CHEMISTRY COMMUNITY

Rami_Z_AbuQubo_2K

Yes for rate of consumption it is going to be $\frac{-[R]}{\Delta T}$ where as rate of formation
for the product will be $\frac{[P]}{\Delta T}$ .

Rami_Z_AbuQubo_2K

If you know that the reaction is a pre-equilibrium reaction, then the second step will be the slow one. That first step is going to be a fast reaction and it is considered to be at equilibrium where there is a bottle necking effect and that leads into your second step which is going to be the rate d...

Rami_Z_AbuQubo_2K

Bi molecular is a description of how many molecules are participating on the reactants side of an equation. In every case, you would look at the reaction before intermediates or catalysts are removed. If there is only one molecule reacting, then you know it is unimolecular. If you see that there are...

Rami_Z_AbuQubo_2K

In this class, you will need to round up, but they do exist as decimals. For the purpose of the class and the exam, just round everything up or down accordingly.

Rami_Z_AbuQubo_2K

Why do we have to use 2H2O-->O2 + 4H + 4e- in this problem. Also where do they get the value of 1.23 for Enaught and why is Ni2+ half reaction the cathode and not the anode even though we have typically tried to create the most positive E.

Rami_Z_AbuQubo_2K

Can someone explain to me the reasoning behind this problem? I do not understand what specifically causes water to be reduced over metal and vise versa.

Rami_Z_AbuQubo_2K

What did he explicitly talk about for the inert electrodes other than their existence.

Rami_Z_AbuQubo_2K

Are we not supposed to use the Appendix for such a question?

Rami_Z_AbuQubo_2K

If Ecell is negative, does that meant that the reverse reaction is favorable since the forward is not?

Rami_Z_AbuQubo_2K

yes, you do need to know how to do that, as well as how to balance in an acidic solution. That will likely be one of the questions on the exam.

Rami_Z_AbuQubo_2K

I believe it is all of Gibbs, which includes some of the thermo and everything until nest, but not including it.

Rami_Z_AbuQubo_2K

When balancing reactions, you:
1. Balance the elements
2. balance the O
3. Balance the H
4. Balance the electrons

Rami_Z_AbuQubo_2K

In a general sense, why are we looking at the K value when solving problems with Gibbs free energy?

Rami_Z_AbuQubo_2K

In what situations would we need to calculate different temperatures?

Rami_Z_AbuQubo_2K

What exactly is Gibbs free energy, and how does it related to entropy and enthalpy without strictly looking at the equations.

Rami_Z_AbuQubo_2K

It is when energy is released from a system to an environment, meaning the reaction is going to be exothermic. Depending on the problem, it may make you calculate q, and that just means energy is released.

Rami_Z_AbuQubo_2K

It is because fusion is the opposite of freezing.

Rami_Z_AbuQubo_2K

There are few general problems in the text book which have you analyze systems and determine which express more disorder. I believe it is section 4H.

Rami_Z_AbuQubo_2K

Most of the time it will given an indicator that pressure has not changed, but in the case that it doesn't, I think it is safe to assume that there is not pressure change. With that being said, just read the problem carefully.

Rami_Z_AbuQubo_2K

Do more orientations mean a greater residual entropy?

Rami_Z_AbuQubo_2K

When pressure is constant, q becomes q_p and this is equal to $\Delta h$ .

Rami_Z_AbuQubo_2K

Does the standard enthalpy of 0 only apply to diatomic molecules like H, O, F, Cl, Br, I, N?

Rami_Z_AbuQubo_2K

Work is being done on they system by compressing it. Following the equation work= Pressure*area*distance, it will give you a positive value.

Rami_Z_AbuQubo_2K

It is both conceptual and applicable. There were two ways to look at it, the specific direct bonds that were broken and subtract the energy needed to form new specific bonds from it. Or you can look at all the energies that form bonds in a molecule and subtract all the energies that form bonds in th...

Rami_Z_AbuQubo_2K

The reason that temperature stays the same is that, energy required to break bonds is greater than the energy needed to raise the temperature. Since the temperature is already high enough in the process of changing phases, the rest of the energy is put into breaking the bonds of the substance to cha...

Rami_Z_AbuQubo_2K · Re: Due date for Discussion Questions They are due Sunday at 11:59 pm

They are due Sunday at 11:59 pm

Rami_Z_AbuQubo_2K

When deltaH is positive, that mean that the reaction is going to absorb heat in order to proceed and is endothermic. When deltaH is negative, that means that the reaction is going to release heat into the environment and is exothermic.

Rami_Z_AbuQubo_2K

Catalysts only speed up the rate of the reaction. They do not have anything to do with the favoring of one side of the reaction or another. They only get you to the products that you want faster and nothing more.

Rami_Z_AbuQubo_2K

-If the pressure increases, the side with less concentration/mols will be favored.
-If the pressure decreases, the side with more concentration/mols will be favored.

Rami_Z_AbuQubo_2K

You ignore pure liquids because only a small amount ends up reacting, meaning the amount in the products and reactants are practically the same and cancel out when you put the liquid in both the denominator and numerator of the equilibrium ratio. As for solids, there is no such thing as a concentrat...

Rami_Z_AbuQubo_2K

From what I know, if you see that an acid has a really small Ka value, then it will be weak. Same goes for Kb, if it is really small, generally the 10^-3 is a good marker for both, then it will be weaker.

Rami_Z_AbuQubo_2K

Q looks at the system when it is not at equilibrium(when it ratio of products to reactants is equal to Kc). Q is compared to Kc to see whether the reaction proceeds to the right or the left. If Q<Kc then the reaction will "shift" to the right. If Q>K then the reaction will "shift"...

Rami_Z_AbuQubo_2K

If it was solids and liquids, then there would not be an equilibrium constant. There is no such thing as a concentration of a solid. Additionally, for pure liquids, they do not react significantly in the reaction. This means that the amount of pure liquid in the reactants is almost the same as that ...

Rami_Z_AbuQubo_2K

K is the actual equilibrium constant that is a specific given value for a certain reaction. However, when you are calculating whether a reaction has reached equilibrium, you use Q, and compare it to the equilibrium constant K. If Q is equal to K, then the reaction is at equilibrium. Since K is a fix...

Rami_Z_AbuQubo_2K

The only thing included is aqueous solutions and gases but not liquids or solids. You cannot have a concentration of a liquid and pure liquids do not really react much so that amount that existed in the reactants will almost be the exact same in the products which will just cancel out in the calcula...

Rami_Z_AbuQubo_2K

Simply put, look at the molar coefficients on each of the molecules and using that number, it will be the value of X in the ICE table. If you are looking at the reactants, you will subtract X. However, if you are looking at products, then you are adding X.

Rami_Z_AbuQubo_2K

Solids and liquids are not included in the K expression because there is no such thing as a concentration of a solids. For liquids, the molar concentration of a pure substance, like water, does not change in a reaction. The amount of liquid that reacts in a reaction is so small that both amounts of ...

Rami_Z_AbuQubo_2K

Because there is more ability for resonance to stabilize the anion after the proton is donated. This means that there is a better distribution of charge and makes the acid more strong.

Rami_Z_AbuQubo_2K

porphyrin is a chelate that binds in 4 places, thus making it tetradentate

Rami_Z_AbuQubo_2K

Since there is an equal amount of oxygens, you look at which is more electronegative (Br or Cl) and since Cl is more electronegative, it is the stronger acid.

Rami_Z_AbuQubo_2K

Bigger london dispersion forces do lead to higher boiling points when a molecule exhibits no hydrogen bonds, is not polar or when the polarity is almost the same between two molecules. The bigger the atom, the greater the shift in electrons to create a temporary dipole and makes H2Se a stronger acid

Rami_Z_AbuQubo_2K

Dont memorize but understand. There are 3 general rules that you can look at that the book describes intensively. 1. The more 0, the stronger the acid 2. The more electronegative the molecule, the more acidic. 3. The easier to break up a bond, the more electronegative and that looks at things like H...

Rami_Z_AbuQubo_2K

Since the difference in electronegativity between Se and S is negligible, we are considering their dipoles as almost the same. The only thing we can look at now is london forces and since Se is bigger, it is going to create a bigger temporary dipole and have stronger london forces.

Rami_Z_AbuQubo_2K

In this case, you are looking at the HF bond length vs the HCL bong length. Since Florine is more elctronegative than Cl, it will have a stronger pull on the H make it harder to donate the H. The longer the bond, the easier it is to remove the H. It follows one of three rules in the book for stronge...

Rami_Z_AbuQubo_2K

To know if something is a polydentate ligand, you need to see if it has multiple places where it can bind to through multiple lone electron pairs. The only way I know of finding this out is by drawing out the lewis structure and looking at is electron geometry to determine if it is polydentate or not.

Rami_Z_AbuQubo_2K

Polydentate ligands are ligands that have multiple areas where they can bond. However, ligands that only have one binding site, like NH³ would be monodentate.

Rami_Z_AbuQubo_2K

He just wanted to illustrate that the central atom can connect to any one of the elements in that particular compound.

Rami_Z_AbuQubo_2K

Memorize the ones that you see come up in the problem set. Those are the ones that are usually brought up during the test but if you can memorize anymore, I would definitely do it to play it safe.

Rami_Z_AbuQubo_2K

I would think of it as areas that are bonded around the central atom. It is similar to the electron geometry structures of compounds, but rather than looking at area of electron density, you look at areas with atoms and ions.

Rami_Z_AbuQubo_2K

A chelate is a made up of many complex compounds where there is a central metal atom that binds to other ligands in a ring/cyclical type structure. I believe the ligand is also polydentrate.

Rami_Z_AbuQubo_2K

So coordination number corresponds to the areas bonded to the center of a ligand similar to the electron geometry structures?

Rami_Z_AbuQubo_2K

When something is polydentrate it means that the electron lone pairs are found to be distributed among different atoms on a particular ligand rather than having the electron lone pairs on one atom, where it would not be polydentrate. Hope this helps

Rami_Z_AbuQubo_2K

A ligand is polydentrate when the electron lone pairs are found to be distributed among different atoms on a particular ligand rather than having the electron lone pairs on one atom, where it would not be polydentrate. Hope this helps

Rami_Z_AbuQubo_2K

It means that you must be able to write out the exact order of a compound from left to right. Knowing the general rules for naming them is helpful, but it is more about the application of those rules on the exam. There are many helpful examples provided in the book to give you a better picture of wh...

Rami_Z_AbuQubo_2K

Memorize the ones that are provided in the book. Also, we will not be given the names on the exam. You must memorize them.

Rami_Z_AbuQubo_2K

What do you mean they are found between the p orbitals? Also, what about d orbitals then?

Rami_Z_AbuQubo_2K

What do you mean by "end to end" or "side to side"?

Rami_Z_AbuQubo_2K

Aside from looking at the regions of electron density in the Lewis dot structure, when you find Be on the periodic table, you see that has 2 valence electrons. When looking at its hybridization, this would give it an s2 denotation. However, we need to have two sites of bonding and through hybridizat...

Rami_Z_AbuQubo_2K

They are not going to test us based on our knowledge of a molecule. It will almost always be provided to us on the exam.

Rami_Z_AbuQubo_2K

Do you remember the box method that was taught in high school? Each box with one electron inside of it opens itself up to bonding. When looking at the boxes, the s has only one box which allows two total electrons, the p has 3 boxes which allows 6 total electrons, the d has 5 boxes which allow for 1...

Rami_Z_AbuQubo_2K

No because hybridization only tells us how atomic orbitals combine to form hybridized orbitals. These orbitals allow us to determine bonding properties of a molecules which contributes to the overall geometric structure. It is not solely to create the structure of a molecule, mainly because were are...

Rami_Z_AbuQubo_2K

Hybridization helps in generally understanding of how orbitals are made suitable in order to pair electrons so that chemical bonds can form with other atoms. Just know the concept behind it because this topic is just to test your knowledge of how bonds can be made. It will be a supplement to a more ...

Rami_Z_AbuQubo_2K

The only way for you to know the shapes is to simply memorize them using the table above or one that you can search up online. By knowing AXE, that should in turn help you figure out the different geometric shapes of each molecule.

Rami_Z_AbuQubo_2K

My TA also said that we might be tested on this too, but I think for the test, Dr.Lavelle will not be asking us about it. Regardless, prepare for it anyways in case.

Rami_Z_AbuQubo_2K

All though hybridization does relate to VSEPR, its main focus is not about helping you draw the model. All hybridization does is illustrate how many areas of electron density there are in a specific molecule and those fields can either be a boding pair or a lone pair. You can use it to help with sma...

Rami_Z_AbuQubo_2K

All though hybridization does relate to VSEPR, its main focus is not about helping you draw the model. All hybridization does is illustrate how many areas of electron density there are in a specific molecule and those fields can either be a boding pair or a lone pair. You can use it to help with sma...

Rami_Z_AbuQubo_2K

You should look up the general table for all the different structures and memorize or try to picture the bonds themselves online. I find it super helpful too see all of them right next to each other and see how the structures change

Rami_Z_AbuQubo_2K

We do not need to memorize polyatomic ions as of right now for this course.

Rami_Z_AbuQubo_2K

Is there a general formula on how to name compounds?

Rami_Z_AbuQubo_2K

Are there other ligands that we must know too?

Rami_Z_AbuQubo_2K

VSEPR essentially uses the ideas from the Lewis dot structure to model the geometry of molecules using electron pairs that are around the central in order to minimize the repulsion found in between the valence electrons of that specific molecule.

Rami_Z_AbuQubo_2K

From my understanding, VSEPR theory says that repulsion from the lone electron pair exhibits greater repulsion than a bonding pair. Additionally, when there are multiple bonds, they exhibit a greater repulsive force on near by electron areas than do single bonds.

Rami_Z_AbuQubo_2K

To figure out the shape, model out the Lewis dot structure of the molecule, figure out the number of bonding pairs and lone pairs around the main/central atom, use VSEPR to figure out the electron pair shape of the molecule and the VSEPR shape to then figure out the angles between that are found bet...

Rami_Z_AbuQubo_2K

Since P is the least electro-negative, it will be the central element. Considering that anything from group 3 to 7 in the p block has an expanded octet, which includes P, that means that it will break the octet rule.

Rami_Z_AbuQubo_2K

Anything from group 3 to 7 in the p block, which are the most common type of used elements, have an expanded octet.

Rami_Z_AbuQubo_2K

The exam will specify which formal structure it wants, which can include all 3. I believe that the more 0's there are, the lower the energy of that molecule but just know how to find the charges in general.

Rami_Z_AbuQubo_2K

If you remember the box method for an orbital, you might remember starting with spin up in each individual box before you start putting in spin down but it does not matter really. As long as you are consistent, say you start spin down, do not mix up spins in each box.

Rami_Z_AbuQubo_2K

The further right you go, the more electro-negative an element is and the further up you go the more electro-negative an element becomes. Elements in the top right corner are the most electro-negative while those in the bottom left are the least electro-negative. Hope that helps..

Rami_Z_AbuQubo_2K

It is basically denoting the orientation. Think of it like a x,y,z graph and imagine the orientation falling on each of the 3 different lines.

Rami_Z_AbuQubo_2K

Yes all of the constants will be provided on the test. You should go to the website and find the formula sheet and get familiar with it.

Rami_Z_AbuQubo_2K

Yes the test will be during the discussion.

Rami_Z_AbuQubo_2K

He said anything that is less than 10^-18 is not detectable. He will make it very obvious and it will be an extreme either direction.

Rami_Z_AbuQubo_2K

Most of the equations will be given on the exam sheet during the exam which you can find on the website if you want to make sure. Rather than memorize them, just make sure you understand each part and know how to manipulate them because that is much more important for exam 2.

Rami_Z_AbuQubo_2K

It is too small because the amplitude of the wave and its frequency is so minuscule to detect that it almost seems like it does not exist. Using a number multiplied by 10^-15 or 10^-18 is just the smallest we can actually detect and any smaller is just theory.

Rami_Z_AbuQubo_2K

I believe we had to remember just a general range and not the actual frequencies of each color. Knowing the frequencies of violet, green and red are probably the most that will be needed and everything else will most likely be given.

Rami_Z_AbuQubo_2K

I think finding the theoretical yield for both reactants that are considered is much easier to get a grasp for what is the limiting reactant but I was curious if there was a faster way that always works.

Rami_Z_AbuQubo_2K

Dived the number of atoms on each side (meaning 9) by the two respective molar masses and look at the individual weight of each atom for each side. That should help you understand it better on how to balance out each side.

Rami_Z_AbuQubo_2K

If I were you, I would memorize the few that we need for the upcoming test. They aren't that hard, especially if you really understand how the units cancel out. But definitely email the TA about this.

Rami_Z_AbuQubo_2K

Wait, so when it is talking about formula units, it does not specifically mean atoms/molecules, right?

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