CHEMISTRY COMMUNITY

Matt F

The key to understanding the graphs is to look at the axis labels, especially if the plot is linear. As mentioned above, graphs of [A], ln[A], and 1/[A] vs t correspond to zero order, first order, and second order reactions when the plots are linear

Matt F

A is considered the frequency factor or "pre-exponential" factor and includes the number of collisions with correct orientation. The Arrhenious equation is typically used with problems involving activation energy

Matt F

Would it be correct to say that the activated complex is part of the transition state (or vise versa)? Or are they completely separate from each other

Matt F

^To add onto this, adsorption occurs with a heterogeneous catalyst where the catalyst is in a different phase as the reactants (usually solids), causing them to sit on the surface

Matt F

We will never need to use steady-state as he never went over examples and pretty much said explicitly we won't need to. Be sure to be familiar with pre-equilibrium though as we did go over examples for it

Matt F

The integrated rate law is the differential rate law written as a function of time. In other words, rather than rate being a function of concentration, rate is a function of time. This can be seen with the graphs of zero order, first order, and second order reactions, in which [A], ln[A], and 1/[A] ...

Matt F

Both versions are listed on the equation sheet. However, understanding how to derive them (at least understanding why/what is done) is definitely beneficial and it provides a better understanding of graphs given

Matt F

Derivations can also be helpful to know if you need to manipulate formulas on a test only given certain information. I think knowing the relationship between [A], ln[A], and 1/[A] and time (t) is also useful (specifically the graphs and how they relate to slope)

Matt F

Usually written as mol.L^-1.s^-1, which makes sense given that the rate can be represented by d[A]/dt, where [A] is in mol per liter over the time (dt)

Matt F

So A will always be given for a reaction?

Matt F

I don't believe H₂O shows up anywhere in the cell diagram in that form, as only Hg can show up as a liquid. However, if in redox half reactions it is broken up into H⁺ and OH^- ions then those are obviously included in their aqueous forms

Matt F

Does anyone know when you would flip the sign in a half relation? I know that if you flip the equation the sign flips, but then do you not do this when calculating E(cell)? like if they give you the E value in the problem and the equation has to be flipped do you flip the sign? The Ecell values giv...

Matt F

Balancing the redox half reactions will give you n for the nernst equation, since you will balance the charges to determine how many electrons to add to both sides. There are no moles in the equation aside from the concentrations (mol.L^-1) or in the Gas constant R

Matt F

Galvanic and Voltaic cells are both sensitive to changes in [R] and [P], so they can be made with the same components at different concentrations. Concentration cells consist of the same species on both sides but differ in concentrations, like the example from lecture with Ag at 1M and 0.1M

Matt F

Convention is for the anode to be written on the left side and the cathode written on the right side, hence why Enaught(cell)=E(right)-E(left). As mentioned above, you can always check to see what reaction is happening where, since oxidation occurs at the anode and reduction occurs at the cathode

Matt F

The way I thought of it was similar to potential and kinetic energy. When there is no movement/work done, potential energy is at a max and kinetic energy is 0. However, once something happens, potential energy is converted into kinetic energy, therefore being used and not staying at a maximum. Simil...

Matt F

John Liang 2I wrote:1) identify the half rxns and look up E for each:

Do you think there would be any cases where E is not given and we would have to calculate it given one of the formulas we have?

Matt F

A little bit of cheat too is that MnO₄^- lost its Oxygen, while H₂SO₃ gained an Oxygen. Therefore, the MnO₄^- is reduced while the H₂SO₃ is oxidized. I'm not sure if this is true in every reaction, but it makes sense

Matt F

If you're using OH- to balance an equation, I'm going to assume the reaction is occurring under basic conditions. First, balance all elements except for hydrogen and oxygen. Then, add H2O to balance the oxygen on the opposing side. Then, balance the hydrogen by using H+, and then the charges by usi...

Matt F

We are still dealing with concentrations like Al and Al3+ or Zn and Zn2+. Therefore, the same principles we learned in the equilibrium unit apply here. If there's too much concentration of one of these ions, the equilibrium will shift to satisfy LeChatelier's. This would then affect Q in the Nernst...

Matt F

So in other words, pressure affects Gibbs Free Energy because it affects the value of Q?

Matt F

Earlier in the quarter we talked about how among changes in temperature, pressure, and concentration, temperature is the only one that can affect the K value for a reaction. Depending on if a reaction is endothermic or exothermic (value of delta H), the K value can be permanently increased or decrea...

Matt F

I think it could be reasonable to have to solve for delta H naught or even one of the temperatures depending on the wording of the problem. The main use for the equation though is calculating K at different temperature values given a value for delta H naught, so ideally you would just be focusing on...

Matt F

^You could say that. I think combining half reactions is less about cancelling out reactants/products and more about fitting them on both sides of the reaction, but it's the same concept

Matt F

In addition, you can tell by looking at the reaction and observing the oxidation numbers of reactants and products. If the oxidation number increases, the reactant was oxidated, and if the oxidation number decreases, the reactant was reduced

Matt F

There can be a delta (or change in) for enthalpy or entropy since both are state functions. The one equation where delta isn't used is the Boltzmann equation, S=k B lnW, since it is calculating the entropy of the system for that many energy states as opposed to the change in entropy due to a change ...

Matt F

Entropy increases as a system gets bigger, meaning that the opposite is also true (a smaller volume means entropy decreases). When we are talking about pressure and temperature we are usually talking about a closed system with a piston, so if a piston compresses and increases pressure, volume is dec...

Matt F

The shape affects entropy because it affects the molar heat capacity, which is part of the equation for a change in temperature (delta S = C*ln(T2/T1)). Depending on the shape the molar heat capacity can be 3/2*R, 5/2*R, or 7/2*R

Matt F

As mentioned above, the Boltzmann Equation is helpful when we are given or can calculate degeneracy (W). The Boltzmann equation also just solves for S and not delta S, whereas the other equations solve for delta S or a change in S

Matt F

Extensive properties like entropy depend on the amount of material present, which is why entropy increases as the size of the system increases. Intensive properties like density do not depend on the amount of material present, which is why something like density remains constant regardless of the si...

Matt F

The monday of week 6 (week of the midterm) Dr. Lavelle said he would be reviewing past midterm questions to review, so that means we have 3 lectures this week to finish the material. I'm fairly certain we will get through the rest of thermodynamics by then

Matt F

As mentioned above, delta U will be equal to only q when work (w) is equal to zero. This can be seen in a closed system with constant volume, as work is typically done due to compression or expansion (changes in volume).

Matt F

The First Law of Thermodynamics says that for a closed system, internal energy (delta U) changes as a function of heat (q) and work (w). This is laid out in the equation delta U = q + w. Note however that this does not fully describe a system, since we also need entropy to know the probability of be...

Matt F

q of the system + q of the surroundings always equals 0. Since an isolated system has no exchanges with surroundings, q of the surroundings is equal to 0, meaning that q of the system must also be 0.

Matt F

In other words, the process is reversible since the system is at equilibrium, correct? So irreversible would mean the change is large enough to not maintain equilibrium?

Matt F

If you remember in lecture Dr. Lavelle gave the example of the hikers climbing up a mountain. They both took different paths but still ended up on the same point on the mountain. Distance traveled, or work in this case, would not be a state property since it differed depending on the path taken. Alt...

Matt F

Dr. Lavelle gave an example in class with the heating curve for water. The graph showed heat absorbed on the x-axis and temperature on the y-axis. As heat constantly increased, the temperature remained constant and only increased once certain thresholds were reached (melting or boiling points). The ...

Matt F

The energy change in the reverse reaction will always be the opposite of the forward reaction since the heat curve is flipped. So the reverse reaction for something endothermic such as vaporization would be exothermic

Matt F

Something also of note is that work is not a state property, meaning that it depends the "path" taken to obtain it. It cannot be added or subtracted like other properties such as enthalpy or pressure.

Matt F

The process of water becoming a gas (vaporization) takes much longer than the process of ice becoming water (melting). Since this process is longer, more heat is supplied throughout and the enthalpy is greater, meaning that the steam will release much more energy/heat as opposed to the liquid

Matt F

The 5% rule can be applied after making an approximation for x. You calculate x/[initial concentration given] and if the value is less than 5% (0.05) then the approximation is valid. However, if the value is greater than 5% then you must reject the approximation and recalculate x

Matt F

Changing the concentration or pressure can cause the equilibrium to "favor" the products or reactants, but in both cases the equilibrium constant K will stay the same. Only when temperature is changed is K changed because of a permanent change in the system's dynamics (depending on if it's...

Matt F

If we are required to convert between pressure and concentration or vice versa, you should be safe using the Ideal Gas Law. Ideal gases were not heavily discussed in class so it should be fairly safe to use the equation when necessary to convert

Matt F

As mentioned above he usually draws them so that we can better visualize what is happening during the reaction. I would be surprised if we had to draw one for the test, but I could see one being given for us to use as part of a question.

Matt F

Most solids and liquids are pure substances, meaning their molar concentration does not change in the reaction. Therefore, it will not affect the equilibrium concentrations of other reactants/products

Matt F

^So are they the same in the case of ideal gases then? Or whenever P=n/V is true?

Matt F

The tests will be given starting Tuesday of week 3, so I imagine whatever we cover in these first two weeks will be on there. I would pay attention to how far he gets in lecture and whatever your TA reviews in discussion

Matt F

Is it appropriate to consider Le Chatelier's Principle as a ratio? I.e. determining the change of one value would aid in determining the other values and thus, the equilibrium? I believe so. Since the equilibrium constant K does not change regardless of changes in concentration or pressure, the cha...

Matt F

The reaction quotient Q is similar to K since they both involve products over reactants ([P]/[R]), with the difference being that Q is taken at any point in the reaction while K is that ratio taken at equilibrium. Therefore, if you observe that Q<K, you know that there is a greater concentration of ...

Matt F

As others mentioned, brackets denote concentration (in mol.L^-1), so you wouldn't use them when calculating partial pressure for Kp. However, just because you are given all gases does not necessarily mean you will be calculating Kp, so pay attention to what the question asks and what values are given

Matt F

When writing the compound itself I don't believe order matters, so long as everything inside the brackets in inside and everything that should be outside is outside. Typically the transition metal cation is written first in the formula, but I'm not sure if that is absolutely required or just a forma...

Matt F

Amphiprotic means that the molecule can both receive and donate protons. Amphoteric means that the molecule can act as both an acid and a base. All amphoteric molecules are amphiprotic, but not all amphiprotic molecules are amphotertic

Matt F

^Correct. Strength for oxoacids is based on the stability of the resulting anion, which is influenced by electronegativity (withdraws electron density) and resonance. Typically I would just look at which acid contains the more electronegative atom

Matt F

The specifics of titration are more important for 14B and 14BL. For the final I would understand the stoichiometric point and its significance

Matt F

From my understanding, the scores themselves are not curved. However, based on the total grades at the end of the quarter, course grades will be adjusted accordingly.

Matt F

I would be familiar with those elements involved, but I don't know if you need to fully memorize them. To determine if an element is amphoteric, you can look chemical reactions involving them to see if they are capable of acting as both an acid and a base (able to donate or accept a proton when nece...

Matt F

Looking at chemical reactions involving the compound and seeing what the resulting products are will probably be the most useful. I would think that a reaction would be given or they would tell you if it is acidic, basic, or amphoteric to begin with. I would also just be familiar with typical acidic...

Matt F

Is there notation to show the direction a reaction is favoring, or do you still just write the double pointing arrow?

Matt F

^I agree. Memorizing isn't always the best method, but I think in this case you just need to know the values as it will definitely save you some time

Matt F

Typically you want to find it by process of elimination. See what the overall charge of the coordination compound is, and then determine what components inside are that make it up. From there, the roman numeral will take on whatever value it needs to be so that the components equal the total charge....

Matt F

Knowing the biological function of both is important to understanding their purpose, but since this is a chem class I think knowing their structures and the roles of coordination compounds is more significant

Matt F

I have in my notes that they are equal to 90 and 120, but I think using the lone pair as reasoning you wouldn't be marked wrong for writing less than 90 and less than 120

Matt F

I think we only have to worry about molecular geometry for this class, just be aware of how many regions of electron density there are in general

Matt F

I would also be familiar with what molecular shapes correspond to what hybridized orbitals. For example, shapes such as Tetrahedral or Trigonal Pyramidal have 4 regions of electron density, therefore having the sp^2 hybridized orbitals

Matt F

I would double check with your TA but doing hybridization and sigma/pi bonds problems should be fine since they were recently covered. There also aren't many problems for coordination compounds so I would think it's okay

Matt F

Classes are only cancelled on Thursday and Friday, so unless Lavelle himself gives us the day off, we should have lecture as usual

Matt F

Great, thank you

Matt F

From what I've been told you will usually be able to tell which one to compare. The molecules given will usually be similar in either electronegativity or size, so use that as a baseline and look at the one that's different

Matt F

Is this the same for melting points? So stronger IMF's also lead to higher melting points?

Matt F

In class he mentioned that we can just draw a normal Lewis Structure as long as we name the shape I think. Do we have to worry about where we arrange the atoms around the central atom (example: drawing a tetrahedral shape vs square planar) or can we just draw it normally and name the shape after?

Matt F

Will we go into more detail in lecture about this concept? I also remember it being mentioned but I don't have much written down either.

Matt F

From my understanding, you really only have to worry about the octet rule with 2nd period elements (mainly N, O, F). As others mentioned, once you reach 3rd period elements, d-orbitals can accommodate extra electrons.

Matt F

Partial points can never hurt, I would only not write it out if I'm in a crunch for time

Matt F

Since we don't have lecture tomorrow, I would assume it's fine to do homework on last section's problems. We haven't covered enough of the new section to do many of the newer problems.

Matt F

Test 2 is in week 8, right? I'm sure some of the prior material will be on it (at least the stuff on chemical bonds) since otherwise it'll only be covering a week or so of material.

Matt F

If I remember correctly Dr. Lavelle said that the final slide he went over in lecture on Friday (the one describing interactions that give rise to attractive forces) would not be on the midterm. However, everything else up to that point would be fair game. I don't think you need to know specific val...

Matt F

I'm not sure if this is the same as what you're describing, but the shortcut I was given was just to count the electrons around the drawn atom and see if they match the number of valence electrons that atom should have. You erase the middle of any bonds, showing that each atom is giving one electron...

Matt F

So the brackets are drawn as opposed to the atoms being bonded since they're ions, correct? So we'd be marked wrong for including a bond between them?

Matt F

Are sigma and pi bonds shown in Lewis Structures or only in describing bond character?

Matt F

Can sigma and pi bonds be found in single or triple bonds too or only double bonds?

Matt F

Anna Chen 4L wrote:The resonance structure with the lowest formal change is the most stable.

I thought the structure with its formal charge = 0 or closest to 0 was the most stable?

Matt F

In class I remember Phosphorus, Sulfur, and Chlorine being given as examples. Are other elements beyond these 3 capable of having an expanded octet. I have written in my notes that atoms in Period 3 or higher have d-orbitals in valence shells that accommodate additional electrons, so does that apply...

Matt F

It's still a Lewis Structure, just seems different because Phosphorus is capable of having an expanded octet

Matt F

I know we haven't covered it yet, but will we eventually need to know how to calculate electronegativity for a specific element? Or will we just need to know the trend for the table

Matt F

Technically we've moved on to the Chemical Bonds section but I imagine it's fine, similar to how he allowed us to turn in homework from the first section because we had a test coming up. You can always email your TA just to be sure.

Matt F

I think Dr. Lavelle explained in class that 4s is written after 3d since 3d has lower energy than 4s. Typically the orbitals are listed in ascending order of energy

Matt F

Basically the question is asking how many possible values of m l there are for the given set of quantum numbers. Remember, the value of m l ranges from l to -l. For b) and d), the answer is 1 since they give you m l already. For a), the answer is 3 since m l can be 1, 0, or -1 given l=1. For c), the...

Matt F

On the Learning Outcomes outline for Chemical Bonds (the next topic) there is no mention of watching modules for that section. He could update it later, but those are likely set in stone as outlines for the rest of the quarter. As mentioned above, the modules were likely just to help ease people bac...

Matt F

The mass of protons, neutron, and electrons will always be given on a formula/constants sheet. There's a link on the class website called "Constants and Equations" that you can use for homework

Matt F

m could represent the mass of an object, like a baseball or car as others mentioned, or a particle such as a proton, neutron, or electron. The only thing it can't represent is a photon since photons have no mass. If the problem involves a photon/light, use c=wavelength*frequency.

Matt F

As others mentioned you'll almost always see wavelength measured in meters (m) since it's distance. However, it's usually an extremely small number, so I'd recommend knowing the different prefixes (nano, pico, etc) in case it's given like that.

Matt F

The modules were very beneficial for me when it came to studying for the first test on High School topics. If you don't have time to watch them before the lectures, I would at least recommend taking the pre/post module test when reviewing as another way to practice.

Matt F

Aside from the lectures, what's helped me understand the material most has been the video modules posted and doing extra homework problems. When I do the problems, however, I try not to use my notes or the textbook as help to see if I've really grasped what I need to do. Reviewing notes is definitel...

Matt F

Nothing has been mentioned about extra credit, so I doubt turning in extra HW would directly help your grade. However, as others mentioned, it's definitely still helpful to do more than the required 5, especially when studying for tests since a lot of the problems are similar to the ones assigned.

Matt F

I typically use pencil for homework since I tend to erase a lot and it helps me be neater without having to cross things out. If you want to work in pen you can always work out a problem on scratch paper and then transfer it onto the homework so it's neater/more detailed. Up to you, I think it's fai...

Matt F

I generally round out to 2 or 3 decimal places, depending on the numbers given in the problem like those above me mentioned. You should be fine as long as you aren't rounding egregiously (Ex: rounding 16.53 to 17 for atomic weight). It also depends on context though. I typically wouldn't round like ...

Matt F

From my understanding, aqueous refers to an element/molecule dissolved in a solvent, typically water. The molecule is a solid, but it is dissolved in the solvent so it's referred to as aqueous. Liquid, on the other hand, is a "pure" form of the substance, meaning that it isn't dissolved in...

Matt F

^To add on to this, I would definitely convert after since it could make your calculations a little wonky if you converting during. Only convert during the calculations if you need units to cancel out (they usually should regardless), so I would suggest waiting until the end.

Matt F

Since we have to use a pen for the exam, are we allowed to use pencil and/or a piece of scratch paper to work things out? I feel like I would have to cross a lot of things out working in pen.

Matt F

^Thank you for this explanation, I found it very helpful. Will there be problems like this in the future where the molecules in the reaction are given by name as opposed to as the equation? (Ex: if they had given us Al(s) + Cl₂(g) ---> AlCl₃(s))

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