CHEMISTRY COMMUNITY

Danielle_Gallandt3I

The issue with looking at the overall reaction is that some of the species may have been "canceled out" or the number reduced so you have to look at each elementary step to know how many species are in each step (its molecularity). If the equation you gave was an elementary step, then yes ...

Danielle_Gallandt3I

I think you might actually need to know the elementary steps I dont think you can determine it from the overall reaction.

Danielle_Gallandt3I

I think it is because it is actually proportional to the rate constant k (so it is dependent on the environment, temp activation energy, etc). Another way to think about this is that a second half life takes the same amount of time as the first half life would, even though there is half the original...

Danielle_Gallandt3I

There is no exact percentage, but the information we learned after the last test will be focused on more

Danielle_Gallandt3I

There might be differences but it shouldn't be drastically different. These differences could arise from the delta G changing for products and reactants based on the environment, or because the delta H might be slightly off because its being used for a wide variety of temps

Danielle_Gallandt3I

They key difference is that an instantaneous rate is specific to each reactant or product within a reaction, whereas a unique rate is the same for all reactants and products within a reaction

Danielle_Gallandt3I

This typically occurs when the reactant concentration is already very high, so adding more reactant doesn't change the rate

Danielle_Gallandt3I

While I do think it is helpful to know so that if you don't remember an equation you can derive or integrate it yourself, I think in the context of this class it is not incredibly necessary. However, this does mean that you will have to memorize more equations.

Danielle_Gallandt3I

Yes, because typically if the temp changes then entropy will change

Danielle_Gallandt3I

As far as I know it is not required to know, as Lavelle typically tests only what he addresses in class

Danielle_Gallandt3I

No, based on the half-life equations it is not affected by initial concentration, only the rate constant. This means that the half life is actually more dependent on other factors, such as temperature

Danielle_Gallandt3I

A first order reaction occurs when any change to the concentration results in an equal change in the rate. A second order reactions occurs when any change in the concentration results in 2x the change in the rate

Danielle_Gallandt3I

A zero order reaction means that any change in the concentration does not affect the rate. A first order reaction means that any change in the reactant concentration will lead to the same change in the rate (concentration doubles means rate will double)

Danielle_Gallandt3I

Another way to look at this is to consider how a rate changes as a reactant's concentration changes. If you add more of a reactant and the rate does not change it is a zero order reaction. If you double the concentration and the rate doubles, it is a first order reaction. If you double the concentra...

Danielle_Gallandt3I

It is important to remember that for the majority of problems you will actually have to use the delta G equation to determine the spontaneity of the reaction, because if if delta H is positive, the combination of the (negative) entropy and the temp must be large enough to overcome the enthalpy value...

Danielle_Gallandt3I

I think the test will probably incorporate some of the elements of thermochem and thermodynamics we learned before the midterm, but the focus will be on topics we were not tested on in the midterm. However I am pretty certain some of the things we are still learning require us to use and build on to...

Danielle_Gallandt3I

Microstates are used when calculating the degeneracy (W) value when you are finding the statistical entropy. Additionally, if you are comparing to molecules, if one has more microstates it is very likely it will have a greater level of entropy.

Danielle_Gallandt3I

If the Br2 to 2Br stayed in the same phase then you would not have to consider the vaporization of Br, but since there is a phase change, this must be considered on top of the bonds of the molecule being broken

Danielle_Gallandt3I

I think for many of them you just have to memorize, but some are pretty apparent and easy to remember, like an O2 molecule and many other diatomic molecules

Danielle_Gallandt3I

The last 3 types of entropy you mentioned relate to the position and possible movement available to a molecule. The first type you mentioned, residual, relates to the residual entropy left in a molecule when it is at 0 K. At that point there is no entropy due to movement, but if the molecule has mor...

Danielle_Gallandt3I

Yes, that is correct. The bonds that form spontaneously release energy because the molecule formed is more stable (and therefore at a lower energy) than each of the elements separately. The molecule therefore releases energy by forming a bond, so q (or delta H) is negative

Danielle_Gallandt3I

If something has a higher heat capacity that means it can absorb more energy before it increases in temperature. Therefore, it should have a smaller slope because the temperature is increasing at a lower rate due to the higher heat capacity of the liquid.

Danielle_Gallandt3I

I think it depends, because the standard enthalpy of formation is in J/mol, so I think it changes with the "type" of enthalpy you are considering

Danielle_Gallandt3I

Since work is a form of energy transfer the units will be in Joules. And the units for pressure and volume will vary depending on the problem but the most common units tend to be atm and liters.

Danielle_Gallandt3I

I don't think there is a specific rule but I think it might jus be what makes more sense for that problem. For example, if it is a really big value it would make more sense to have units in kJ rather than J. I think unless it specifies it in the question it will probably be up to your own discretion.

Danielle_Gallandt3I

Work is a type of energy transfer, so if work is being done TO as system the system is being transferred energy so it will be a positive value. If work is being done BY a system then energy is being transferred out of a system so it will be a negative value

Danielle_Gallandt3I

Q uses the same equation as K, but it can be calculated at any point of the reaction, whereas K can only be calculated at equilibrium. You can then compare Q to K to find out where the reaction is relative to equilibrium and which way the reaction might favor.

Danielle_Gallandt3I

A good rule of thumb is that a K value that is 10^-3 or less usually means that the x value will not cause a significant difference in the initial concentration value and can be disregarded when subtracting from the initial concentration. You can also check to see if this assumption is correct if th...

Danielle_Gallandt3I

If they do not specify I think it is safe to say that it would be Kp (partial pressure). From my understanding the partial pressure K is actually the more "traditional" K value that then has to be converted to Molar concentration.

Danielle_Gallandt3I

The powerpoints are not posted but the audiovisual topics tend to be almost exactly the same thing he teaches in class, even the powerpoints in the videos. There is not a video to go with each lecture, but they do cover a good portion of what he teaches in class if you missed something or need addit...

Danielle_Gallandt3I

A good rule of thumb for strong bases is that the cation of the base is typically in the first or second group within the periodic table.

Danielle_Gallandt3I

Dr. Lavelle did mention however that the pH does not change very significantly so a small change in temperature will mean the water is basically still at a pH of 7

Danielle_Gallandt3I

The K value of a reverse reaction is always the inverse of the K value for the forward reaction.

Danielle_Gallandt3I

So during a reaction, a pure substance (like water) does not have a big change in concentration. The other reactants and products that are not pure substances will have a change in concentration as the reaction is reaching an equilibrium, which is why they remain in the equation to find the equilibr...

Danielle_Gallandt3I

I don't know for sure but it might be caused by the change in energy level and movement/speed of the molecules and also their potential change in state.

Danielle_Gallandt3I

What Lavelle means by "they only exist for a short time" is that because they have an unpaired electron, they are extremely unstable and very likely to bond with another atom/molecule to crete a more stable molecule. That means that no compound can ever really "support" a radical...

Danielle_Gallandt3I

I think one of the ways that could be a good signifier is if there is an unbalanced dipole moment in the Lewis structure. For example, water is an amphoteric compound with a negative dipole on the oxygen and a positive dipole on the hydrogen side. That is why it can act as an acid and give off a hyd...

Danielle_Gallandt3I

A strong acid will have a weak conjugate base and a strong base will have a weak conjugate acid. Weak acids tend to have stronger conjugate bases, and weak bases tend to have stronger conjugate acids. What this basically means is that with a strong acid, its conjugate base is unlikely to be a good p...

Danielle_Gallandt3I

It's polar if it's got a dipole moment which means the little arrow with a plus sign points toward the more electronegative element (usually oxygen, occasionally carbon...etc) This is only polar if the dipole moments do not cancel out. You can have a situation where there are dipole moments and it ...

Danielle_Gallandt3I

Not necessarily. It depends on the overall molecular geometry. If the dipole moments cancel out due to molecular geometry then it is nonpolar. This is the case if the molecular geometry is linear, because the dipole moments would go toward the center so no one part of the molecule has a significant ...

Danielle_Gallandt3I

That is what I thought but in discussion my T.A. said that covalent bond strength increases with atomic radii

Danielle_Gallandt3I

A good way to remember this is that these two characteristics will come with a pair of atoms, one being an anion and one being a cation. When you think of polarizing power think of how having this power creates action, so a cation is creating action by moving this electrons away from their original ...

Danielle_Gallandt3I

There are pros and cons to each shape, so it is really just dependent on what the intended purpose is of the molecule.

Danielle_Gallandt3I

Pretty much! The oxidation number is the number of electrons removed from the atom to make it a cation, so if a cation has a 2+ charge that means it has an oxidation number of 2 because 2 electrons have been removed from the original atom

Danielle_Gallandt3I

This concept is similar to how when an atom has a 4s orbital filled and a 3d orbital with 9 electrons, it is actually at a lower energy to remove one of the electrons from the 4s orbital and add it to the 3d orbital to create a full 3d orbital. It is basically a matter of an atom having an electron ...

Danielle_Gallandt3I

It depends on the situation. It is better to have them concentrated if one molecule is much more electronegative than the other, because it would not make sense for electrons to be more electrostatically attracted to something like hydrogen over something like oxygen. Another thing to consider,, how...

Danielle_Gallandt3I

They also relate to bonding and the strength of bonding. If something has permanent dipole moments (like H20) that will result in stronger dipole-dipole bonds, compared to molecules with temporary dipole moments (aka van der waals forces) that are weaker bonds because they do not last as long.

Danielle_Gallandt3I

Why is it that as atomic radii increases the strength of the covalent bond also increases? Wouldn't you want the atomic radii to be smaller to decrease bond length and strengthen the bond?

Danielle_Gallandt3I

As was said above, dipole-dipole interactions are a general term, and hydrogen bonding is a specific term for dipole-dipole interactions that occur when a hydrogen bonded to an O, N, or F atom forms a dipole-dipole interaction with another molecule. So to determine whether or not this specific form ...

Danielle_Gallandt3I

One way to think about it is that with more bonds and/or with stronger bonds, you need to supply more energy to break said bonds. That is why with something like a boiling point, the compound needs to be at a much higher temperature if it has a lot of strong bonds because it requires more energy to ...

Danielle_Gallandt3I

Based on the way our past tests have been structured, we have only been tested on information from that particular unit, not an accumulation. I would guess that test 3 will be similar and only be information given after the midterm.

Danielle_Gallandt3I

An additional element to consider here is how many double bonds there are and how many single bonds. If having more double bonds lowers the energy of the molecule, then you can assume that structures with more double bonds are contributing more to the resonance structure. In that case, the resonance...

Danielle_Gallandt3I

From my understanding, as you go down a family you see that the electrons are held less strongly because the radius is getting bigger. Therefore, when an anion is bonded to a cation, as the anion gets bigger it cannot hold onto its electrons as strongly so the electrons gravitate to a "shared r...

Danielle_Gallandt3I

For resonance structures it also depends on which structures are contributing more to the overall resonance structure. So if the structures with double bonds have lower energy than the structures with single bonds, they will be contributing more to the overall structure, so the bond length is probab...

Danielle_Gallandt3I

The elements you would typically see with an expanded octet are going to be in Period 3 or below and most likely in the p-block (right side of periodic table and mostly nonmetals) because they have an empty d-orbital.

Danielle_Gallandt3I

Ok so each row on the periodic table actually tends to correspond to a particular shell of an atom. So the first row corresponds to the first shell (with only an s-orbital) so there are only 2 e- within that shell. So you know that H has 1 valence e- and He has 2 e- because those are the number of e...

Danielle_Gallandt3I

There is not an explicit electronegativity trend because at some point the atoms with more protons (meaning a greater electron affinity) end up having so many electrons that their valence electrons are far away from the nucleus and so their electron affinity and electronegativity decreases. At that ...

Danielle_Gallandt3I

The way this can be calculated is that d-orbitals exist in n=3 or above, which means that l=2 (so there are s, p, and d orbitals). When l equals 2, ml can equal -2, -1, 0, 1, or 2 (5 orbitals). In each of these orbitals there can be a maximum of 2 electrons, where one of the electrons has an up spin...

Danielle_Gallandt3I

So I think part of the reason might be that when there are two electrons in the same orbital (one with a spin up, one with a spin down) that is at a higher energy state because the electrons repel each other. By that logic, once the 4s orbital is filled (with the two electrons repelling each other) ...

Danielle_Gallandt3I

Yes, the same rule applies to those elements as well. The Mo configuration is 4d^5 5s^1 and the Ag configuration is 4d^10 5s^1. I would recommend that you look at a Periodic Table that provides the electron configuration (such as the one in our textbook) because those are not the only two that have ...

Danielle_Gallandt3I

So you have to first determine what the ground state electron configuration should be for each of these different elements. If the electron configuration does not match that of the configuration in a ground state, it will be in the excited state. For example, the ground state of Carbon is 2s^2 2p^2,...

Danielle_Gallandt3I

From my understanding I think the ways in which we would be using the equation are more theoretical than practical, so I don't think we will be using it to find values (unless he shows us how to in the next few classes) but rather have a good understanding of what the equation means and what it will...

Danielle_Gallandt3I

Does anyone know what we will be tested on for the Week 4 test? Thanks!

Danielle_Gallandt3I

Another thing to note about the photoelectric effect is that since it showed that large wavelengths (and low frequencies) did not have enough energy, but small wavelength values (and high frequencies) did, one of the components that influences the energy of a photon is the frequency of light. That i...

Danielle_Gallandt3I

When you are plugging in the E value for the equation E=hv the energy input is the energy of a photon. When you use the equation E= -hR/n^2 you are calculating the energy an electron lost when going down one or more energy levels. The energy the electron lost is emitted as a photon. Since the electr...

Danielle_Gallandt3I

So interference requires at least two separate waves interacting with each other to occur. Constructive interference is when two waves have the same phase length, so their waves will combine to a larger total wave. Destructive interference occurs if two waves go out of phase with each other, so they...

Danielle_Gallandt3I

So when an electron moves down an energy level it is releasing energy (typically in the form of light). In that case, we are trying to calculate the energy lost from the electron, which should equal the energy of a photon. The equation you would use is E= -hR/n^2, where h and R are constants and n=t...

Danielle_Gallandt3I

A good way to balance equations is to start with the elements that appear the least within the equation, and then move on to the elements that appear with greater frequency. In this case, you would probably want to start with balancing carbon and end with balancing oxygen. Another thing to note is t...

Danielle_Gallandt3I

If you look at the outline for week 2 there are 4 additional suggested video assessments, but I think that is all the videos that are suggested for the quarter (unless he uploads more to the site later on). I think these videos are just to help start us off at the beginning of the year and cover mos...

Danielle_Gallandt3I

In this case, another way of getting to that answer would be to take the smallest number in that ratio and divide all numbers within the ratio by that number. When you do that you get:
1/0.13= 7.7
0.13/0.13= 1
2.23/0.13=17.15
When rounded, the ratio will then be: 8:1:17

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