CHEMISTRY COMMUNITY

aTirumalai-1I

Yes, both K and k/k' work, but k/k' would be the more accurate answer.

aTirumalai-1I

Correct me if I'm wrong, but as far as I know, you are correct. The cell potential/voltage can never be larger than the standard cell potential because the standard cell potential describes the maximum potential difference.

aTirumalai-1I

Also remember that Dr. Lavelle outlined 2 methods to write the rate law of a mechanism: steady-state approximations and pre-equilibrium conditions. He told us that we would be using the pre-equilibrium method in 14B, so I think approaching this problem with that method will make it a bit easier.

aTirumalai-1I

Are there any homework problems to go with this topic? I couldn't find a lecture outline for organic chemistry on the class webpage.

aTirumalai-1I

Yes, NO(g) was the catalyst. It was used up in the first step of the mechanism, but then re-produced in the second step of the mechanism.

aTirumalai-1I

Yes, Dr. Lavelle probably just wants us to know that the concepts between unimolecular, bimolecular, and termolecular reactions are similar. I also don't think we'll need to know this for test 3, because it falls under "Reaction Mechanisms."

aTirumalai-1I

I also think you can do problems 83, 99, 103, and 105 for more practice!

aTirumalai-1I

For these kinds of problems, use the chart for standard reduction potentials. The larger (more positive) the standard potential, the easier it is to be reduced. The species that is reduced is also the oxidizing agent. Hope this helps!

aTirumalai-1I

I was wondering about this too, but the textbook doesn't mention anything about pH as an aspect of standard conditions for electrochemical cells.

aTirumalai-1I

Deriving the integrated rate law from the differential rate law is more useful to us because we can plot ln[A] (and the natural logs of any other reactants present) against time and figure out the overall order of a reaction. As we discussed today, if the graph is linear, you know you are dealing wi...

aTirumalai-1I

Keep in mind that we calculate unique rates because the average rates can be different depending on the time interval you are calculating for. When you calculate the unique rate, the coefficient is taken into account, so you have a more accurate answer.

aTirumalai-1I

In terms of concepts, I'm sure we will have to know how the Gibbs free energy relates to the cell potential. Section 14.4 is especially helpful for this part.

aTirumalai-1I

I think O₂ has an oxidation number of 0 because it is in its elemental form. This is the same case for other elemental forms such as H₂ or Fe.

aTirumalai-1I

Dr. Lavelle usually gives us the molecular formulas for complicated molecules, but don't depend on it for every reactant/product. We may have to find the compound by its name (covered in 14A).

aTirumalai-1I

You can calculate the deltaG with the Gibbs Free Energy of Formations in Appendix 2 only for questions involving standard conditions. In other words, for 11.83, you can only use Appendix 2 for 25 degrees.

aTirumalai-1I

This is what I got when I tried the question: 1. True only when heat isn't transferred to/from the system (q=0) 2. Always true (because volume isn't changing) 3. Always false (work=0 because delta V=0) 4. Always true (because work=0) 5. True only when heat isn't transferred to/from the system (q=0);...

aTirumalai-1I

Conceptually, that would make sense, but when you're solving the problem, I don't think it really makes a difference (because you can technically just multiply both sides of your equation by -1 and get the other version).

aTirumalai-1I

Dr. Lavelle confirmed in lecture today that the midterm would cover only chapters 8, 9, and 11. Electrochemistry will not be on the midterm.

aTirumalai-1I

Salts that are poorly soluble are not effective in a salt bridge because the ions would react to form a precipitate, instead of balancing the charges.

aTirumalai-1I

Without a salt bridge to manage the unequal charges, the electrons flowing to the cathode will eventually stop because the buildup will cause electron-electron repulsion. Likewise, the anode would have such a high positive charge that it will want to hold on to the negative electrons.

aTirumalai-1I

To add some foundation, entropy is an extensive property (it depends on the number of particles we are calculating for), so when the number of particles increases, the degeneracy increases as well (W = 2^n). Thus, based on the entropy equation, S = k_b (ln W), as W increases, so does S.

aTirumalai-1I

Hi,

Can someone explain the relationship between ΔG⁰ AND K? In other words, how do we know that we can relate the equilibrium constant directly to changes in enthalpy and entropy?

Thanks in advance!

aTirumalai-1I

(ΔG) = (ΔH⁰) - (T)ΔS is the equation you use to calculate Gibbs free energy under standard conditions. You use this equation when the free energy change associated with the formation of the substance from the elements in their most stable forms exist under the "standard conditions" of 1 at...

aTirumalai-1I

TarynLane2J wrote:I don't think so because I don't remember going over it in class but I could be wrong

I know for a fact that we didn't go over the formula in Dr. Lavelle's 11am lecture, but I'm not sure if it was brought up in the 1pm lecture.

aTirumalai-1I

For example 8.8 in the textbook, can someone explain why the answer has 4 significant figures? Since the temperature we use (298 K) has the least number of significant figures (3), shouldn't the answer have 3 sig figs as well?

aTirumalai-1I

Hi, I was reading the textbook, and on page 287, it says "The steeper the slope of a heating curve, the lower is the heat capacity." Can someone explain why this relationship is true?

Thanks in advance!

aTirumalai-1I

Take a look at this response; it really helped me understand the conceptual reasoning behind it:

viewtopic.php?t=2341

aTirumalai-1I

Hi, I know Dr. Lavelle mentioned that students tend to understand the chapter 8 material if they flip the order of the chapter and read the thermochemistry sections of the text before the thermodynamics sections. Which specific section of chapter 8 starts the thermochemistry part?

Thanks so much!

aTirumalai-1I

Here's a haiku on Aluminum:

Spent kindergarten
endlessly writing your name.
One i or two i's?

aTirumalai-1I

Can someone explain what Dr. Lavelle was trying to clarify with the tire example and the compression of the bike pump he was talking about in class today? I don't think I completely followed it.

Thanks in advance!

aTirumalai-1I

I'm pretty sure the problem will give us whatever values we need, since Dr. Lavelle usually doesn't expect us to memorize specific values from tables.

aTirumalai-1I

The way my discussion worked in chem 14A was that the TA would put up specific homework problems on the board for us to work on in small groups. So I agree that the PDF is good enough, since you'll just be using it to refer to problems in the textbook!

aTirumalai-1I

I don't see aqueous equilibria on the Chem 14B syllabus, so I don't think we'll go over it this quarter.

aTirumalai-1I

In lecture one, I believe Dr. Lavelle said that for Week 1, we turn in homework during Friday's lecture.

aTirumalai-1I

The value of K for the forward reaction is 0.23, so the K value for the reverse reaction would be the reciprocal of that (1/0.23), which equals 4.3.

aTirumalai-1I

Could someone please summarize all the exceptions we need to be aware of for chemical bonds? I know of: 1. S, P, Xe, and the halogens can have an expanded octet 2. Be and B can have less than 8 3. Radicals are unpaired electrons Is there anything I am missing? Another thing to remember is that Hydr...

aTirumalai-1I

I believe Dr. Lavelle said that all values we need will be given to us, and do not have to be memorized.

aTirumalai-1I

The syllabus states that we only have to read 11.1 to 11.7. I don't think we will be tested on this topic because diamagnetism and paramagnetism falls under 11.8.

aTirumalai-1I

The answer is D, because at equilibrium, the reactants are still forming products, and the products are still forming reactants. In other words, at equilibrium, the ratio [P]/[R] doesn't change, but the individual concentrations are still changing at equal rates. Hope that makes sense.

aTirumalai-1I

You can use Q for any reversible reaction that has not reached equilibrium yet. This value of Q can be compared to the value of K to decide whether the forward or the reverse reaction is favored. If Q>K, the reverse reaction is favored, and if Q>K, the forward reaction is favored. Finally if Q=K, th...

aTirumalai-1I

I think you only solve for Kp if the compounds in the given reaction are all gases, or if you are given the partial pressures (instead of concentrations) of every compound. Also, solids and liquids are the only states that are omitted from K calculations (Kc and Kp); thus the aqueous reactant/produc...

aTirumalai-1I

Nope, a molecule with 4 regions of electron density, 2 bonded and 2 lone pairs, would be bent.

aTirumalai-1I

That statement would only be true given that the frequency is high enough to eject the electron from the metal surface in the first place. For example, it could be possible that the increase in frequency (and thus energy) of the incident light is still less than the work function of the metal.

aTirumalai-1I

CF4 has the shortest bond because F is the most electronegative among F, Cl, and Br. This basically means that F pulls on its surrounding electrons with more strength than Cl or Br can, making it so that in CF4, the atoms are all very close together (aka short bond lengths).

aTirumalai-1I

I think Chlorine gets the unpaired electron because Oxygen is more electronegative (meaning Oxygen wants its octet more than Chlorine does). Thus you would first complete the octet for Oxygen before distributing the electrons around Chlorine.

aTirumalai-1I

Both are technically correct, but the reason we usually write it the second way (neon = [He]2s2 2p6) is so that we can see how many valence electrons are in the outermost shell.

aTirumalai-1I

Nope it does not matter, although you have to be consistent. For example, if you start filling spin-ups in the 2px orbital, you have to continue filling spin-ups in the 2py and 2pz orbitals until you need to pair all the spins in the p-orbitals to complete the configuration.

aTirumalai-1I

[Xe] 4f^14 5s^2 5p^6 5d^4 6s^2 wouldn't make sense in this case because the "[Xe]" already accounts for the electrons in the 5s and 5p orbitals. Thus the electron configuration would simply be [Xe] 4f^14 5d^4 6s^2.

aTirumalai-1I

To clarify, the textbook says that when the 3d subshell is empty, its energy is greater than than that of the 4s subshell; thus for elements like Potassium (K), the s orbital is filled first. However, when the 3d orbital is occupied by one or more electrons, its energy becomes lower than the energy ...

aTirumalai-1I

I know we don't have to know Wien's law for Test 2, but I had a question regarding significant digits for the textbook's example problem 1.3 displaying Wien's law: "The maximum intensity of solar radiation occurs at 490. nm. What is the temperature of the surface of the Sun?" The only valu...

aTirumalai-1I

When I asked my TA, she said that the problems from Chapter One can still be turned in for credit this week. :)

aTirumalai-1I

I'm a bit confused at the basics. Let's say an electron has been excited to the n=3 state from its original n=1 state. When it's returning to its original energy level, does it stop at n = 2 to give off radiation and then go to n = 1 to give off radiation? Or does the electron go directly from n = 3...

aTirumalai-1I

I think I understand the concept of the photoelectric effect, but I still don't completely understand why it can occur only in a vacuum. Can someone clarify that for me? Thanks!

aTirumalai-1I

From my understanding, you have to work backwards from the third reaction. It is given that you have 2.50 t of NaBr. Using stoichiometric coefficients, you can find how much Fe3Br8 was needed to produce the 2.50 tons of NaBr. Once you find out how much Fe3Br8 you have, you can turn your attention to...

aTirumalai-1I

If no values are specified in the problem, I would use the molar masses and round to 4 sig figs, since the periodic table allows for that precision. AKA whenever possible, it's best to be more precise.

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