bme post

[QofQuimica]

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

Acceptable topics:
-general, MCAT-level gen chem.
-particular MCAT-level gen chem problems, whether your own or from study material
-what you need to know about gen chem for the MCAT
-how best to approach to MCAT gen chem passages
-how best to study MCAT gen chem
-how best to tackle the MCAT physical sciences section

Unacceptable topics:
-actual MCAT questions or passages, or close paraphrasings thereof
-anything you know to be beyond the scope of the MCAT

*********

If you really know your gen chem, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current members of the General Chemistry Team:

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

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

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

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

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[harrypotter]

Ksp is the salt solubility equilibrium constant; it is a measure of how much of the salt is in the ionized form. Molar solubility is the "x" that you are solving for when you set up a Ksp expression; it tells you how much of that particular ion dissolves in the solvent. The units of molar solubility will depend on how many ions are formed for that salt. Each concentration that you multiply in the Ksp expression contributes an extra "M" to the units.

Thank you for defining the terms! Can you compare Molar Solubilities that end in differing units: M^2 vs M^3?

When asked which salt is the most soluble, do you look at Ksp? And when asked about a specific ion look at Molar Solubility which you multiply by how many of that specific ion is in the salt? (AgF vs Ag2CrO4)

 

[googlinggoogler]

Hi, I tried doing a search on this question on this site, but I couldn't find the answer..I might have missed it if it was posted before. In my TPR book, it says that Keq increases with temperature for an endothermic reaction and decreases with temperature for an exothermic reaction. I understand why Keq would increase with temperature for an endothermic reaction (increased temperature would be used like a reactant, forming more product, thus making the Keq larger). However, I can't seem to understand why Keq would decrease with temperature for an exothermic reaction. I would think that decreasing energy would be favorable for an exothermic reaction, and therefore Keq would increase (because more product is formed). Can someone explain where my logic goes wrong? Thanks in advance

 

[QofQuimica]

Thank you for defining the terms! Can you compare Molar Solubilities that end in differing units: M^2 vs M^3?

When asked which salt is the most soluble, do you look at Ksp? And when asked about a specific ion look at Molar Solubility which you multiply by how many of that specific ion is in the salt? (AgF vs Ag2CrO4)

1) Yes, because you'll take that root. In other words, if you have x^2, you'll take the square root; if you have x^3, you'll take the cube root, and so on.

2) Yes. A smaller Ksp means that you have a lower concentration of ions, so it's a less soluble salt. For the molar solubility, you need to get the value of x. So you won't be multiplying; you'll be taking the nth root to get x.

 

[QofQuimica]

Hi, I tried doing a search on this question on this site, but I couldn't find the answer..I might have missed it if it was posted before. In my TPR book, it says that Keq increases with temperature for an endothermic reaction and decreases with temperature for an exothermic reaction. I understand why Keq would increase with temperature for an endothermic reaction (increased temperature would be used like a reactant, forming more product, thus making the Keq larger). However, I can't seem to understand why Keq would decrease with temperature for an exothermic reaction. I would think that decreasing energy would be favorable for an exothermic reaction, and therefore Keq would increase (because more product is formed). Can someone explain where my logic goes wrong? Thanks in advance

This whole business of considering heat as a product or reactant for exothermic and endothermic reactions respectively is a bit simplistic, because the value of Keq itself is changing with temperature. So you actually have two separate issues here: for endothermic reactions, heat is like a reactant, so you would predict that heating them up makes them go to the right. But, at the same time, heating the reaction up also increases Keq's value (i.e., at the new equilibrium, you have a greater concentration of your products at the old equilibrium). For example, the Kw for water autoionization is 1 x 10^-14, but only at 25 C. If you raise the temperature to about 65 C, now Kw is 1 x 10^-13, and a neutral solution won't have a pH of 7 any more! (Neutral pH will be the -log of the square root of 1 x 10^-13.) So you are actually changing the equilibrium position itself when you play around with the temperature of a reaction, which doesn't happen when you mess with the concentrations of reactants or products.

 

[googlinggoogler]

This whole business of considering heat as a product or reactant for exothermic and endothermic reactions respectively is a bit simplistic, because the value of Keq itself is changing with temperature. So you actually have two separate issues here: for endothermic reactions, heat is like a reactant, so you would predict that heating them up makes them go to the right. But, at the same time, heating the reaction up also increases Keq's value (i.e., at the new equilibrium, you have a greater concentration of your products at the old equilibrium). For example, the Kw for water autoionization is 1 x 10^-14, but only at 25 C. If you raise the temperature to about 65 C, now Kw is 1 x 10^-13, and a neutral solution won't have a pH of 7 any more! (Neutral pH will be the -log of the square root of 1 x 10^-13.) So you are actually changing the equilibrium position itself when you play around with the temperature of a reaction, which doesn't happen when you mess with the concentrations of reactants or products.

Thanks again QofQuimica! I agree with what you've posted, and that's an important point you've made that Keq actually changes with temperature.

Just in case I might have confused anyone else who was reading my post and studying for the mcat, I just looked back at my gen chem textbook (from 4 years ago! ), and it says that Keq decreases with increasing temperature for an exothermic reaction. So I think maybe that was a typo from the TPR book or I may have interpreted what they wrote in the wrong way.

This subforum is excellent. I've gotten a lot of help here and otherwise would be struggling like this .

 

[QofQuimica]

Thanks again QofQuimica! I agree with what you've posted, and that's an important point you've made that Keq actually changes with temperature.

Just in case I might have confused anyone else who was reading my post and studying for the mcat, I just looked back at my gen chem textbook (from 4 years ago! ), and it says that Keq decreases with increasing temperature for an exothermic reaction. So I think maybe that was a typo from the TPR book or I may have interpreted what they wrote in the wrong way.

This subforum is excellent. I've gotten a lot of help here and otherwise would be struggling like this .

That's why we started it. If you feel so inclined to give back to the SDN community, please drop in and help us answer questions for the January testers after you take the test this month. to you.

 

[pezzang]

I have two questions regarding thermodynamics:

1. Why is that for an ideal gas, any state function can be described as a function of temp and vol only?

2. For an ideal gas, internal energy is only dependent upon temperature. How come it isn't dependent on volume for an idea gas but for a real gas? How does the behavior of real gas (intermolecular attractions and size of molecules) affect these deviations?

Thank you for the help!

 

[ssa915]

On a certain exam, "CMAA #9", in the physical science section, # 29, I don't understand how we know that HC is in the slow/rate-determining step based on the fact that its a nonlinear plot?

 

[H and D]

When determining Ecell from two given electrode potentials why do you not double the electrode potential if 2 moles of it are needed in the Ecell?

For instance if one electrode potential is:
Cu2+ + 2e- goes to Cu(s) E=0.34V

But in the Ecell reaction it is written:
2Cu2+ goes to 2Cu(s)
why don't you double the given E to determine the Ecell (ie 0.68V instead of 0.34V)

Does the amount (moles) not matter when determining the properties of electric cells?

I hope I have explained this clearly enough.

 

[QofQuimica]

When determining Ecell from two given electrode potentials why do you not double the electrode potential if 2 moles of it are needed in the Ecell?

For instance if one electrode potential is:
Cu2+ + 2e- goes to Cu(s) E=0.34V

But in the Ecell reaction it is written:
2Cu2+ goes to 2Cu(s)
why don't you double the given E to determine the Ecell (ie 0.68V instead of 0.34V)

Does the amount (moles) not matter when determining the properties of electric cells?

I hope I have explained this clearly enough.

http://forums.studentdoctor.net/showpost.php?p=2862982&postcount=9

 

[QofQuimica]

On a certain exam, "CMAA #9", in the physical science section, # 29, I don't understand how we know that HC is in the slow/rate-determining step based on the fact that its a nonlinear plot?

I don't have any AAMC tests, so sorry, I have no earthly idea what you're talking about. Don't post the actual question, but can you at least give me a general gist of what the topic is?

 

[goinverted]

What is the concentration (M) of sodium ions in 4.57 L of a 0.847 M Na3P solution?

 

[pezzang]

What is the concentration (M) of sodium ions in 4.57 L of a 0.847 M Na3P solution?

3 x 0.847M would be the conc of sodium because for every Na3P that dissociates, 3Na+ and 1P3- are formed. So the ratio of Na3P to Na+ is 1:3.

 

[icyfate]

Can I know how the reaction between a monobasic acid (HCl) with Iron and dibasic acid (H2SO4) with Iron will be different? I'm actually trying to find the enthalpy change of reaction for both reactions and compare, but I need a sound explanation as to why dibasic acid released more heat

 

[pezzang]

How does increasing the partial presure of a gas increase the solubioity of gas? I am not really getting this concept... any help?

 

[orient03]

#156) The students were careful to keep the temperature of the reaction flask under 90C to prevent:
Answer) the unreacted cyclohexanol from distilling

The passage indicates that cyclohexanol has a much higher boiling point of 161C. Is this still relevant?

Granted, the other answers probably make less sense...

What would provoke a rearrangement of the carbocation intermediate?

 

[goinverted]

PCl5 <-> PCl3 + Cl2

Initially, 0.84 mol of PCl5 was placed in a 1.0 L flask. At equilibrium, 0.72 mol of PCl5 was present. What is the value of Kc for this reaction???

 

[nope80]

Hi everyone,

I'm having trouble understanding the meaning/point of galvanic and electrolytic cells. I'm pretty confident with solving the problems but I dont understand anything beyond the mechanics. If someone could help me out with this, that would be great

 

[jimmyl930]

For both oxidative and reduction potentials, is more positive always the stronger one? Thanks.

 

[nope80]

A problem that I came in across an MCAT book that I dont understand

Question: Hydrogen peroxide H2O2 can behave as a reducing agent or an oxidizing agent as shown by the 2 half reactions below labelled R1 and R2. Which of the following is true?

R1: O2g + 2H+ + 2e- ----> H2O2 E0 = .68V

R2: 2H+ + H2O2 + 2e- -----> 2H20 E0 = 1.78V

a) R1 demonstrates that hydrogen peroxide behaves as a reducing agent in basic solution.
b) R2 demonstrates that hydrogen peroxide behaves as a reducing agent in basic solution.
c) R1 demonstrates that hydrogen peroxide behaves as a oxidizing agent in basic solution.
d) R2 demonstrates that hydrogen peroxide behaves as a oxidizing agent in basic solution.

A is the correct answer but I dont understand why and how the basic/acidic solution comes into play. Any help would be greatly appreciated.

 

[orient03]

PCl5 <-> PCl3 + Cl2

Initially, 0.84 mol of PCl5 was placed in a 1.0 L flask. At equilibrium, 0.72 mol of PCl5 was present. What is the value of Kc for this reaction???

PCl5 ---> PCl3 + Cl2
.82 mol/L 0 0
-x +x +x
.72M .10 .10

.82 - x = .72
x = .10

Kc = [PCl3][Cl2] / [PCl5]
= (.10)(.10)/(.72)
= .014

 

[orient03]

Hi everyone,

I'm having trouble understanding the meaning/point of galvanic and electrolytic cells. I'm pretty confident with solving the problems but I dont understand anything beyond the mechanics. If someone could help me out with this, that would be great

A galvanic cell undergoes a spontaneous reaction whereas an electrolytic cell undergoes a nonspontaneous reaction (which requires electrical energy). The reverse of a galvanic reaction would be an electrolytic one.

 

[orient03]

For both oxidative and reduction potentials, is more positive always the stronger one? Thanks.

On a Reduction Potential Table, the most negative number is the strongest reducing agent, the most positive number is the strongest oxidizing agent.

 

[orient03]

A problem that I came in across an MCAT book that I dont understand

Question: Hydrogen peroxide H2O2 can behave as a reducing agent or an oxidizing agent as shown by the 2 half reactions below labelled R1 and R2. Which of the following is true?

R1: O2g + 2H+ + 2e- ----> H2O2 E0 = .68V

R2: 2H+ + H2O2 + 2e- -----> 2H20 E0 = 1.78V

a) R1 demonstrates that hydrogen peroxide behaves as a reducing agent in basic solution.
b) R2 demonstrates that hydrogen peroxide behaves as a reducing agent in basic solution.
c) R1 demonstrates that hydrogen peroxide behaves as a oxidizing agent in basic solution.
d) R2 demonstrates that hydrogen peroxide behaves as a oxidizing agent in basic solution.

A is the correct answer but I dont understand why and how the basic/acidic solution comes into play. Any help would be greatly appreciated.

Since H+ and H2O2 are on the same side in R2, that reaction occurs in acid.

H2O2 acts as a reducing agent in base. To check, you can add an OH- to both sides of R1, and are left with
O2g + 2H+ + 2e- ----> H2O2 E0 = .68V
O2g + 2H+ + 2OH- + 2 e- -----> H202 + OH-
O2g + 2H2O ----> H2O2 + OH- (peroxide in base)
Oxygen is being reduced, thus H2O2 acts as a reducing agent.

 

[orient03]

How does increasing the partial presure of a gas increase the solubioity of gas? I am not really getting this concept... any help?

Henry's Law
C1/P1 = C2/P2
C = concentration of gas dissolved in liquid
P = partial pressure of gas over surface of liquid

If P is increased, C increases, thus solubility increases.

Pressure pushes gas particles into the liquid.

 

[QofQuimica]

PCl5 <-> PCl3 + Cl2

Initially, 0.84 mol of PCl5 was placed in a 1.0 L flask. At equilibrium, 0.72 mol of PCl5 was present. What is the value of Kc for this reaction???

You need to set up the equilibrium expression and figure out what your concentratrions are. Then solve.

 

[QofQuimica]

Hi everyone,

I'm having trouble understanding the meaning/point of galvanic and electrolytic cells. I'm pretty confident with solving the problems but I dont understand anything beyond the mechanics. If someone could help me out with this, that would be great

http://forums.studentdoctor.net/showpost.php?p=3777346&postcount=12

 

[QofQuimica]

For both oxidative and reduction potentials, is more positive always the stronger one? Thanks.

Yes; if you have something with a very low negative reduction potential, and you turn it around, you're going to have a high positive oxidation potential.

 

[orient03]

On a certain exam, "CMAA #9", in the physical science section, # 29, I don't understand how we know that HC is in the slow/rate-determining step based on the fact that its a nonlinear plot?

The passage says that when HC is in excess:
ln[M/M'] = -k[HC]t = -kt
so ln [M] vs t is linear (and the amount of HC doesn't matter)

Since it isn't linear, HC must not be in excess, and the [HC] must matter. Therefore the reaction involving HC is the slow, rate-det. step.

 

[nope80]

Since H+ and H2O2 are on the same side in R2, that reaction occurs in acid.

H2O2 acts as a reducing agent in base. To check, you can add an OH- to both sides of R1, and are left with
O2g + 2H+ + 2e- ----> H2O2 E0 = .68V
O2g + 2H+ + 2OH- + 2 e- -----> H202 + OH-
O2g + 2H2O ----> H2O2 + OH- (peroxide in base)
Oxygen is being reduced, thus H2O2 acts as a reducing agent.

Can someone explain a little further? I still dont get it-how you know that the reaction occurs in acid or base.

 

[googlinggoogler]

PCl5 ---> PCl3 + Cl2
.82 mol/L 0 0
-x +x +x
.72M .10 .10

.82 - x = .72
x = .10

Kc = [PCl3][Cl2] / [PCl5]
= (.10)(.10)/(.72)
= .014

That looks good, except the starting material is 0.84 so I think Kc is probably 0.02

 

[googlinggoogler]

Can someone explain a little further? I still dont get it-how you know that the reaction occurs in acid or base.

BUMP

 

[googlinggoogler]

That's why we started it. If you feel so inclined to give back to the SDN community, please drop in and help us answer questions for the January testers after you take the test this month. to you.

thanks, most definantly will!

 

[okcomputer]

Question:

A strip Cu placed in HNO3 (aq) evolves a gas. Which is it?

a) NO
b) CO2
c) H2
d) O3

Answer is NO

I am not sure how to approach this problem (even though I can rule out B) and I think I'm missing some basic concept. Could someone tell me how to solve it?
why couldnt it be H2?

Thanks

 

[QofQuimica]

Question:

A strip Cu placed in HNO3 (aq) evolves a gas. Which is it?

a) NO
b) CO2
c) H2
d) O3

Answer is NO

I am not sure how to approach this problem (even though I can rule out B) and I think I'm missing some basic concept. Could someone tell me how to solve it?
why couldnt it be H2?

Thanks

Since you have dilute HNO3, that answer is correct. Here is the balanced equation for the reaction:

3Cu(s) + 8HNO3(aq) ——> 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)

Basically it is a decomposition (and a redox) reaction. The copper atoms are being oxidized to Cu2+, and they donate their electrons to the nitrate ions, two of which are being reduced to NO. In general, it is always best to start by writing out the equation and balancing it when you solve gen chem problems. This one is pretty complex though. I wouldn't expect you to have to come up with it on your own for the MCAT.

 

[Cloudcube]

Hi there,

Say I have a compound. K2Cr2O7, for example. How would I figure out how many moles of Cr ions there are? I'm not sure I'm approaching problems like that correctly.

Thanks so much!

 

[H and D]

I am still a little confused on the difference between PV work and non PV work. EK seems to stress this topic a lot but it was never a subject I covered in my undergrad classes so I am a little lost.
I take it both systems have the ability to do work but I thought some energy in the non PV work system goes to heat since it cannot expand. However galvanic cells do Non PV work and now I am confused.
Very lost, please help.

Thanks.

 

[QofQuimica]

Hi there,

Say I have a compound. K2Cr2O7, for example. How would I figure out how many moles of Cr ions there are? I'm not sure I'm approaching problems like that correctly.

Thanks so much!

Assuming you know how many moles of the entire compound you have, you will have twice as many moles of Cr ions as you do of the entire compound, since Cr appears twice in the formula.

 

[QofQuimica]

I am still a little confused on the difference between PV work and non PV work. EK seems to stress this topic a lot but it was never a subject I covered in my undergrad classes so I am a little lost.
I take it both systems have the ability to do work but I thought some energy in the non PV work system goes to heat since it cannot expand. However galvanic cells do Non PV work and now I am confused.
Very lost, please help.

Thanks.

I've never read EK, so I'm not sure what the main point they are trying to make to you is. But it seems to me from what you've said that they are trying to distinguish work done by an expanding gas (PV work) from other work (such as work done by an electrochemical potential). In any real system, you will lose some energy to heat, but you may see problems that tell you to neglect that loss in order to simplify the calculations. Maybe BME has some thoughts; I'll PM him and ask him if he can add anything.

 

[Cloudcube]

Assuming you know how many moles of the entire compound you have, you will have twice as many moles of Cr ions as you do of the entire compound, since Cr appears twice in the formula.

Well that certainly gets me the right answer, but how can one portion of a compound have more moles than the moles of the entire compound? I realize it appears twice, but shouldn't the number of moles of the Cr ions be just one portion of the entire thing?

 

[QofQuimica]

Well that certainly gets me the right answer, but how can one portion of a compound have more moles than the moles of the entire compound? I realize it appears twice, but shouldn't the number of moles of the Cr ions be just one portion of the entire thing?

No. Remember, a mole is a count of something; it's not a measure of the mass of stuff that you have. So I can have more moles of chromium than I do of dichromate. Let's use a silly example and see if this helps make the concept clearer: Say I have a dozen eggs. That means I have 12 of them, right? Each egg has one yolk, so I have a dozen yolks, too. But if you ask me how many albumin molecules I have in a dozen eggs (albumin is the major protein in egg whites), each egg contains many, many molecules of albumin. Let's say for the sake of this example that every egg contains 1000 albumin molecules. In that case, if I have a dozen eggs, and each one has 1000 albumins in it, then I have 12,000 albumins, right?

Ok, so analogously, let's say I have 10 molecules of dichromate. Each molecule of dichromate has 2 chromium atoms in it. So for my ten molecules, I would have 20 chromium atoms, right? Ok, so now think bigger. Let's say I have an entire mole of dichromate. (Remember that a mole is 6.022 x 10^23 molecules.) Still, every dichromate molecule contains two chromiums. So I must therefore have 2 x 6.022 x 10^23 atoms of chromium, which is equal to two moles.

 

[QofQuimica]

When energy is transferred between a system and its surroundings it can be transferred in the form of work or heat. To understand how work is involved a derivation may help. Consider a gas which is confined to a cylinder with a movable piston. Furthermore, consider a lead shot of some weight on top of the piston so that at rest the weight is balanced by the pressure of the gas. If we remove some lead shot, the gas will push upward through some differential displacement. We can relate this to the work by:

dW = Fds
= (pA)(ds)
= p(Ads)
=pdV
​

dV represents the differential change in volume. Integrating from an initial to final state to solve the integral yields the following equation which relates pressure and volume changes to work:

W = P [V final - V initial]​

Don't worry about having to derive this equation, just know the equation. This derivation tells us that when a gas expands the work done is positive and when the gas is compressed the work done is negative. Some books actually say the opposite. However, the important thing to remember is to use one sign convention and to assign positive and negative quantities appropriately--the answer will come out this way.

So how do we use this equation in thermodynamics?​

There are many different processes but for MCAT purposes know the difference between a closed process and a open process. In a closed process, in the example above the piston would be locked to prevent the gas from expanding, no work can be done because the volume is held constant. Because the volume is held constant energy can be transferred in the form of heat. This means that the change in internal energy, from the first law of thermodynamics, is equal to the heat changes that occur. Furthermore, since no work is taking place, enthalpy changes are also equal to zero. However, in a open system, work does take place. A open system is one at constant pressure. Now a material, such as a gas, can expand or compress leading to changes in work. Furthermore, at constant pressure enthalpy changes can take place--this is because a gas can expand or compress (not only a gas but any material).

Ultimately, the difference between a open and a close system is the work done.​

Energy is transferred between the system and surroundings in both.

For the question about galvanic cells. I am not too sure what you mean by non-PV work. I think you might be confusing the work done by a expanding gas and the work done in a electric potential. Remember, galvanic cells convert chemical energy into electric energy. Since there is a flow of electrons, than work is being done from the anode to the cathod. But that is different work than work done by an expanding gas--it is work done in a electric field.​

I hope I have helped and have answered your question. I just thought that a derivation of the equation might help you to understand what it means and how it is applied. Good luck

This is a good post. I'm going to copy it and add it to the gen chem explanations thread. Thanks for writing it.