CHEMISTRY COMMUNITY

KSong_1J

Catalysts do not affect the equilibrium constant. They create a different pathway and, since the equations for K is k over k prime, there shouldn't be a change in K

KSong_1J

Since Lavelle said he's making the final short and straightforward I don't think he's going to have third order reactions on the exam

KSong_1J

An overall rate law is made up of elementary rate laws (which I believe represents what actually happens in a reaction at a molecular level).

KSong_1J

It can be used to see how activation energy and/or temperature affects the rate constant

KSong_1J

I believe Lavelle makes the final

KSong_1J

Yes, you need a solid metal on that side

KSong_1J

Yes, usually a larger k results in a faster rate of reaction

KSong_1J

I believe in a concentration cell usually oxidation occurs at the cell with the lower concentration (so this will be the anode) and reduction occurs at the cell with higher concentration (cathode).

KSong_1J

k is the rate constant for the forward reaction, while k prime is the rate constant for the reverse reaction

KSong_1J

No, you would use OH- to balance basic solutions instead of H+. However, you do use H2O for balancing in both acidic and basic solutions

KSong_1J

I think Lavelle said in lecture that we will not be using the steady state method, which is why he didn’t assign any problems where we needed to use that method!

KSong_1J

It’s given in the problem; the first sentence says “each of the following second-order reactions.”

KSong_1J

Catalysts lower the activation energy of a reaction by creating an alternate pathway. Since a higher activation energy leads to a slower rate of reaction, by lowering the activation energy, the catalyst increases the rate of reaction.

KSong_1J

You plug the initial mass into [A]initial! I suggest rearranging the equation to [A] = [A]initial*e^-kt to make it easier to solve (since you’re solving for [A]) and don’t forget to convert 5.0 hours into minutes (since the units for k is given as min^-1)! Hope this helps :)

KSong_1J

F is Faraday's constant (96,485 C·mol^-1)!

KSong_1J

The final is cumulative so it's going to cover outlines 1-6!

KSong_1J

I think quantum from 14A was my favorite and I love chemical equilibrium from 14B!

KSong_1J

Question 7.17E asks whether a catalyst accelerating the third step of a three step reaction would increase the overall rate of reaction. Why is the answer no? I think the answer is no because the rate of the overall reaction depends on the slowest step of the reaction and the third step is the fast...

KSong_1J

I think as long as you have 100 or more posts you should be fine (50 from 14A and 50 from 14B)!

KSong_1J

He just sent an email saying it will be a take home final!

KSong_1J

I think this equation is used for the system at equilibrium, which is why K (equilibrium constant) is in there!

KSong_1J

I think 5G and 5J from thermodynamics, since 5G talks about the relationship between delta G and K, and 5J talks about the van’t Hoff equation (both of which we covered in lecture on Friday)!

KSong_1J

Regarding the reaction, the van’t Hoff equation is basically just telling us that an increase in temperature favors the formation of products if it’s an endothermic reaction, and it favors the formation of reactants if it’s exothermic. Page 433 of the textbook explains this more thoroughly!

KSong_1J

Yes fusion is the same as melting; it’s the phase change from a solid to a liquid

KSong_1J

Doing all the practice problems he gives us is very helpful, as well as making sure you completely understand all the learning objectives! I also recommend watching organic chemistry tutor on YouTube if you don’t understand a concept

KSong_1J

The midterm will cover equilibrium, acids/bases, thermochemistry, and thermodynamics (only the first page of the thermodynamics learning outcomes will be covered)

KSong_1J

Definitely go over all of Lavelle’s learning outcomes on his website! It might also be helpful to create a list of all the equations that tells you under what conditions to use them/in what type of problems. Also, at the end of each section right before the problems there’s a section that says “the ...

KSong_1J

A closed system can exchange energy with its surroundings but not matter, while an isolated system can’t exchange matter or energy with its surroundings. An example of an isolated system would be a thermos (insulated, neither matter nor energy can be exchanged) and an example of a closed system coul...

KSong_1J

Yes because at higher temperatures there’s more heat and since endothermic reactions absorb heat, the reaction will occur/tend towards the products

KSong_1J

I think you use Kelvin because the gas constant R always has Kelvin in the denominator so they cancel out!

KSong_1J

Unfortunately I don’t believe Dr Lavelle offers extra credit opportunities

KSong_1J

Delta H is positive because energy must be used to break the bonds to turn ice into a liquid

KSong_1J

A negative delta H is exothermic because when energy leaves a constant pressure system as heat, the enthalpy of the system decreases by that amount.

KSong_1J

Delta U is the change in internal energy, while Delta H is the change in enthalpy. The internal energy of a system is the total energy in the system (sum of potential and kinetic energy) while enthalpy is the heat energy that is absorbed or released in a chemical reaction. They’re related in that th...

KSong_1J

Work is a form of energy; when work is done on a system, that means that energy is transferred into the system by its surroundings. The textbook definition of work states that it’s “the process of achieving motion against an opposing force” and is equal to the product of an opposing force and distan...

KSong_1J

In regards to what we covered in class,4C goes over the heating curve/the basics of enthalpy and 4D goes over hess’s law so for practice problems I’d say both are good but I’d focus more on 4D!

KSong_1J

I think sections 4C and 4D cover all the recent material we’ve done in class so far so I’d do problems from those sections :)

KSong_1J

In all the examples Lavelle has showed in class so far they were given to us so I think the values will be given

KSong_1J

Section 4D.4 covers Hess’s Law! 4C and 4D cover pretty much everything we’ve gone over in class so far

KSong_1J

The textbook states that enthalpies of fusion and vaporization are always positive and I’m pretty sure the same goes for the enthalpy of sublimation (since it’s the conversion of a solid to vapor so it’s endothermic)

KSong_1J

I’m not sure if this is the right way to do it but what I did was look at the Kc values for both and compared them. Since both are diatomic molecules, they’re more stable in the form F2/Cl2 instead of by themselves, so the one that produces more F2/Cl2 is more stable. Since the equation for Cl2 has ...

KSong_1J

For 5H.1, when you modify a reaction, the K value is also affected. So for part a, since the reaction was flipped backwards, you take the reciprocal of K (1/41) to get 0.024; for Part b, since the reaction was multiplied by 1/2, you take the square root of K to get 6.4; for Part c, since the reactio...

KSong_1J

Dr. Lavelle said that we won’t be using the equation in 14b so I don’t think you need to worry about it!

KSong_1J

Endothermic reactions absorb energy to create products while exothermic products release energy when creating products. An example that combines this idea with Le Chatelier’s principle is if you raised the temperature in a reaction; if it’s an endothermic reaction, a higher temperature will cause th...

KSong_1J

Test #1 begins on Tuesday, January 21 in our discussion sections and yes it’s chemical equilibrium and acids and bases (basically everything we’ve learned so far)!

KSong_1J

If you increase the N2, then the equilibrium will want to shift towards the right to create more ammonia. Creating more ammonia requires using more N2 and H2, thus the amount of H2 will decrease

KSong_1J

If you decrease the volume in the system, then the reaction will favor the direction that has fewer moles of gas (since, according to Le Chatelier’s principle, when stress is applied to a system in equilibrium, the equilibrium will shift to reduce the effect of the stress). By reducing the volume, y...

KSong_1J

Did you mean to write something like “if volume decreases and there’s more moles of gas on the left, then the reaction shifts to the right” and “if volume decreases and there’s more moles of gas on the right, then the reaction shifts to the left”?

KSong_1J

Q is larger than K in situations where there is more products than reactants, in which case the reverse reaction will be favored so that Q = K

KSong_1J

The ratio will be the same because (PO2)^3/(PO3)^2 is the formula for its equilibrium constant, K, which always stays the same whether more moles are added or not

KSong_1J

I think you can use aqueous solutions and gases to solve for Kc but you don’t include solids or liquids

KSong_1J

According to the textbook (Toolbox 9C.1),if it's not a negatively charged complex, you use iron instead of ferrate (ex. [FCl(OH2)5]+ is a pentaaquachlorido(II) ion

KSong_1J

Ka is equal to [H+][A-]/[HA] and, for this equation, you can rewrite it as Ka = [H+]^2/[HA] because H+ and A- have the same concentration. To find the pH, you need to take the negative log of the H+ ions. Since they give you Ka and you know the concentration of the acid, you can just plug them in an...

KSong_1J

There's a table in the textbook (Fundamentals J) of the strong acids and bases that is pretty helpful! It's Table J.1 on page F75

KSong_1J

When it comes to oxoacids with the same number off oxygens attached to the central atom, the greater the electronegativity of the central atom, the stronger the acid. This is due to the fact that a greater electronegativity means it's more electron withdrawing. Therefore, HClO4 is stronger than H3PO...

KSong_1J

The solutions are on his website! Here's the link: https://lavelle.chem.ucla.edu/wp-conten ... rm_ans.pdf

KSong_1J

I don't think this class has recorded lectures/bruincast. The only recorded lectures he has are the audio-visual focus topics on his website that cover fundamentals and some topics of the quantum world.

KSong_1J

Organic Chemistry Tutor and Professor Dave Explains on Youtube both have great videos for all those topics!

KSong_1J

The textbook covers acids and bases in Fundamentals J and Chapter 6. In Lavelle's learning outcomes it says to look at Fundamentals J, Focus 6A.1-6A.3, 6B.1, 6C.3-6C.4, 6D.3, 6E.1 and Box 6E.1

KSong_1J

^I agree, Organic Chemistry Tutor is a great resource as well as Professor Dave Explains on Youtube!

KSong_1J

I think we do because it was one of our homework problems (9C.3)

KSong_1J

A polydentate is basically a ligand that can bind to the central atom at multiple sites (multiple atoms). An example of a monodentate would be H20 (the oxygen makes one bond with the central atom), a bidentate would be oxalate (the two oxygens that are not double bonded to the carbons can each form ...

KSong_1J

-ate is used if the complex is an anion (is overall negatively charged), so that's when colbatlate would be used. If the complex has an overall positive charge, you would use cobalt.

KSong_1J

You use those prefixes when a ligand is a polydentate or if it already has a prefix

KSong_1J

I believe we still have to cover the acids and bases section before the final

KSong_1J

You use brackets for anything within the coordination sphere (basically the complex)

KSong_1J

I believe it’s everything we learned after the midterm up to/including sigma and pi bonds

KSong_1J

More lone pairs results in more electron-electron repulsion which pushes the atoms away. This results in smaller bond angles because the atoms move closer together.

KSong_1J

Yes you should memorize them for the upcoming test!

KSong_1J

Lone pairs affect molecular geometry because the electron-electron repulsion causes the atoms to move further away

KSong_1J

I don’t think hybridization is going to be covered on Test 2

KSong_1J

You put a negative charge on the atom that’s more likely to gain the electron(s) in a bond!

KSong_1J

I think it’s just everything we learn after the midterm!

KSong_1J

I think I remember him saying we get our midterms back on Wednesday!

KSong_1J

You can find sigma bonds in single, double, and triple bonds while pi bonds are only found in double bonds (where there’s one sigma and one pi bond) and triple bonds (one sigma and two pi bonds).

KSong_1J

N, O, and F form hydrogen bonds!

KSong_1J

They’re held together by electrostatic attraction (the attraction between atoms of opposite charges)

KSong_1J

Definitely do the homework problems and also search up more problems online if you need more practice! To understand the topic better conceptually I recommend reading the textbook and watching videos about the topic online (the organic chemistry tutor on YouTube has really good ones)

KSong_1J

You want to remove electrons from higher energy levels before moving onto lower energy levels, so you’ll want to remove them from the highest orbital. They’re also usually easier to remove because they’re further away from the nucleus

KSong_1J

The magnetic quantum number describes the number of orbitals within a subshell and the orientation of the orbital; s will have a magnetic quantum number of 0, p can have one between -1 and 1, d can have between -2 and 2, and so on

KSong_1J

I don’t think we’ll specifically asked to identify if the element is a metal, nonmetal, etc. but I think it’s still good information to know because you need it for many things (like naming, bonding, etc)

KSong_1J

Anish Patel 4B wrote:Do you know if this also includes the material from the first test since it was high school stuff?

I think material from the first test is also included in the midterm!

KSong_1J

I know it's still weeks away but I was wondering, does anyone know about the format of the midterm? Is it comparable to Test 1 but just longer in length? What types of questions will be on it and which units does it cover? Where is the midterm located? Will that be announced closer to the midterm d...

KSong_1J

Yes, covalent bonds are only between two nonmetals! If you have a metal-nonmetal bond it's an ionic bond and a metal-metal is a metallic bond.

KSong_1J

I think you have to use the dots to represent the two electrons on the oxygen

KSong_1J

For studying I definitely recommend reading the textbook chapters (which helps a lot with conceptual understanding of topics), doing all the homework problems, and just searching up youtube videos of topics you don't fully understand!

KSong_1J

I think we can still do questions under quantum world since it falls under the category of “recently covered material”

KSong_1J

I think it can be from both. In the syllabus it says to hand in homework from recently covered material so I believe either should be fine (it also could depend on your discussion section; for example if you had yours on Tuesday, we haven’t really gone over Chemical bonds yet so I don’t think they’d...

KSong_1J

Since s-block metals have lower ionization energies, they’re more likely to give away their electrons which makes them more reactive

KSong_1J

n is related to the size and energy of the orbital; as n (aka the energy level) increases, the mean distance of an electron from the nucleus increases, which means the radius of the atom gets larger.

KSong_1J

In general, you would follow the order of s, p, d, f but I think there are some special cases in which you don’t follow that order (for example, copper)

KSong_1J

I don’t think he posts his slides, but the audio-visual focus topics do go over what he talks about in class (though they only seem to be available for fundamentals and quantum world).

KSong_1J

Photons are basically particles of energy that make up light

KSong_1J

To calculate the different wavelengths, use de Broglie’s equation (λ = h/mv) and you just search up the mass of the proton and neutron because they’re set values and plug those in for m, plug in 2.75 x 10^5m/s for v, and plug in Planck’s constant for h: Proton: (6.626 x 10^-34Js)/(1.673 x 10^-27kg)(...

KSong_1J

Light is made up of photons and the photon’s energy is dependent on its wavelength; longer wavelengths of light have less energy than shorter wavelengths, which is why the shorter wavelengths are able to eject the electrons (since they have more energy and you need to reach a certain threshold to ej...

KSong_1J

No, the surface must be metal to exhibit the photoelectric effect because you have to be able to completely remove the electron in order to achieve the effect. Since metals are able to donate/spare electrons while non-metals can’t, only metals will work.

KSong_1J

Sig figs are basically like a way to round. Khan Academy is a great resource to learn more about the basics of sig figs :)

KSong_1J

I think we have to use scientific calculators for tests!

KSong_1J

For sig fig practice I like using Khan Academy or if you just search up “sig fig practice,” some good websites for practice will pop up (like sciencegeek.net)!

KSong_1J

I know this was kind of addressed in a question someone asked earlier but I just wanted to ask if we would be marked down if we left the stoichiometric coefficient as a fraction? I know different professors prefer different things so I just wanted to clarify to see if it would necessarily be seen as...

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