CHEMISTRY COMMUNITY

MaggieHan1L

k=Ae^{-Ea/RT} is the equation and doesn't have anything that is affected by order.

MaggieHan1L

It is present at the beginning and reappears at the end as opposed to an intermediate which forms and disappears during the reaction

MaggieHan1L

Also when you are looking at activation energy

MaggieHan1L

I think it's helpful in the future when you are working in the lab and you need to choose a catalyst.

MaggieHan1L

Its not commonly seen though because the chances of 3 molecules hitting simultaneously is super low which is why we don't cover third order

MaggieHan1L

Second order half life is 1/k[A]0 and it is used.

MaggieHan1L

Activation energy is the amount of energy needed before a reaction could proceed. There is a numerical value and it's important because a larger activation energy tends to have a slower reaction. Also this is important when it comes to enzymes

MaggieHan1L

I think he could give us a question with a plot and ask us what order the reaction is.

MaggieHan1L

Dr. Lavelle said he would show us an example Monday but yeah since there is a buildup of B the B reverts back to A so thus we can treat it as if the step is in equilibrium.

MaggieHan1L

I found this on a website and it feels like an apt summary

Zero order: No effect
First order: Rate doubles when reactant doubled
Second order: Rate quadruples when reactant doubled
Third order: Rate increases eightfold when reactant doubled.

MaggieHan1L

I think they might ask you to try to find the order of reactions if they give you experimental data.

MaggieHan1L

If you're talking about kinetics n is the order of the reactant.

MaggieHan1L

you're trying to get a rate of M/s so if you set up the rate calculations you can see that for first order you get k times M. Thus k would need units of s^-1 to obtain M/s. So you can calculate the units of k like this for each order. Its easier to be able to calculate then remembering.

MaggieHan1L

It's all about how the rates are calculated for products its d[P]/dt while for reactants its -d[R]/dt. Its just easier to think about rates in terms of being positive.

MaggieHan1L

I think that this was when Professor Lavelle was talking about the switch being opened

MaggieHan1L

It lets ions flow freely and thus the reaction won't be influenced by all of the free ions.

MaggieHan1L

Remember OIL RIG or LEO says GER. Look at oxidation states of elements. When oxidation states get more positive they are losing electrons and thus being oxidized.

MaggieHan1L

If you look at it, it doesn't make sense for it to go from Cl2 to Cl2. Even if I didn't know its a typo, I would change the products because it feels wrong to have the reactant also be a product. Or at the very least ask about it if its on an exam.

MaggieHan1L

The n stands for the number of moles of electrons. Look at the oxidation states of the reactions and figure out how many moles of electrons are exchanged.

MaggieHan1L

K2 is K at one temperature while K2 is K at another temperature. K means equilibrium constant. Delta H is enthalpy and delta s is entropy. R is the gas constant and T is temperature.

MaggieHan1L

I think it’s harder to relate K to temperature since the equation compares the two Ks as a ratio of ln(K2/k1)

MaggieHan1L

The Van’t Hoff equation is to compare Ks at different temperatures. Thus if you know K at one temperature you can find K at another temperature as long as you know delta H, delta S, and R

MaggieHan1L

I think that there are certain caveats, but the assumption is that the two temperatures wouldn’t be different enough to change the delta S

MaggieHan1L

I think the problems usually use -G=RTlnK more, but you would probably use G=G*+RTlnK to compare G at non standard conditions

MaggieHan1L

Try writing down the tidbits given to you in problem and looking for equations that have the variables that you know. Sorry if this isn't helpful

MaggieHan1L

W= #microstates ^ # of particles. There are 12 micro states and 6.02*10^23 is the amount of particles in one mole

MaggieHan1L

When there is a degrees symbol and it's not attached to C, it means standard so under standard conditions.

MaggieHan1L

You can also tell by the reactions. For example a solid going to liquids or gas. You need heat to provide energy to break up intermolecular forces and thus the reaction is endothermic. Try thinking about the forces and bonds and how you need energy to break them

MaggieHan1L

I think it has to do with the fact that it is a monatomic gas compared to b and c which are diatomic gases. Because each particle is not bound to another, the monatomic gas probably has more available positions and entropy.

MaggieHan1L

The negative sign is there to help signify the relationship between the two temperatures and it doesn't really matter because you just want to have the right amounts of negatives on both sides.

MaggieHan1L

Whenever there is phase change you add the heat of fusion/vaporization to the amount of energy it takes to raise the substance a certain amount of degrees. If you look at the phase change curve you can see it's flat then it starts increasing in temperature. The flat area is the phase change.

MaggieHan1L

We haven't gone over kinetics but is more about rate while thermodynamics is more like the transference of energy.

MaggieHan1L

Where are you getting 3.314?

MaggieHan1L

when you use -p delta v you get units of 1 L atm so you use 1 L atm = 101.33 J to get to J

MaggieHan1L

Standard enthalpy of reaction can be any type of reaction but standard enthalpy of formation is specifically forming one mole of a compound using elements in standard states.

MaggieHan1L

standard enthalpy changes will be in a certain phase and sometimes you have to add enthalpy of phase changes to calculate enthalpy of reactions.

MaggieHan1L

02 is already the most stable form of oxygen and so if you look at 02 to 02 the enthalpy of formation would be enthalpy of 02-enthalpy of 02 equaling 0

MaggieHan1L

All you care about is final and initial so enthalpy of product minus enthalpy of reactants

MaggieHan1L

There are tables where all of the enthalpies have been found.

MaggieHan1L

When volume decreases, the reaction favors the side with less moles of gas.

MaggieHan1L

Lower pKa means a less negative Ka which means that there is a high amount of dissociation and thus a stronger acid.

MaggieHan1L

It's not a part of the problem. It's most likely put there to trick you. You don't need to know about the reactants since it is a solid and thus wouldn't go into calculations. Since you know the mass of CO2 then you know molarity of CO2 and molarity of NH3. CO2 would be equivalent to X and NH3 would...

MaggieHan1L

Acids/Bases usually require a lot of ICE tables since weak acids/bases don't dissociate completely

MaggieHan1L

If a reaction is endothermic than it requires heat to get to the products so adding heat is favoring the products. If a reaction is exothermic, that is saying the same thing as the reverse reaction is endothermic so increasing temperature favors the reactants.

MaggieHan1L

A hint to do an ICE might be when you get only a few concentrations and you get K

MaggieHan1L

Solids and liquids have a basically constant concentration so it won't change K because it would just cancel out if you write it out.

MaggieHan1L

K is the ratio of products and reactants. Things like increasing temperature and adding a catalyst speed up the rates.

MaggieHan1L

If you included it since the concentration doesn't change, the liquid and solid would just cancel out.

MaggieHan1L

Yes when we see gas we assume Kp and when we see aqueous we assume Kc.

MaggieHan1L

The rule is if it ends in ide it turns to ido

MaggieHan1L

neutralization is usually strong acids and bases I think

MaggieHan1L

That's the only way to get a number without knowing the concentrations. Since it's -log (0.12H+)-(-log (H+), you can change it to log (H+)-(log 0.12(H+)) and thus you can take log 1/0.12 since the H+ will cancel.

MaggieHan1L

My TA mentioned that usually the organic molecules were the weak ones.

MaggieHan1L

Amphiprotic is always amphoteric but amphoteric is not always amphiprotic.

MaggieHan1L

It depends on whether or not they dissociate completely in water.

MaggieHan1L

I think the easiest way when you're told that the molecule is amphoteric is to just add a proton and minus a proton. So you would end up with PO4 (3-) and H2PO4 for HPO4.

MaggieHan1L

If you want to calculate directly [H3O+][OH-]=10^-14.

MaggieHan1L

My TA explained it as those molecules having more resonance usually and thus without the proton they are more stable so they are more willing to be dissociated. At least this is how I understood it.

MaggieHan1L

Usually acids that have an atom which is more electronegative will be stronger

MaggieHan1L

I don't think you actually have to memorize them because he probably just wanted to show us the importance of these metals.

MaggieHan1L

Doesn't it have to do with alphabetical order?

MaggieHan1L

Yes cation before anion

MaggieHan1L

Coordination numbers depend on the number of bonds.

MaggieHan1L

You calculate oxidation numbers based on charges of the rest of the compound.

MaggieHan1L

A for the central atom. X for the bonded atoms. E for the lone pairs.

MaggieHan1L

I don't know if the angle has a difference that is significant. 90 and 120 should be fine.

MaggieHan1L

He said that the ones we discussed in class in Friday we need to know and we also need to know Monday's lecture.

MaggieHan1L

Electron focuses more on how the electrons are placed and bonding than molecular. So molecular tends to be more general focusing on regions of electron density while electron focuses on lone pairs and where they are placed because of the electron electron repulsion.

MaggieHan1L

I think he's going to cover this more in depth on Monday

MaggieHan1L

H bond needs a hydrogen bonded to a highly electronegative atom such as F, O, and N.

MaggieHan1L

Think of ionization energy as what it takes to make an anion and electronegativity is how badly it wants to be negative. That's how I remember it. Don't know if that helps

MaggieHan1L

You just have to be aware of the proportionality and that r is to the power of 6 so it has a pretty big impact.

MaggieHan1L

London forces are created when there are slight e density fluctuations and these are just in all compounds

MaggieHan1L

It also has to do with size. Larger atoms results in longer bond length and a weaker bond.

MaggieHan1L

Friday's lecture. The last slide was going to be on it but then we got class canceled on Monday.

MaggieHan1L

I think that formal charge also comes into play.

MaggieHan1L

My TA said that it has to do with energies and hybridizations. An easy way is just to think about it in terms of being able to draw from the d subshells which only start when n=3

MaggieHan1L

Because they are exceptions to the octet rule, they are highly reactive and unstable. They're very dangerous because they can modify DNA and other biologic molecules. They usually don't remain in that state for long.

MaggieHan1L

You can think of bigger polarity as one element wanting and pulling on electrons more. Thus ionic is just when there is such a big difference that one element "gets" the electron instead of sharing it.

MaggieHan1L

The preferred formal charge is 0. When he gave the example of sulphate SO4, the lewis structure was technically correct but the formal charges weren't ideal. We use formal charge to give us better lewis dot structures. When you're drawing lewis dot structures and you can't get all of the formal char...

MaggieHan1L

When he was talking about Benzene, he was pointing out that the bonds are neither the length of a single or a double bond because of the resonance. The bond lengths depend based on the elements and what type of bond it is. For example N-O in nitrate was 1.24 A while C-C in benzene was 1.39 A.

MaggieHan1L

Other elements such as N and O and F can't because they are in N=2 meaning they only have access to s and p and can only form an octet. In general just remember n=3 as what makes it an exception.

MaggieHan1L

Positive means that they lost electrons so that's why you subtract from the ground state. If they were negative you would add because they gained electrons.

MaggieHan1L

Losing 1 electron in the p orbital doesn't lead to an octet because there are still 2 electrons in the s orbital

MaggieHan1L

The uncertainty of the electron would be diameter and you used radius.

MaggieHan1L

The part where they say "Model the atom as a one-dimensional box with a length equal to the diameter of the actual atom." just means make delta x = to diameter. That's basically what you always do when you see diameter and uncertainty. Then you get speed by changing delta p into m times de...

MaggieHan1L

It has to do with electron repulsions. The electrons make sure every orbital has one electron first because two electrons in the orbital has repulsions. At least that's what I remember from high school.

MaggieHan1L

Hung Indy Bui 3H is right, it all depends on how you label it and look at it. You can go any way you want.

MaggieHan1L

Dr. Lavelle said he wouldn't go into the deriving of the equation because that was out of the scope of the class.

MaggieHan1L

Even though we didn't need it, I'm pretty sure on the last test Dr. Lavelle printed out the mass of an electron on the formula sheet

MaggieHan1L

I think that many things have wavelike properties but they aren't really detectable such as when they are smaller than 10^-15 and then you say they have more particle like properties

MaggieHan1L

Everything is quantized. It's not like a normal equation where you can plug in any number. There are only specific energies.

MaggieHan1L

For week 3, we definitely have to do quantum.

MaggieHan1L

Showing your work can also make it easier for you to double check your answers too!

MaggieHan1L

The rydberg equation I think is just an equation that combines a few steps of the Einitial - Efinal method.

MaggieHan1L

Empirical formula is the lowest ratio so molecular formula should not be greater than the empirical.

MaggieHan1L

For some problems where you have to find the molar mass of a sample with a lot of isotopes just multiply the isotope percentage by the isotope weight and add all of them together.

MaggieHan1L

One thing we talked about today in discussion is that when you see oxidized that also means that it is combustion. Thought that would be useful in decoding problems.

MaggieHan1L

One thing that stuck with me throughout high school chemistry is the mole tunnel. If you're having trouble just always remember grams to moles to moles to grams.

MaggieHan1L

Thanks so much for taking the time to help out all of us who are so clueless. You gave such great and meaningful advice.

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