CHEMISTRY COMMUNITY

Aayushi Jani 3A

The slow step is the rate-determining step since the rate cannot go faster than the slowest rate, and as a result, the reaction depends on this step.

Aayushi Jani 3A

Enzymes are a kind of catalyst that lower the activation energy for a reaction, which lowers their energy barrier, allowing for the reaction rate to increase.

Aayushi Jani 3A

As others have said, I don't think we would be tested on how to graph but we may be tested on how to interpret the graphs. Also, even if the graphs won't necessarily show up on the test, I would still spend some time trying to understand them since it'll help you understand how it all conceptually w...

Aayushi Jani 3A

As others have said, it is possible if all the individual reactants are zero-order as well. However, I haven't seen any problems like that in the textbook or Sapling, so I assume it is rare and doesn't really occur. I would assume that the rate of the reaction would be based on the rate constant only.

Aayushi Jani 3A

Intermediates are not included in rate laws since they are products of one reaction which immediately become reactants in the next reaction.

Aayushi Jani 3A

Yes, it will be cumulative and will include everything from Week 1 to Week 10.

Aayushi Jani 3A

Yes, the problem should specify since there's no way for us to know otherwise.

Aayushi Jani 3A

As others have said, you use Pt(s) when there is no solid metal conductor. You only need to add it to the side that does not have a solid metal conductor (you may not need it for both sides).

Aayushi Jani 3A

Yes, n is the moles of electrons transferred. After you balance the half-reactions, you should be able to see how many electrons were transferred (which is your value for n).

Aayushi Jani 3A

Yeah, I was having some trouble with that question as well, but once I read it, I realized that we just have to use the equation that is given (and therefore, since it is not spontaneous, the cathode & anode you end up with make sense).

Aayushi Jani 3A

Yeah, for certain problems on Sapling, they required states of matter. In general, for Sapling, I would include all aspects unless it specifically says you don't have to.

Aayushi Jani 3A

I think that the questions on the exam would be framed in a way in which we would have to be able to gather information from drawing out a cell diagram (similar to 14A where we didn't have questions about what the correct Lewis structure was but for many problems, we had to draw out the Lewis struct...

Aayushi Jani 3A

Yes, it is the moles of electrons being transferred after balancing the redox reaction.

Aayushi Jani 3A

An oxidizing agent is being reduced (it gains electrons and oxidizes another substance) while a reducing agent is being oxidized (since it loses electrons and reduces another substance).

Aayushi Jani 3A

Gibbs free energy determines whether or not a reaction is spontaneous (or thermodynamically favorable) in a given direction. A reaction with a negative gibbs free energy is said to be spontaneous while a reaction with a positive gibbs free energy is non-spontaneous.

Aayushi Jani 3A

I had a problem with sapling earlier in the quarter where it wouldn't let me edit or delete anything. I just emailed Dr. Lavelle and he contacted Sapling and it was fixed the next day so I could redo the problem and get full points for it.

Aayushi Jani 3A

You would first have to calculate delta Gr, using the delta Gf values found in the appendix. Then, you would plug that into the equation delta Gr = -RTlnK. Remember, in order to find K, you would have to do e^-(Gr/RT).

Aayushi Jani 3A

qrev refers to the amount of heat transferred in a reversible reaction.

Aayushi Jani 3A

Yes, so a larger molecule (heavier molecule) has higher entropy since the energy levels are closer together, meaning they are more accessible. Complex molecules also have higher entropy (as seen through W since they would be able to occupy more orientations/positions). Ranking from highest entropy t...

Aayushi Jani 3A

I was having the same issue with that question, but I think that the temperature, pressure, volume, etc will be made more clear on the exam (or it will just say standard conditions).

Aayushi Jani 3A

If the reaction is irreversible, then yes that is the correct equation. If it is not (it's an isothermal reversible expansion), use w=-nRTln(V2/V1). However, based on textbook problems, I think it will be pretty clear whether or not it's irreversible or reversible.

Aayushi Jani 3A

An increase in volume increases entropy since the molecules are able to occupy more states. An increase in temperature will also increase entropy since the movement of particles increases, which increases disorder.

Aayushi Jani 3A

I don't think there's a difference. I tried to search online to see if there was a difference, but it seems like k and Kb are used interchangeably, but feel free to correct me if I'm wrong.

Aayushi Jani 3A

Yes, as others have said, I think both kJ and J are fine, but just make sure when you're doing the problem, it's all in kJ or J.

Aayushi Jani 3A

I would be familiar with those equations even if they aren't provided on the equation sheet since there are a couple of textbook problems that require you to use them (and some of the exam questions come from the textbook problems).

Aayushi Jani 3A

As others have stated, the conversion for celsius into kelvin is C + 273 = Kelvin.

Aayushi Jani 3A

Like the others have said, W takes into account all the different possibilities of orientations that can occur. In the lecture slide, it showed how for one CO molecule there are two possibilities for its positions: C on the left and O on the right or C on the right and O on the left. Then, because w...

Aayushi Jani 3A

I don't think degeneracy has any units because it is just the number of different states that could exist for the particular situation you are presented with.

Aayushi Jani 3A

The total entropy (S total) is S(system) + S(surroundings)

Aayushi Jani 3A

Like others have said, the difference is in the wording. If work is being done BY the system, then it is negative, but if work is being done ON the system, it is positive.

Aayushi Jani 3A

Yeah, I have the same issue. Since Week 5 isn't over yet, we only need to have completed 20 posts (5 posts a week for 4 weeks), so it technically should be out of 20 points. Hopefully the grades will be updated at the end of this week to reflect what we actually got out of 25 points.

Aayushi Jani 3A

You should choose the R value for which the units cancel out so that the only units that remain in the overall equation are the ones for the variable you are trying to solve for. It is definitely helpful to write out all the units and then physically cross them out so that you can clearly see which ...

Aayushi Jani 3A

Like the others have stated, they are not interchangeable since molar heat capacity is for 1 mole while specific heat capacity is for 1 gram. Depending on whether the substance is given in moles or grams, you would use the appropriate one.

Aayushi Jani 3A

The change in Kelvin and Celsius would remain the same because they increase/decrease at the same rate. The only difference is that they have different starting points (Kelvin = Celsius + 273). For example, the change in temperature from 0 celsius to 1 celsius = 1, and in the Kelvin equivalent, 273 ...

Aayushi Jani 3A

Yes, I agree with the others, while temperature and heat are closely related, I think thermodynamics mostly focuses on the transfer of heat (which inevitably would have an effect on the surrounding temperature).

Aayushi Jani 3A

The equation for forming N2O5 is N2 + (5/2)O2 --> N2O5. You would have to write one more equation for the formation of NO (N2 + O2 --> 2NO). For this problem, all you have to do is essentially reform the three equations so that they add up to the final one. You would want the products & reactant...

Aayushi Jani 3A

Yes, I agree, the best way to prepare would be to make sure you understand how to do all the textbook & sapling problems related to thermochemistry for this midterm.

Aayushi Jani 3A

Yes, like the others have said, the standard enthalpy of formation of an element in its most stable form is 0. This means that O2 is the most stable form of oxygen that exists in nature, and thus would have a standard enthalpy of formation equals 0.

Aayushi Jani 3A

Yes they are the same thing!

Aayushi Jani 3A

q is the measurement for heat (in joules) and H is for enthalpy. While enthalpy is a state property (not dependent on the path taken to obtain that state), heat is not.

Aayushi Jani 3A

Like the others have said, I would think that all we have to know is that graphite is the most stable form of carbon and the standard enthalpy of formation is 0 kJ/mol (since the standard enthalpy of formation of an element in its most stable form is 0).

Aayushi Jani 3A

Yes, endothermic reactions have a positive delta H since they require energy while exothermic reactions have a negative delta H since they release energy.

Aayushi Jani 3A

Yes, the units for ΔHf are kJ/mol while the units for ΔHrxn are just kJ (since it is representing the total of the overall reaction).

Aayushi Jani 3A

In order to get the desired equation, 2NH3(g)+3I2(g)↽−−⇀N2(g)+6HI(g), you have to add two of the equations that are given. From the four equations given, we can see that adding the first and third one will get us our desired equation. However, realize that the first equation must be multiplied by 3 ...

Aayushi Jani 3A

pKa & pKb tells you how strong a particular acid/base is, while pH and pOH simply tell you the concentration of OH- and H3O+. However, both are used to describe acids and bases.

Aayushi Jani 3A

Yes, the stronger the base, the larger the Kb value and lower the pKb value.

Aayushi Jani 3A

I would look at outlines 1 & 2 on his website because those are the ones we've done/are doing so far. The bullet points on the outlines tell you all the material you need to know.

Aayushi Jani 3A

Q is the reaction quotient (taken at any time during the reaction). If you let the reaction go until it reaches equilibrium, Q should equal K (since K is the equilibrium constant).

Aayushi Jani 3A

Yes, as others stated above, you would need to know whether the reaction is endothermic or exothermic (given by delta h).

Aayushi Jani 3A

I think that these values are just chosen as they are both clearly very small or large values, so if the value of K was greater/less than that, it would definitely imply that it sits to the left/right.

Aayushi Jani 3A

I think that that is possible looking at it mathematically (for example if the equation is 2R --> P where [R] = 2 mol/L and [P] = 4 mol/L, then K = 2^2/4 = 1). However, I think it is still rare for this to occur, but feel free to correct me if I'm wrong.

Aayushi Jani 3A

If you have two positive values, chances are one of them is larger than the initial concentrations. If x is larger than the initial concentration, then the equilibrium concentration value becomes negative, which is impossible.

Aayushi Jani 3A

We use Kelvin.

Aayushi Jani 3A

There would be no change as the moles on both side are equal (the reaction is not favored to go to any side).

Aayushi Jani 3A

They will be updating the points in a few days (since we were given an extension).

Aayushi Jani 3A

Yes, as Dr. Lavelle mentioned in the lecture, the most common prefixes are mono-, di-, tri-, tetra-, penta-, hexa-.

Aayushi Jani 3A

HF is not a strong acid as F is so electronegative, which means that HF does not easily dissociate into H+ and F- easily in water. This is why, as atomic radius increases, the acids become stronger as they are able to dissociate much easier.

Aayushi Jani 3A

Bronsted Acids are proton donors (donate H+) while Bronsted Bases are proton acceptors (accept H+). Typically, Bronsted Acids have hydrogen (for example: HCl) which dissociates into H+ ions in water to form H3O+.

Aayushi Jani 3A

You would first find the pOH = -log[OH-]. Then, using the value for pOH, you can find the pH by doing 14-pOH = pH. Once you've converted all of your values to pH, you can see which is more acidic and which is more basic (acids have pH < 7, bases have pH > 7).

Aayushi Jani 3A

I think that an amphoteric compound tends to be a weak acid/base as it can be either acid or base. I would imagine that it would be difficult for a strong acid to behave as a base and vice versa.

Aayushi Jani 3A

I think that the pH scale goes from 0-14 (as it is found by finding -log[H+]). In order for the pH to be negative, the concentration of H+ would have to be greater than 1M.

Aayushi Jani 3A

Yes, the pOH scale also goes from 1-14, where 7 is neutral. Also, pOH + pH = 14.

Aayushi Jani 3A

Organic compounds (or at least most of them) contain C-H bonds (an example is C6H12O6 which is glucose). Inorganic compounds do not have C-H bonds.

Aayushi Jani 3A

There are four regions of electron density (three single bonds and one lone pair). Thus, the hybridized orbital that corresponds to four regions of electron density is sp3 (as there are four orbitals, one for each region of electron density).

Aayushi Jani 3A

I agree with the answer given above. The Cl2 that is outside of the brackets is not a part of the coordination sphere, so you would write it separately instead of grouping it with the chlorine inside the brackets. Thus, instead of it being trichloro, it would be simply chloro to represent the chlori...

Aayushi Jani 3A

A Bronsted Acid is a proton donor while a Bronsted Base is a proton acceptor. It is just another way to define/distinguish between acids and bases (another way acids and bases are described is Lewis Acid (electron pair acceptor) and Lewis Base (electron pair donor).

Aayushi Jani 3A

I think we would need to memorize it. However, it is pretty simple to remember (<7 is an acid, >7 is a base, 7 is neutral).

Aayushi Jani 3A

It can be written either way, they both work and represent the same thing.

Aayushi Jani 3A

Yes, I agree with the other answers given. OH2 and H2O are the same compound, however, it was written as OH2 to show the bonding order.

Aayushi Jani 3A

Yes, I think that we're expected to have the molecular shapes & bond angles memorized. However, there are many great infographics & charts online that concisely explain everything and I think Dr. Lavelle also does a really good job going in depth in his lectures.

Aayushi Jani 3A

Yes, I think that might have been an error since I also think that ethene has two sp2 hybridization orbitals.

Aayushi Jani 3A

With lone pairs, they are more delocalized than bonding electrons. As a result, lone pairs have greater electron-electron repulsion compared to bonding electrons, so they would want to be farther away.

Aayushi Jani 3A

Yes, all single bonds are sigma bonds while double bonds have 1 sigma and 1 pi bond and triple bonds have 1 sigma and 2 pi bonds.

Aayushi Jani 3A

Yes, I agree with the others. Sigma and pi bonds occur with all covalent bonds. However, single bonds only have a sigma bond while double & triple bonds have sigma & pi bonds.

Aayushi Jani 3A

Technically electronegativity is quantifiable (each element has a specific electronegativity number) so if those numbers were given to you, you would be able to find the exact difference. However, for now, we just need to know the trend that electronegativity follows and we don't need to do exact ca...

Aayushi Jani 3A

Helium is a noble gas, so it does not want more electrons as its valence shell is already full. Fluorine, on the other hand, is a halogen, and only needs one more electron to form an octet. Thus, it has the higher electronegativity.

Aayushi Jani 3A

Yes, LDF are found in all molecules and they are the weakest of all the intermolecular forces.

Aayushi Jani 3A

Yes, it is called bent as there are two lone pairs of electrons and two pairs of bonded electrons.

Aayushi Jani 3A

I think that we can expect it to take around the same time as last time (I'm guessing we'll get our scores sometime next week, after everyone has taken the midterm this week).

Aayushi Jani 3A

So for all of these questions, you would have to use the expected charges. For a, we know that magnesium has a +2 charge in order to be stable and arsenic has a -3 charge to be stable. To form the chemical formula, you would use the charges to decide how many of each ion is needed to make a stable c...

Aayushi Jani 3A

Some species have an odd number of valence electrons, and therefore cannot form a full octet, and may have an unpaired electron (known as a radical). For this problem, I would draw each Lewis structure to see which structures have an unpaired electron.

Aayushi Jani 3A

Radicals have one unpaired electron (for example, in NO). Radicals form when species have an odd number of valence electrons (for example, N has 5 valence electrons).

Aayushi Jani 3A

For this problem, I first determined the number/moles of each atom/molecule. So for example, for part a, which is ammonium chloride, you know that ammonium is NH4 and chloride (chlorine) is just Cl. In order for both of them to be stable, NH4 needs to get rid of one electron and Cl needs to gain an ...

Aayushi Jani 3A

In the lecture, it was mentioned that high polarizability means higher attractive force. The greater the number of electrons, the more polarizable it is. When there is high polarizability, and therefore more attraction, instead of it being a liquid, it would form a more waxy solid (as seen in the le...

Aayushi Jani 3A

London dispersion forces occur between all molecules and are also known as Van der Waals forces and induced dipole-induced dipole forces in the lecture. In this question, you would look at only nonpolar molecules as they would only exhibit London dispersion forces (others would exhibit London disper...

Aayushi Jani 3A

Yes this is true as the Lewis base is the one donating electrons while the Lewis acid gains the electrons.

Aayushi Jani 3A

Yes, I believe that is correct. You would have one electron in each of the px, py, and pz orbitals to have the lowest energy.

Aayushi Jani 3A

A stable structure is one that has a formal charge of 0. This is because it would have the lowest energy.

Aayushi Jani 3A

I think that finding the formal charge is a good way to check that your Lewis structure is the most stable. I agree, I think that it is mostly used for structures with resonance as they have multiple Lewis structures for the molecule.

Aayushi Jani 3A

This is also a double bond related question. How do we exactly know when there has to be a double bond when creating the Lewis Structure? We know what bonds to use based on the amount of available valence electrons and the octet rule. For example, in NO3-, if you try to use only single bonds, you w...

Aayushi Jani 3A

Yes, the charge represents the charge of the polyatomic ion/molecule. NO3 has a -1 charge, hence the -1 sign at the top right. This is because the Lewis structure drawing itself cannot showcase the specific charge of the molecule, so it is shown in this way.

Aayushi Jani 3A

Yes I agree, it can be on any side. What matters is that the number of electrons add up to the available valence electrons and that the appropriate bonds are used (single, double, triple).

Aayushi Jani 3A

Ionic bonds form between a metal and a nonmetal atom (an example from the lecture: KCl, potassium chloride). In an ionic bond, the electron transfer results in an an octet of electrons for each atom. The cations lose electrons and the anions gain electrons. In covalent bonds, however, cations cannot...

Aayushi Jani 3A

h-bar is h/2pi. You can use either equation, but personally, I prefer just remembering h and using ∆p ∆x ≥ h/4π.

Aayushi Jani 3A

The uncertainty in position refers to the position of the electron (which is in meters), so it would also be in meters.

Aayushi Jani 3A

En refers to the energy at an energy level, n. The E in E=hv refers to the energy of the incoming photon, while Ek in Ek = 1/2 mv^2 refers to the energy of the electron.

Aayushi Jani 3A

l could be 4, but I think that in this class, we'll only deal with elements up to l=3.

Aayushi Jani 3A

p represents momentum, which can be found by multiplying mass and velocity.

Aayushi Jani 3A

For this question, l refers to the orbital angular momentum, which is found by finding n-1. However, the l values that are possible are everything from 0 up to n-1. Therefore, when n=7, the possible values can be 0, 1, 2, 3, 4, 5, 6 which are 7 total values.

Aayushi Jani 3A

We haven't learned how to utilize it mathematically yet, however, conceptually, it essentially describes the concept that an electron (which has wavelike properties and indeterminacy in momentum) can be described by a wave function.

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