrong please correct me....good luck.

I did most of what you said, but I used n2/n1 instead of V because I was trying to find the number of moles. My first thought was to find volume as well, but it doesn't work out to the right answer that way. I got 29% when I did it with moles, which is supposedly the right answer.

Now on part B, I'm having one hell of a time because the gage pressure is whatever the overall pressure is minus atmospheric pressure and for some reason, I get stuck.

Thank you so much for your help!

I did most of what you said, but I used n2/n1 instead of V because I was trying to find the number of moles. My first thought was to find volume as well, but it doesn't work out to the right answer that way. I got 29% when I did it with moles, which is supposedly the right answer.

Now on part B, I'm having one hell of a time because the gage pressure is whatever the overall pressure is minus atmospheric pressure and for some reason, I get stuck.

Thank you so much for your help!

For part a, that was a mistake on my behalf. I thought it was asking for the volume of air leaked out, but looking back at your question, it is moles. For part B, remember, total pressure is the sum of the atmospheric pressure plus the gauge pressure. It is very unusual to see c.g.s units on the MCAT---if I remember correctly, the MCAT uses SI units only. 14.7 pounds per square inch is equivalent to 760 mmHg, which is 1 atm. For part b, set n2=0.5n1. This way, you eliminate n1 from the overall equation. I assume you can use the initial parameters (P1, and T1) from part a of the question and T2 from part a. This time, you will get a value of P2 which should represent the overall pressure. I assume this part builds off part a, so you can use the value of T2, T1 and P1. Now, assuming it builds of part a, subtract the overall pressure you get from your calculation from the atmospheric pressure(14.7 lb per square inch) [overall pressure - atmoshperic pressure] and that value should be your gauge pressure. I hope this helps!

For part a, that was a mistake on my behalf. I thought it was asking for the volume of air leaked out, but looking back at your question, it is moles. For part B, remember, total pressure is the sum of the atmospheric pressure plus the gauge pressure. It is very unusual to see c.g.s units on the MCAT---if I remember correctly, the MCAT uses SI units only. 14.7 pounds per square inch is equivalent to 760 mmHg, which is 1 atm. For part b, set n2=0.5n1. This way, you eliminate n1 from the overall equation. I assume you can use the initial parameters (P1, and T1) from part a of the question and T2 from part a. This time, you will get a value of P2 which should represent the overall pressure. I assume this part builds off part a, so you can use the value of T2, T1 and P1. Now, assuming it builds of part a, subtract the overall pressure you get from your calculation from the atmospheric pressure(14.7 lb per square inch) [overall pressure - atmoshperic pressure] and that value should be your gauge pressure. I hope this helps!

I can't believe how many careless mistakes I make. I got the cross-multiplying wrong when I was trying to isolate P2, which is what threw off my answer. Thank you so much for your help! The gage pressure is 13.8, if you care. LOL!

One last question. I know this is basic, but I've never had chemistry in my life and I'm taking Chemistry I this summer and I'm overwhelmed with the material, so please help me understand how you knew it wanted moles instead of volume? I immediately thought it was asking for volume when I initially read it and only changed the unknown to moles when the answer wasn't coming out the way it was supposed to.

I can't believe how many careless mistakes I make. I got the cross-multiplying wrong when I was trying to isolate P2, which is what threw off my answer. Thank you so much for your help! The gage pressure is 13.8, if you care. LOL!

One last question. I know this is basic, but I've never had chemistry in my life and I'm taking Chemistry I this summer and I'm overwhelmed with the material, so please help me understand how you knew it wanted moles instead of volume? I immediately thought it was asking for volume when I initially read it and only changed the unknown to moles when the answer wasn't coming out the way it was supposed to.

Hey, no problem---that is why I am here to help. I figured it was moles and not volume (my mistake earlier) because I assumed the gas we were dealing with was ideal. Since the atmospheric pressure is the same for the input and output of the process, the only thing changing then, the temperature and pressure, is the internal energy of the gas. From the numbers, the temperature is decreasing as is the gage pressure. This means the total pressure is also decreasing. So if we think of the gas as concealed in the container, some amount escapes as it expands. Since the internal energy of the system is changing, that means some amount is leaving---and this is moles. With ideal gas laws, a frequent topic on the MCAT, if the internal energy is changing than volume changes are negligent. Volume changes take into effect with a change in phase--something not indicated in the problem. Best of luck....feel free to post any other questions/issues you might have :)

I've got a thermochemistry question if anyone's up for it. I've gone over it a number of times and I can't figure out why the solutions manual did what it did for this problem.

I'm working on Hess's Law.

We have to calculate the enthaply change for the reaction:

N2H4 (l) + 2H2O2 (l) ----> N2 (g) +4H2O

From these equations:

N2H4 (l) + O2 (g) -----> N2 (g) + 2H2O (l) H= -622.2

H2 (g) + 1/2 O2 (g) ----> H2O (l) H = -285.8

H2 (g) + O2 (g) -----> H2O2 (l) H = -187.8


So I kept the first equation as is because the N2H4 is on the left in both the first equation and the original. I didn't need to multiply or anything because it's just one mole in the original equation.

I changed the third equation so that the H2O2 was on the left since it's on the left in the original. I also multiplied by two since it's 2H2O2 in the original. I also changed the sign.

Now for the second equation -- I kept the second equation as is, but multiplied by four because in the original equation, it's 4H2O on the right. However, the solutions guide only multiplied it by two. Why is that? I thought you use the original equation as a guide.

P.S. If anyone can clarify something for me, I'd be forever grateful. Is the answer always negative if the question is asking for how much heat was absorbed? Is the sign always positive if the question is asking for how much heat was evolved or released? I'm having trouble keeping the signs straight and my professor has contradicted the solutions manual a few times. He said the solutions manual is wrong and that just confuses me more because that's what I depend on to make sure I'm doing the problems correctly.

I've got a thermochemistry question if anyone's up for it. I've gone over it a number of times and I can't figure out why the solutions manual did what it did for this problem.

I'm working on Hess's Law.

We have to calculate the enthaply change for the reaction:

N2H4 (l) + 2H2O2 (l) ----> N2 (g) +4H2O

From these equations:

N2H4 (l) + O2 (g) -----> N2 (g) + 2H2O (l) H= -622.2

H2 (g) + 1/2 O2 (g) ----> H2O (l) H = -285.8

H2 (g) + O2 (g) -----> H2O2 (l) H = -187.8


So I kept the first equation as is because the N2H4 is on the left in both the first equation and the original. I didn't need to multiply or anything because it's just one mole in the original equation.

I changed the third equation so that the H2O2 was on the left since it's on the left in the original. I also multiplied by two since it's 2H2O2 in the original. I also changed the sign.

Now for the second equation -- I kept the second equation as is, but multiplied by four because in the original equation, it's 4H2O on the right. However, the solutions guide only multiplied it by two. Why is that? I thought you use the original equation as a guide.

P.S. If anyone can clarify something for me, I'd be forever grateful. Is the answer always negative if the question is asking for how much heat was absorbed? Is the sign always positive if the question is asking for how much heat was evolved or released? I'm having trouble keeping the signs straight and my professor has contradicted the solutions manual a few times. He said the solutions manual is wrong and that just confuses me more because that's what I depend on to make sure I'm doing the problems correctly.

Hey. For the first part, you have done all the correct steps until the second equation. You are right to keep the first equation untouched, and to reverse the second equation and to multiply it by a factor of two. However, the third equation, you multiply by 2 and not 4. The first equation already has 2H20. That means, the second equation also needs to have 2H20. Therefore, you multiply the second equation by a factor of 2. You also multiply the change in enthalpy by two since it a state function. You do the same for the third equation. When you solve all three, you should get a change in enthalpy of -818.2 units. For the second question--it all depends on how you view the system and surroundings. My college chemical engineering professor gave me a good analogy---When you take out of the bank it is negative and when you put into the bank it is positive. Therefore, when heat has to be put in than the sign is positive, and when heat is released the sign is negative. If you use this convention, make sure you keeps the signs straight or you will get a different answer. Most endothermic reactions have a positive change in enthalpy and exothermic reactions have a negative change in enthalpy. Hope this helps!! Good luck :luck:

Hey. For the first part, you have done all the correct steps until the second equation. You are right to keep the first equation untouched, and to reverse the second equation and to multiply it by a factor of two. However, the third equation, you multiply by 2 and not 4. The first equation already has 2H20. That means, the second equation also needs to have 2H20. Therefore, you multiply the second equation by a factor of 2. You also multiply the change in enthalpy by two since it a state function. You do the same for the third equation.

Oh, so you don't always have to go by the original equation? The equations just have to add up to the original? So if the original equation had 4H20s and the first equation had just one H20, then I'd multiply the second equation by 3? That way the first and second equation together would give me 4H20s? Is that right or did I misunderstand?

My college chemical engineering professor gave me a good analogy---When you take out of the bank it is negative and when you put into the bank it is positive. Therefore, when heat has to be put in than the sign is positive, and when heat is released the sign is negative. If you use this convention, make sure you keeps the signs straight or you will get a different answer. Most endothermic reactions have a positive change in enthalpy and exothermic reactions have a negative change in enthalpy. Hope this helps!! Good luck

I'm sorry to be such a pain, but I'm confused because that's exactly the opposite of what my professor said. He said if heat is coming out (exothermic), it'll have a positive charge, but endothermic will have a negative charge. Does it not matter in the final answer as long as you choose a way and stick to it? What happens if my answer is correct, but I make it a negative instead of a positive like the solutions manual did on a couple of problems?

Oh, so you don't always have to go by the original equation? The equations just have to add up to the original? So if the original equation had 4H20s and the first equation had just one H20, then I'd multiply the second equation by 3? That way the first and second equation together would give me 4H20s? Is that right or did I misunderstand?[/QUOTE]

Problems with Hess Law involve manipulation of the reactions given to you. You have to make sure you understand the original equation before you try to manipulate the reactions given to you. Some people see how to manipulate the reactions right away and for some it involves practice. What I did was I wrote all the reactions down one by one. First, I try to match the reactants side of the original equation. Usually, when I figure that out, I can tackle the products side. But, you always have to make sure the reactions add up to the original. It is like solving a puzzle, once you get some to fit everything else falls into place.



I'm sorry to be such a pain, but I'm confused because that's exactly the opposite of what my professor said. He said if heat is coming out (exothermic), it'll have a positive charge, but endothermic will have a negative charge. Does it not matter in the final answer as long as you choose a way and stick to it? What happens if my answer is correct, but I make it a negative instead of a positive like the solutions manual did on a couple of problems?[/QUOTE]

Different text books give different signs but it all depends on how you view the system and the surroundings. The final answer doens't matter as long as you stick with one system--the numbers will comply accordingly. If you say exothermic is negative then whenever you see negative or release heat put a negative sign. The same holds true for endothermic reactions. The whole sign issue comes into play when you have to choose a way and stick with it. My professor told us that you will get a answer but you can change the sign accordingly depending on what your sign convention is. Just make sure to stick with only one sign convention---otherwise it can get tricky and troublesome. Good luck :luck:

Great! Thanks for all your help!

You know, it just clicked. It isn't dependent on that first equation but on the reactions side of the second equation and since there's an H2 and a 1/2 O2 there, that's why you multiply by two. I misunderstood you the first time. Duh. Thank you again! You saved me on this problem!

Great! Thanks for all your help!

Hey, its no big deal. I love to help others with questions they might have in physics or chemistry because I know how stressful studying for the exam is :laugh:

What bonds are the strongest and would have the highest boiling point? What are the order of the bonds from the strongest to the weakest?

What bonds are the strongest and would have the highest boiling point? What are the order of the bonds from the strongest to the weakest?
From strongest to weakest (and correspondingly highest BP to lowest BP):

Triple bonds > Double bonds > Single bonds

And if you're looking at bond length from longest to shortest:

Single bonds > Double bonds > Triple bonds

From strongest to weakest (and correspondingly highest BP to lowest BP):

Triple bonds > Double bonds > Single bonds

And if you're looking at bond length from longest to shortest:

Single bonds > Double bonds > Triple bonds

Sorry, I wasn't more specific, I meant between intramolecular bonds.
Polar, nonpolar, ionic, dipole-dipole, hydrogen bonds. I don't know the order of strongest to weakest. Do you?

Sorry, I wasn't more specific, I meant between intramolecular bonds.
Polar, nonpolar, ionic, dipole-dipole, hydrogen bonds. I don't know the order of strongest to weakest. Do you?
I think you are having trouble distinguishing intermolecular interactions versus intramolecular bonds. There are two posts about them in the gen chem explanations thread that you might want to read if you haven't already. For intermolecular interactions, H bonds are the strongest, then dipole-dipole, then dispersion forces. For intramolecular bonds (ionic versus covalent), it depends. You can't make a blanket statement that ionic are always stronger than covalent or vice versa.

Hi
Sorry - this a long question. I am getting terribly confused with pH and pKa problems.

I tend to think that strong acids dissociate fully and so pH = -log[H+].

For weaker acids, I calculate pKa = -log ([H+][A-]/[HA])
where HA is the initial acid concentration and A- is the conjugate base conc.
If x moles of HA are added to a 1 L solution then y moles of HA dissociate to produce y moles of H+ and y moles of A-.
So I calculate Ka = y^2/(x-y)

I recently came a across the following situation: 0.0001 moles of benzoic acid (Ka = 6.6 x 10^-5) in 1L of water.
Solving this problem using the above method results in a quadratic equation. Should I assume that insignificant amount of acid will dissociate and so I can ignore the value y in the denominator? If I do this then y turns out to be approximately 8 x 10^-5. Is this correct?

Also, how do I know if the acid is weak or strong? If the question states a Ka or pKa value, can I assume the acid is weak?
Thanks a lot.

Hi
Sorry - this a long question. I am getting terribly confused with pH and pKa problems.

I tend to think that strong acids dissociate fully and so pH = -log[H+].

For weaker acids, I calculate pKa = -log ([H+][A-]/[HA])
where HA is the initial acid concentration and A- is the conjugate base conc.
If x moles of HA are added to a 1 L solution then y moles of HA dissociate to produce y moles of H+ and y moles of A-.
So I calculate Ka = y^2/(x-y)

I recently came a across the following situation: 0.0001 moles of benzoic acid (Ka = 6.6 x 10^-5) in 1L of water.
Solving this problem using the above method results in a quadratic equation. Should I assume that insignificant amount of acid will dissociate and so I can ignore the value y in the denominator? If I do this then y turns out to be approximately 8 x 10^-5. Is this correct?

Also, how do I know if the acid is weak or strong? If the question states a Ka or pKa value, can I assume the acid is weak?
Thanks a lot.

Hello. Your first statement about strong acids is correct--they do dissociate fully in water. Weak acids do not dissociate completely in water. For your question, set Ka= (moles H+)(moles conjugate base)/acid. What will help the most is the ICE chart (where I is the initial concentrations, c is change and e is equilibrium). Doing that, you see that you have the following: (0.0001 Molar acid, 0 moles H+ and conjugate base as initial concentrations. The changes is: 0.0001-x for acid, +x for H+ and conjugate base.) Now you have (x^2/0.0001-x)=Ka. Since this acid is weak, it dissociate's very moderatly in water so you can assume (this is the 5% approximation when the Ka is slightly greater than 10^-4) that 0.0001-x is so small that it is equal to 0.0001. Therefore, when you solve for x you do approximately get 8 * 10^-5.

For your other question. The strong acids are common (HCL, HBr, H2SO4, HI, HCLO4, HCLO3, and HNO3). These are the most common strong acids for which Ka>1. There are other acids that fit the definition of strong (thiocyanic acid, selenic acid, and permanganic acid) but are so uncommon that it is very unlikely to appear on the MCAT. I would suggest committing the above list to memory or that 3 are from the same group--the halogens. On the test, just because you are given a Ka value you cannot assume the acid is weak. You have to judge this by the magnitude of the Ka value. Strong acids will have a Ka value >>1 since they completely dissociate but weak acids will have a Ka value<<1 since they do not completely dissociate. Otherwise, I suggest to look at the identity of the acid--if it is from the list above, then it is a strong acid. I hope this helps and good luck :luck:

Hi
Sorry - this a long question. I am getting terribly confused with pH and pKa problems.

I tend to think that strong acids dissociate fully and so pH = -log[H+].

For weaker acids, I calculate pKa = -log ([H+][A-]/[HA])
where HA is the initial acid concentration and A- is the conjugate base conc.
If x moles of HA are added to a 1 L solution then y moles of HA dissociate to produce y moles of H+ and y moles of A-.
So I calculate Ka = y^2/(x-y)

I recently came a across the following situation: 0.0001 moles of benzoic acid (Ka = 6.6 x 10^-5) in 1L of water.
Solving this problem using the above method results in a quadratic equation. Should I assume that insignificant amount of acid will dissociate and so I can ignore the value y in the denominator? If I do this then y turns out to be approximately 8 x 10^-5. Is this correct?

Also, how do I know if the acid is weak or strong? If the question states a Ka or pKa value, can I assume the acid is weak?
Thanks a lot.
1) Yes. You won't have a calculator on the MCAT, so always make the approximation that y is very small compared to x and can be neglected in the denominator. That way, you will not have to ever use the quadratic equation. I got the same answer that you did.

2) You should memorize the list of strong acids for the test. Go to the gen chem explanations thread and read the post about strong acids there. Any acid not on the list should be assumed to be weak.

Hi, I was wondering if someone can help me answer this question:

A 3:1 mixture of H2 and N2 was rapidly pumped into a Haber apparatus until the internal pressure was 60atm at 350 degrees celsius. A catalyst was then added and the system was allowed to reach equilibrium. The internal pressure was 45atm. What was the partial pressure of NH3 in the final mixture at 350 degrees celsius?

Answer - 15atm.

Thank you!

So I was wondering,

a question asks, which of the ions is expected to be most resitant to a reducing agent during reverse osmosis.

Choices:

Ni 2+
Cu 2+
Zn 2+ (This is the answer)
Ga 2+

In the passage it states tht reverse osmoiss, contains a substance that reduces the contaminants and that the ion easily gains electrons

No sure why the right answer is right

Thanks

So I was wondering,

a question asks, which of the ions is expected to be most resitant to a reducing agent during reverse osmosis.

Choices:

Ni 2+
Cu 2+
Zn 2+ (This is the answer)
Ga 2+

In the passage it states tht reverse osmoiss, contains a substance that reduces the contaminants and that the ion easily gains electrons

No sure why the right answer is right

Thanks

Hey. I'm curious, do they provide a table of reduction potentials in the passage? If they do, this problem becomes a little bit easier. A reducing agent is a substance that gets oxidized during the course of the reaction. Normally, you would be give a table of reducing potentials in the passage. The more negative the reducing potential, the stronger the reducing agent the metal is. The best way I memorized the trend was: (and it may not be right but it helped me memorize it regardless) the more positive the reduction potential, the more favorable the species to picking up electrons. The more negative the reduction potential, the opposite reaction dominates. If you look at a table of reduction potentials, zinc has the most negative potential and thus is the more favorable reducing agent. I hope this helps!! Good luck.

Hi, I was wondering if someone can help me answer this question:

A 3:1 mixture of H2 and N2 was rapidly pumped into a Haber apparatus until the internal pressure was 60atm at 350 degrees celsius. A catalyst was then added and the system was allowed to reach equilibrium. The internal pressure was 45atm. What was the partial pressure of NH3 in the final mixture at 350 degrees celsius?

Answer - 15atm.

Thank you!

Hey. The reaction between hydrogen gas and nitrogen gas produces ammonia. The balanced reaction looks like this:
3H2 + N2 --> 2NH3.
Since you are given molar quantities of two species, you know you are dealing with a limiting reactant problem. The stoichiometric ratio of H2 to N2 is 3:1 and the mixture is 3:1. That means nitrogen gas is the limiting reactant in the problem. One mole of nitrogen gas will produce 2 moles of ammonia. The reactant mixture contains a total of four moles. The partial pressure of H2 is (3/4)60 and N2 is (1/4)60. Since nitrogen is limiting, then this will give rise to a partial pressure of 15atm of ammonia gas in the product since all nitrogen gas is used up in the reaction. The remainder is hydrogen gas. I hope this helps. Good luck!

1) Yes. You won't have a calculator on the MCAT, so always make the approximation that y is very small compared to x and can be neglected in the denominator. That way, you will not have to ever use the quadratic equation. I got the same answer that you did.

2) You should memorize the list of strong acids for the test. Go to the gen chem explanations thread and read the post about strong acids there. Any acid not on the list should be assumed to be weak.

Hi Everyone,
Thanks for your help. I have one follow up question -

In the above problem, the starting concentration of benzoic acid is 10^-4 moles/L. About 0.8x10^-4 moles/L of the acid dissociates and 0.2x10^-4 moles/L are left. Can this be considered a negligible concentration? Is there a particular cut-off below which the remaining acid concentration can be neglected on the MCAT?

I recently came across the following situation: 0.0001 moles of benzoic acid (Ka = 6.6 x 10^-5) in 1L of water.
Solving this problem using the above method results in a quadratic equation. Should I assume that insignificant amount of acid will dissociate and so I can ignore the value y in the denominator? If I do this then y turns out to be approximately 8 x 10^-5. Is this correct?

Hi Everyone,
Thanks for your help. I have one follow up question -

In the above problem, the starting concentration of benzoic acid is 10^-4 moles/L. About 0.8x10^-4 moles/L of the acid dissociates and 0.2x10^-4 moles/L are left. Can this be considered a negligible concentration? Is there a particular cut-off below which the remaining acid concentration can be neglected on the MCAT?

Hi Everyone,
Thanks for your help. I have one follow up question -

In the above problem, the starting concentration of benzoic acid is 10^-4 moles/L. About 0.8x10^-4 moles/L of the acid dissociates and 0.2x10^-4 moles/L are left. Can this be considered a negligible concentration? Is there a particular cut-off below which the remaining acid concentration can be neglected on the MCAT?
Like BME said, y should generally be 5% or less of x in order for the approximation to hold true. That's the cutoff for problems you do in gen chem class, anyway. But even if y is greather than 5% of x like in this case, always make the approximation anyway on the MCAT. Remember that the MCAT is a timed test. You don't have a calculator, and you don't have all day to solve these problems. So you need to be expedient and estimate here, not get your answer accurate to the thousandths decimal place. You really, really don't want to be trying to solve quadratic equations with no calculator.

Could someone please give me a breakdown of these?

What does a low Ka indicate?
A low Kb
A high Ka
A high Kb
A low pKa
A low pKb
A high pKa
A high pKb

I can't seem to get any of them straight except that a low pKa is similar to a low pH and that means a strong acid. Thanks in advance.

Could someone please give me a breakdown of these?

What does a low Ka indicate?
A low Kb
A high Ka
A high Kb
A low pKa
A low pKb
A high pKa
A high pKb

I can't seem to get any of them straight except that a low pKa is similar to a low pH and that means a strong acid. Thanks in advance.
Ka is the equilibrium constant for an acid dissociation. In other words, if you have an acid dissociating:

HA <-> H+ + A-

Then Ka = {[H+][A-]}/[HA]

If you look at this equation, you will see that Ka is very large if the acid dissociates a lot (i.e., if it is a strong acid). In other words, for a strong acid, [HA] is small (small denominator) or [H+] and {A-] are large (large numerator). A low Ka means that the acid is weak: there is more HA than dissociated products.

Taking the pKa means you are taking the negative log. That means that if your Ka is low, when you take the negative log of that, you will get a large number, and vice versa. So a weak acid will have a larger pKa, and a strong acid will have a smaller one. (if you're not convinced of this, try taking some negative logs of several numbers using a calculator, and you will see that larger numbers have smaller negative logs and vice versa.)

Kb is similar, except that it is the equation for a base accepting a proton (if we used the Bronsted-Lowry definition). So here is the equation:

B + H2O <-> BH+ + OH-

Kb is set up analogously to Ka, and therefore a low Kb means you have a weak base that does not readily dissociate water. As before, if Kb is low, then pKb will be high, and vice versa.

question?

Isn't the more negative the reduction potential the higher liklihood it gets oxidized

Problem

two inert electrodes connected to a power source placed in aqueous solution of NACl, in resulting electrolytic solution , which forms at the anode?
E^0
1.Cl2 = 1.42
2.02 = -1.23
3.H2 =+.83
4.Na = +2.71

Answer says:

oxidation potential of water (–1.23 V) the opposite of its reduction potential is less negative than
the oxidation potential of chloride (–1.42 V). Thus, oxygen will form at the anode, not chlorine.

thanks

question?

Isn't the more negative the reduction potential the higher liklihood it gets oxidized

Problem

two inert electrodes connected to a power source placed in aqueous solution of NACl, in resulting electrolytic solution , which forms at the anode?
E^0
1.Cl2 = 1.42
2.02 = -1.23
3.H2 =+.83
4.Na = +2.71

Answer says:

oxidation potential of water (–1.23 V) the opposite of its reduction potential is less negative than
the oxidation potential of chloride (–1.42 V). Thus, oxygen will form at the anode, not chlorine.

thanks

Usually, when you are given two species, you will be given the reduction potentials. The more negative the reduction potential, the better the species is a reducing agent--it is oxidized.

Why is it that at standard state for an element the standard Gibbs energy is zero?

delta G = delta H + T(delta S)
At standard state, delta H for an element is, by convention, zero. So delta H part becomes zero. As for delta S, it won't be zero becuase it's zero at 0K(absolute temp) and in equilibrium. So how come delta G is zero? Please correct me..... :)

.

Is it possible to say the reaction to be exothermic when a solution of NaCl and another solution of AgNo3 are mixed to form a precipitate? My reasoning would be because that bonds formed from the precipitate reaction is stronger than the bonds broken so the overall heat of reaction should be exothermic. That is, less heat was put into breaking the bonds than released by the formation of stronger bonds, which is the precipiate (in solid state as compared to aqueous).
Am I on the right track? Thanks!!

Why is it that at standard state for an element the standard Gibbs energy is zero?

delta G = delta H + T(delta S)
At standard state, delta H for an element is, by convention, zero. So delta H part becomes zero. As for delta S, it won't be zero becuase it's zero at 0K(absolute temp) and in equilibrium. So how come delta G is zero? Please correct me..... :)
The second term is also zero because it's DELTA S. The entropy isn't changing. :)

Is it possible to say the reaction to be exothermic when a solution of NaCl and another solution of AgNo3 are mixed to form a precipitate? My reasoning would be because that bonds formed from the precipitate reaction is stronger than the bonds broken so the overall heat of reaction should be exothermic. That is, less heat was put into breaking the bonds than released by the formation of stronger bonds, which is the precipiate (in solid state as compared to aqueous).
Am I on the right track? Thanks!!
Assuming that your analysis is correct about the relative bond strengths, what you said makes sense to me.

I agree to what pezzang posted. where did yu learn bout it?

The second term is also zero because it's DELTA S. The entropy isn't changing. :)

Is the delta S = 0 because there is no heat (delta H = 0) absorbed or released by the system? Thanks!

I agree to what pezzang posted. where did yu learn bout it?

What do you mean? :