CHEMISTRY COMMUNITY

Juwon Lim 2A

There are two typos in this question that were corrected in the solution manual errors pdf on Dr.Lavelle's website. First, 37 degrees C should be 310 K and the -0.59 you mentioned should be positive 0.59.

Juwon Lim 2A

First I identified the two atoms that were being reduced/oxidized, which is Cr and C. Then I wrote the half-reactions and balanced them accordingly using H2O and H+ because this is in an acidic condition. Then, I balanced the charges and combined the half-reactions to form the balanced overall react...

Juwon Lim 2A

All metals are good conductors so if there are metal solids then there is no need to add a conductor. If there is an ambiguous solid, I would assume it would be stated in the question in the form of "the non-conductive/conductive solid is oxidized/reduced..."

Juwon Lim 2A

Because the four questions are essentially the same in concept, I will explain a). First, write out the half reactions: Cr3+ + e- --> Cr2+ , Cu2+ + 2e---> Cu Then, find the standard reduction potentials using the Appendix. In the appendix, the reactions may be flipped, but this does not matter, just...

Juwon Lim 2A

When the oxidation numbers are added up, its sum should equal the charge of the molecule. Cl2O7: Its charge is 0, and O is generally 2-. Therefore, we can make the equation 0 = 2(Cl) + 7(O). Then, this becomes 0 = 2(Cl) + 7(-2). FInally, solve for Cl: 14 = 2Cl; Cl = 7. ClO2-: Its charge is -1, and O...

Juwon Lim 2A

When t is close to t=0, this is just saying that there is little product which is important because we want to determine the reactants' effect on the rate of the reaction. If N2 gas leaves the solution, this means there won't be any product in the solution and it follows that the reverse reaction ca...

Juwon Lim 2A

The given rates should be for the forward reaction. We use initial rates to make sure that the reverse reaction does not happen and so that we only have concentrations of the reactants.

Juwon Lim 2A

Use the equation Rate = k[A]^n[B]^m to find Rate2/Rate1. Then, k cancels and you should be left with Rate = [A]^n or Rate = [B]^m. Then, use the given intial rates to solve for n or m

Juwon Lim 2A

Order tells us how many reactants determine the rate of the reaction. First order means one reactant determines the rate, second order means two, etc.

Juwon Lim 2A

I believe if the rate is doubled n and m = 1 but for other rates, we need to use the mathematical analysis.

Juwon Lim 2A

Recharging batteries is not a spontaneous process so delta G will be negative. Then, using the delta G = -nFE equation, we can deduce that E will be negative, which would be impossible under normal circumstances. I don't think this would affect the cell diagrams, although the electrons would now flo...

Juwon Lim 2A

A nonspontaneous reaction cannot occur spontaneously, and reverse reactions for batteries cannot occur spontaneously. If we visualize the cell diagram, the reaction goes from an anode, which loses electrons, to the cathode, which gains the electron. The reaction cannot occur starting from the cathod...

Juwon Lim 2A

In a voltage, the reaction spontaneously occurs in the forward direction. If we start with more products, the reaction will not be favorable.

Juwon Lim 2A

I think in this scenario, we can refer to the definition of reversibility learned in thermodynamics. In thermodynamics, we referred to reversible work as maximum work, which was calculated by summing work done in infinitely small volume changes. In each of these steps, the external pressure matched ...

Juwon Lim 2A

To figure this out, first, we must identify which elements are being oxidized or reduced in the general reaction. The oxidized element and its half reaction will be the one that is reversed.

Juwon Lim 2A

I believe ideal gases, because they do not follow the forces of attraction that real gases do. Therefore, ideal gases are more free in their movement than real gases, which increases entropy.

Juwon Lim 2A

It can be used for both. ∆S = nRlnV2/V1 is the same equation as ∆S = qrev/T.
In a reversible reaction, qrev = -w. The equation for w = -nRTlnV2/V1. Thus, qrev = nRTlnV2/V1. Then, using the equation ∆S = qrev/T, we get ∆S = nRlnV2/V1.

Juwon Lim 2A

In an isothermal reversible reaction, the temperature of the system does not change because every time the system does work, the energy lost is replaced by heat from the surroundings, hence q = -w. Then, if we calculate entropy using the equation ∆S = q/T, we find that ∆Ssys = q/T and ∆Ssurr = -q/T....

Juwon Lim 2A

c) We have established that the system will dissolve meaning there will be an increase in entropy and that the enthalpy of the system does not favor this process. Therefore, the increase in entropy can be attributed to positional disorder and not thermal. d)If the system requires 34.9kJ of energy fo...

Juwon Lim 2A

A bunch of the textbook questions assumed standard conditions when it was not clear from reading the problem, which is pretty frustrating. For the textbook, I think it is safe to assume standard conditions when not said otherwise, however, for the test, I think we will be provided with the temperatu...

Juwon Lim 2A

It's like heat. An exothermic reaction is represented by a negative sign because heat is released by the system and an endothermic reaction is represented by a positive sign because heat is added to the system. In a similar fashion, when work is done by the system, it is represented by a negative si...

Juwon Lim 2A

Professor Lavelle has provided us with all the equations we have needed for exams in the past so if we were to need these equations he would probably give it to us on the question. However, I would be familiar with those equations because we have seen them in multiple questions previously.

Juwon Lim 2A

Since reversible expansions do not result in an overall temperature change, I would say that all reversible expansions are isothermal.

Juwon Lim 2A

In a reversible reaction, a piston slowly increases the volume of the container, so that work can be done in small increments. In this system, heat will be added to the system as work is being done in small increments so that the overall temperature of the system does not change. However, in irrever...

Juwon Lim 2A

In an isolated system, heat and matter cannot be exchanged with the system and its surroundings. This means that the number of "states" possible is confined to those possible inside this system, which is greater than if the system was not isolated. At equilibrium, these states will be all ...

Juwon Lim 2A

In summary, Exothermic reactions have negative Delta H and endothermic reactions have positive Delta H. In exothermic reactions, heat is transferred out of the system, so the system loses heat. Therefore, Delta H is negative. In endothermic reactions, heat is transferred into the system, so the syst...

Juwon Lim 2A

I would approach this the same way as determining whether a system is absorbing heat (endothermic) or outputting heat (exothermic). When work is done on the system, energy is transferred to the system, and when work is done by the system, energy is transferred out of the system. When something melts...

Juwon Lim 2A

This is correct. The pressure of an open system remains constant throughout reactions because the pressure created does not change the pressure of the external pressure ((which is the atmosphere) in a significant way.

Juwon Lim 2A

Isochoric, isothermal, and isobaric just mean that the net change is 0. Although the net change in temperature might be 0, this does not mean heat was not exchanged which does not make it isolated.

Juwon Lim 2A

Cv = Cp - R. We may learn in coming lectures, but I looked up this relationship while doing this problem.

Juwon Lim 2A

It depends on the calculator, but I would just use parentheses as much as possible even if it gets complicated. Another tip is to use E or the calculator equivalent of 10^x with more parentheses. Being organized takes a lot of work, but I find it better than doing the problem mutiple times.

Juwon Lim 2A

In general, you would calculate the enthalpies for bonds broken and bonds formed and find the sum. This would be done by looking at lewis structures, etc, but for the combustion of methane, all bonds are broken for the reactants and all bonds are formed so you would just add up the bond enthalpies f...

Juwon Lim 2A

I doubt Lavelle will expect us to memorize these. In the lectures, he says he looked up the enthalpies so I don't think we will need to memorize anything.

Juwon Lim 2A

A neutralization reaction occurs when an acid and a base react to form a salt and water. In this example, HCl (Acid) reacts with NaOH(base) to create NaCl (salt) and H2O (water). This process is called neutralization because the pH of the solution becomes neutral at equilibrium since water is created.

Juwon Lim 2A

There are state changes that need to be addressed. You would combine the 2nd and 3rd reactions, multiplying accordingly. Then you would add that reaction to the first reaction, multiplying accordingly. Then, the 4th reaction needs to be reversed so that MCl3 (aq) --> Mcl3 (s) and add that to the pre...

Juwon Lim 2A

I believe so. And to find the overall K of the reactions, we would just multiply the K values we found.

Juwon Lim 2A

This is sort of unrelated to the topic (I don't know how to make a topic on here), but in relation to Friday's lecture if the concentration of the hydronium ion is really small then do you just assume that the pH is 7? What that was Lavelle was saying? Because he gave an example of [H30+] being aro...

Juwon Lim 2A

Acids need to have an H to give off. For example, H2SO3 is an acid because it can give off hydrogens. Generally, the hydrogens that can be given off will be separate from other hydrogens in the molecule such as in acetic acid: CH3COOH. If a molecule doe not have hydrogen, it cannot be an acid Bases ...

Juwon Lim 2A

For strong acids/bases, we are allowed to take the negative log of the given concentration to find pH or pOH because strong acids/bases dissociate completely in water. This means the concentration of H3O or OH ions is the same as the initial concentration of the strong acid/base at equilibrium. Howe...

Juwon Lim 2A

To start, write out the dissociation reaction of lactic acid into lactate in water: CH3CH(OH)COOH(aq) + H2O(l) <-> CH3CH(OH)COO- + H3O+. From here, use the initial concentration of lactic acid (0.1173 M) to set up an ICE table. Then, use the K value given (8.4 x 10-4) and the equilibrium constant eq...

Juwon Lim 2A

It shouldn't matter whether the reaction is heterogeneous or homogenous. Pressure only affects gases, so we just add the moles of gas on each side of the reaction and compare.

Juwon Lim 2A

To add on, pure water is neither acidic nor basic. The acid (H3O+) and base (OH-) exist mostly in similar concentrations, so we say water is neutral. However, water can both act as an acid and a base when molecules are in water because of its amphiprotic properties.

Juwon Lim 2A

Q is calculated using the current reaction concentrations while K is calculated using future equilibrium concentrations. Say we increase the temperature of a reaction that is endothermic. Right now, K would increase because K is calculated using the future equilibrium concentrations. However, right ...

Juwon Lim 2A

Adding a liquid, a solid, or an inert gas does not affect the equilibrium.

Juwon Lim 2A

When we say that the equilibrium favors reactants/products we are saying that at equilibrium there are more reactants than products or vice versa. At equilibrium, reactants and products are both created, but at equal rates, so that the net concentrations of reactants/products do not change overall. ...

Juwon Lim 2A

a) When water is added, we are adding a reactant. Therefore products are favored. b) When the partial pressure of CO2 is decreased, we are decreasing reactants. Therefore, reactants are favored. in terms of the equilibrium composition, for a) [C6H12O6] and [O2] is increased and for b) [O2] is decrea...

Juwon Lim 2A

i. When volume is compressed, pressure increases. In this case, the side with fewer moles of gas will be favored. In this reaction, there are 3 moles of gas on the reactant side to the 2 moles of gas on the product side. Therefore, products are favored. ii. When the reaction is exothermic (which thi...

Juwon Lim 2A

To calculate K, you would have to wait until the reaction reaches equilibrium, and K does not change in this situation. However, Q will change because we are inputting new concentrations for products and/or reactants.

Juwon Lim 2A

The equation for the equilibrium constant is [Product]^coefficent / [Reactant]^coefficient. If all the stoichiometric coefficients are 1 then the expression should be [Product]^1/[Reactant]^1 or just [Product] / [Reactant]. This would mean the answer is B and the ratio would be the equilibrium conce...

Juwon Lim 2A

To solve this problem, we need to find an equation for concentration. We can do this by manipulating PV = nRT. Concentration is equal to n/V. Therefore, the equation for concentration is n/V = P/RT or Concentration = P/RT. From here we just plug in the given measurements and the result should be the...

Juwon Lim 2A

Since H2O is outside the brackets, this means H2O is not part of the coordination sphere/not bonded to the TM cation. Therefore, it is not considered a ligand and we would not use aqua to name the ligand, and instead use hydrate.

Juwon Lim 2A

In addition to the person above, Professor Lavelle said we will not need to distinguish between square planar and tetrahedral shapes for coordination compounds. If there are 6 bonds, the shape is octahedral, and if there are 4 bonds, the shape can be either square planar or tetrahedral.

Juwon Lim 2A

I would assume the number of ligands would equal the number of bonds to the transition metal, but to be safe, use the number of bonds to determine the shape.

Juwon Lim 2A

T-shape geometry would result from a trigonal bipyramidal shape with two lone pairs. A trigonal planar shape would result from a molecule with three bonds and no lone pairs. Because of this, T-shape and trigonal planar are different.

Juwon Lim 2A

The 3 bond angles would add up to 360 if the geometry of the molecule was trigonal planar. This is because, on a flat circle, the angles should add up to 360. However, SO3^2- has a lone pair making the shape tetrahedral. Then, the 360 rule does not apply since the shape is 3 dimensional.

Juwon Lim 2A

I also got the same answers

Juwon Lim 2A

The seesaw shape results due to the lone pair on a trigonal bipyramidal molecule. However, for coordination compounds, there won't be any lone pairs, so the seesaw shape would not be the most stable shape. Instead, the compound will form a tetrahedral or square planar shape.

Juwon Lim 2A

The overall charge of the compound is -1. Cl each have a charge of -1, and since there are two, this would bring the charge of the compound to -2. Therefore, to reach the overall charge of -1, Ag must have a charge of +1, so an oxidations state of 1+

Juwon Lim 2A

A hexagonal planar does have a coordination number of 6. However, the question asks for the geometries that are common for compounds that have a coordination number of 6. Hexagonal planar is very very rare, while octahedrals are much more common.

Juwon Lim 2A

For coordination compounds, I think it is more preferred to write ligands with the atom that binds to the transition metal first. Therefore, nitrito should be written as ONO instead of NO2 which would be called nitro.

Juwon Lim 2A

Everything inside the bracket is inside the coordination sphere, which means everything inside the bracket is bound to the transition metal, while everything outside the bracket is not bound. To distinguish between bound ligands and non-bound ligands, we use different names: chloro for chloride, aqu...

Juwon Lim 2A

An most important aspect of radicals are that they have an odd number of electrons. If you add up the electrons in a molecule and it is odd, that molecule is a radical. Another important aspect is that they are highly reactive. These two should be the only aspects we need to know so far.

Juwon Lim 2A

I remember it by thinking about what we are actually modeling. We are modeling the shape of the molecule which is a result of electron pairs repulsions. Since we only look at valence shell electrons, I fill in the acronym: Valence Shell, Electron Pair, Repulsion Model.

Juwon Lim 2A

Regions of electron density is determined by counting bonds and line pairs. Double bonds and Triple bonds count as 1 region of electron density, since they occupy one region.

Juwon Lim 2A

I believe we just have to memorize it. In this case, the formula is written out as H2CCCCH2, which is indicative of a chain. Benzene, for example, forms a ring, and it is written as C6H6, not H3CCCCCCH3.

Juwon Lim 2A

Either one is fine. I prefer to put the d after s and p because it is in order of increasing energy levels, but take your pick.

Juwon Lim 2A

A tetrahedral results from a central atom with 4 bonded atoms, while a square planar results from a central atom with 4 bonded atoms plus 2 lone pairs. An example of this would be XeF4. Xe has 4 bonded atoms (F) and 2 lone pairs. The lone pairs are as far away from each other as possible, which resu...

Juwon Lim 2A

To determine if a molecule will experience a dipole-dipole interaction, we need to first determine if the molecule is polar. Looking at the shape of this molecule, we can see that we have two, smaller (therefore highly electronegative) atoms on one side of the molecule, and a bigger (therefore less ...

Juwon Lim 2A

A molecule like XeF4 would have lone pairs as far as possible from each other. Professor Lavelle went over this molecule in his 11/18 lecture, if you need further help.

Juwon Lim 2A

The lone pair is found in the equatorial plane and not the axial plane because we want to minimize repulsion. If the lone pair was found on the axial plane, it would repel the 3 bonded pairs in the equatorial plane. If the lone pair was found on the equatorial plane, it would repel the 2 bonded pair...

Juwon Lim 2A

I was having trouble with this as well. What I've figured out so far is I draw the Lewis structure of the molecule. If the atoms attached to the central atom are identical and symmetrical, the molecule is generally non-polar. If there are different atoms attached to the central atom, the molecule is...

Juwon Lim 2A

Double bonds are used in Lewis Structures as a mathematical model. The actual structure of a molecule contain bond lengths that are equal to each other. This means that if there is a molecule that has one single bond and one double bond in the Lewis Structure, the actual molecule will have a 1.5 bon...

Juwon Lim 2A

Oxygen has a slightly higher ionization energy than Chlorine. However, remember that symmetry plays a role in determining the central atom so I doubt the difference in ionization energy will be very important

Juwon Lim 2A

Hydrogen bonds form between lone pairs in N, O, or F and hydrogen atoms bonded to N, O, or F. Therefore, this molecule can form 8 hydrogen bonds: two with the two lone pairs in O, two with the two lone pairs in N, and four with the four hydrogens.

Juwon Lim 2A

Polarizability is related to the attractive force of a molecule. The higher the polarizability, the higher the attractive force. In Professor Lavelle's example, the molecule on the left had less electrons which means low polarizability and low attractive force, compared to the molecule on the right....

Juwon Lim 2A

Hydrogen bonds occur between lone pairs in O,N, or F and Hydrogens bonded to O, N, or F.
In the first option, H4C cannot form a hydrogen bond
In the second option, H in H-CH3 is bonded to C, therefore H-CH3 cannot form a hydrogen bond.

Juwon Lim 2A

I think we represent double and single bonds with one or two lines in the Lewis structures for sake of simplicity. It is easier to draw 2 lines to represent double bonds than to draw two 1.3 lines, even though the 1.3 bond would be more representative of the actual structure of the molecule.

Juwon Lim 2A

Electrons are removed from the highest state. If there are s, p, and d- Orbitals, then electrons will be removed from the d-Orbital first.

Juwon Lim 2A

To have an octet, we need 1s, 2s, 2p orbitals because the outer shells, 2s and 2p, add up to a possible total of 8 valence electrons. When these two outer shells have been filled, we call that an octet. the first element that can have electrons in the p-Orbital is Boron, so elements before boron (H,...

Juwon Lim 2A

If there are unpaired electrons, generally the atom is reactive. For expanded octets, the electrons are paired so they are not reactive. Radicals have unpaired electrons so they are reactive.

Juwon Lim 2A

Yes, this phenomenon applies to other related configurations, but since we are only studying up to the first row of the d-block (3d), we don't have to know the rules for other d-blocks.

Juwon Lim 2A

There are a max of 2 electrons per orbital. So if n = 3, there can be 3s, 3p, 3d Orbitals. There is 1 type of s-Orbitals, 3 types of p-Orbitals, and 5 types of d-Orbitals. Therefore, if each orbital can have 2 electrons max, we can have (3s) + (3p) + (3d) = (2) + (6) + (10) = 18 electrons total for ...

Juwon Lim 2A

Excited electrons refer to when an electron has moved up a shell, while ground state electrons are at the base electron configuration. To solve this problem, first identify the ground state configuration for Oxygen, then look for an electron that as moved up a shell.

Juwon Lim 2A

2d cannot exist because the quantum numbers would be as follows: n = 2, l = 2. However, l = n-1, and 2-1 cannot equal 2. 4g cannot exist for similar reasons because its quantum numbers would be as follows: n = 4, l > 3 (Since f Orbitals are l = 3, g orbitals must be greater than l = 3). Since l = n-...

Juwon Lim 2A

By putting a noble gas, for example Ar (argon) helps to reduce the length of electron configurations. We know that the electron configuration of Ar is 1s^2 2s^2 2p^6, 3s^2, 3p^6, so writing all that out again for Cr (Chromium) would take a while and look messier than it has to be.

Juwon Lim 2A

My TA said Professor Lavelle said we need to know the different wave types and their relative frequencies. I'm assuming this is for the midterm, so good luck!

Juwon Lim 2A

Unless specified, answers should be in SI units, so for wavelength, meters. If the question asks for nanometers or picometers, you would answer in the specified units. This goes for all other measurements.

Juwon Lim 2A

I was also confused on the way he wrote the list of possible ml values at first. What he means by ml = l, l-1, ..., -l is that ml can be any value between l and -l. If l = 2, the possible ml values are ml = 2, 1, 0, -1, -2. On a side note, ml values represent the orientation of the orbital. Therefor...

Juwon Lim 2A

A degenerate electron system simply means all orbitals have the same energy. In multi-electron systems, s-Orbitals have less energy than p-Orbitals due to shielding. However, if there is only one electron, there cannot be such shielding, and therefore, have the same energy.

Juwon Lim 2A

Recall that are there 7 types of f-Orbitals and 5 types of d-Orbitals. ml simply represents one type of these orbitals. Since we know that there can be a maximum of two electrons per orbital, if we are given n, l, and ml, we can conclude that 2 electrons can have those quantum numbers. When we are g...

Juwon Lim 2A

Recall that there can be a maximum of 2 electrons per orbital, and that no two electrons can have identical four quantum numbers. In the question, n = 6, l = 3, ml = -1. This tells us that this electron is in one of the 7 f-orbitals, since l = 3. Since there can be a maximum of 2 electrons per orbit...

Juwon Lim 2A

Measurements must be converted into SI units: kg for mass, m for length, s for time, etc. This is because non-intuitive units such as joules (J) are measured in these units. If we were to plug measurements that were not in SI units, we would not be able to cancel out the necessary units. If you need...

Juwon Lim 2A

Yes.
Just as we are allowed to change negative velocity into positive speed, we are allowed to change negative delta E into positive energy of the photon (Ep)
The negative in velocity just means that we are moving left, and the negative delta E just means that we are emitting light

Juwon Lim 2A

Electrons were ejected in the photoelectric effect experiment. This means the electrons were completely removed from the metal surface.

Electrons were excited in the absorption line spectroscopy experiment. This means the electrons moved up the electron levels but was not completely ejected.

Juwon Lim 2A

Joules are measured in the following units: kg.m^{2}.s^{-2} (Kilograms times meters squared per second squared) Mass (of the electron) is measured in the following units: kg (kilograms) Velocity is measured in the following units: m.s^{-1} (meters per second) Now if we plug just the units into the e...

Juwon Lim 2A

Increasing intensity means increasing amplitude which also means increased number of photons. Essentially, by increasing the amplitude of the wave, we are simply increasing the number of photons. Although this might not be the most correct way of thinking about a photon, I like to visualize photons ...

Juwon Lim 2A

We are allowed to have periodic tables out for tests/quizzes. The periodic table on Professor Lavelle's site has both symbols and names of each element so you will not need to memorize these. Here's the link: https://lavelle.chem.ucla.edu/wp-conten ... _IUPAC.pdf

Juwon Lim 2A

I think Professor Lavelle said in a previous lecture that chemists prefer coefficients to be integers instead of fractions. I know it seems bothersome to convert decimals and fractions into integers, but I think the safe bet is to convert them into integers.

Juwon Lim 2A

Leading zeros do not count as significant figures. 0.034 would be 2 sig figs and 0.0034 would still be 2 sig figs. Trailing zeros do not count as significant figures unless there are decimal points. 100 would be 1 sig fig, however 100.00 would be 5 sig figs. To answer the question, 0.034 would be 2 ...

Juwon Lim 2A

Generally, you want to use the least precise measurement to determine how many sig figs your answer will have. When multiplying/dividing, your answer should have as many sig figs as the measurement given that has the least sig figs. When adding/subtracting, your answer should have as many decimal po...

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