CHEMISTRY COMMUNITY

Kaitlyn Ang 1J

And the ln(k) = -Ea/RT + ln(A) version is useful when you're comparing two temperature's k values. The equation becomes ln(k2/k1) = (Ea/R)((1/T1) - (1/T2))

Kaitlyn Ang 1J

Think of it like a bottleneck. The slowest step is what determines the rate, because the reaction can't get any slower (it can get faster, but what we want is the maximum amount of reactant change possible which comes with the slowest step). The slowest step is the rate determining step.

Kaitlyn Ang 1J

Each approach varies in usefulness depending on the context of the problem, the complexity of the reaction, and what information is provided. The differential rate law approach expresses the reaction rate in terms of changes in the concentration of one or more reactants (Δ[R]) over a specific time ...

Kaitlyn Ang 1J

Another way to say this is that if you have the general rate = k[A]^m[B]^n, the overall order would be m+n
(aka add the exponents of each reactant)

Kaitlyn Ang 1J

You would use "pseudo" rate orders when one reactant is in excess so it doesn't really affect the rate. The pseudo constant k' is k[reactant in excess]^m. Using the pseudo constant, you can determine the order of the reactant not in excess and using experimental data, you can determine k'....

Kaitlyn Ang 1J

You can see this through the other form of the Arrhenius Equation: k = Ae^(-Ea/RT) ln(k1) = -Ea/RT1 + ln(A) ln(k2) = -Ea/RT2 + ln(A) ln(A) = ln(k2) + Ea/RT2 ln(A) = ln(k1) + Ea/RT1 ln(k2) + Ea/RT2 = ln(k1) + Ea(RT1) ln(k2/k1) = Ea/RT1 - Ea/RT2 ln(k2/k1) = Ea/R((1/T1)-(1/T2)) where a high positive Ea...

Kaitlyn Ang 1J

Additionally, changing the mass of the electrode (the solid) doesn't affect anything. Only changing the concentration of aqueous ions will change E cell (because E cell = E standard - (RT/nF)lnQ

Kaitlyn Ang 1J

Thank you for everything you've done these past two quarters, Dr. Lavelle! I know this situation was very unexpected and difficult to manage, but I think you handled it wonderfully. *high five* :) Thank you for all you've done this quarter! I know it must have been really difficult remaking the fin...

Kaitlyn Ang 1J

Abigail Sanders 1E wrote:Will this review session still be held now that all classes are cancelled? Lyndon's sessions are normally the most helpful thing before the finals/midterms.

Facts! If not hosted in person, maybe livestreamed...?

Kaitlyn Ang 1J

Thank you Lyndon for putting so much time into us. I’m not alone in saying that your weekly workshops, your packets, and your review sessions are probably the only reason I passed 14A and (hopefully!) pass 14B.

P.S. Endgame is a suitable name for replacing Gummy Worms

Kaitlyn Ang 1J

After you balance the redox reactions so that the "number" of electrons from the half reactions cancel, that common "number" of electrons is your n

Kaitlyn Ang 1J

The rate law shows the order of each reactant. The order of an individual reactant is the power it's raised to (zero if it's not in the rate law at all), the overall reaction order is the sum of all individual reactant orders. The overall reaction order is N + M in the equation Lavelle showed in cl...

Kaitlyn Ang 1J

Add Pt(s) whenever there is no solid in the electrodes

Kaitlyn Ang 1J

When pKa is smaller than pH, that means that Ka is larger than the H+ concentration, meaning that the products (H+ concentration included) is larger, so since there is a higher H+ concentration than the neutral pH indicates, this means that the acid protonates.

Kaitlyn Ang 1J

I'm not sure it's necessarily easier, but it is standard and lets us know that we're comparing the same thing when looking at different reactions or different concentrations of reactants. If we didn't have this standard, we would be comparing rates from all different points at a reaction, making it...

Kaitlyn Ang 1J

DHavo_1E wrote:Hello,

Will we have to know about entropy and enthalpy? Thank you!

Probably only in connection to the Van'T Hoff Equation.

Kaitlyn Ang 1J

A similar question like this popped up on Test 2 last year, but it clarified which oxidation states were being used (i.e. Fe3+ to Fe2+ instead of Fe3+ to Fe). The UA said that in a scenario like this where you have to order which metal is a better reducing/oxidizing agent, they would clarify, so don...

Kaitlyn Ang 1J

Yes, standard convention is that anode will be on the left and cathode on the right

Kaitlyn Ang 1J

Because you have to balance the S2O3 and SO4, you have to add H2O to get the additional oxygens to form the SO4 product, and then since you added hydrogens as part of the H2O, you need to add H+ to the reactants.

Kaitlyn Ang 1J

For basic solutions, you would use H2O to balance the oxygens, then add OH- (instead of H+) to balance the hydrogens, however, since we added additional oxygens from the OH-, we have to add water again, but this time to the other side, to balance the additional oxygens out. Honestly, basic solutions...

Kaitlyn Ang 1J

Yes, there is a typo. It should be Cl2 -> HClO + Cl-

Kaitlyn Ang 1J

I'm assuming you're talking about 6K.3 a) That one was hard for me too, but basically, you balance the 5H20 on the reactant side with 10H+ on the products side. Because you have 10H+, there are 10e- along side it. However, there is a charge difference from the reactants to the products where S2O3^2-...

Kaitlyn Ang 1J

It's mainly on the overall Ecell voltage. Since Ecell will always be positive, flip the one that will give you that overall positive Ecell.

Kaitlyn Ang 1J

What I did was I found Ka and used the pH given to find the concentration of H+. Since the equation for Ka is [H+][A-]/[HA], you can find the ratio of conjugate base to acid by dividing Ka by the concentration of H+. But how would you know if it stays as a salt in the stomach or becomes something e...

Kaitlyn Ang 1J

You have to think about how Ka=[A][H+]/[CB] (CB being conjugate base). By rearranging the equation, you can get Ka/[H+] = [A]/[CB], which is a ratio. Were we given this equation anywhere? I mean like the original, not manipulated, equation? And I don't recall using it before the midterm but I may b...

Kaitlyn Ang 1J

You can think about the relationship between $\Delta G = -nFE$ and since you know a spontaneous reaction is when $\Delta G$ is negative, this only happens when $E$ is positive (since n and F are always positive).

Kaitlyn Ang 1J

So Dr. Lavelle mentioned this in lecture. You need a solid electrode for the cell diagram and when there isn't for one side you need to have a element that wouldn't interact with the solution . An inert electrode such as Pt is necessary when both oxidized and reduced species are in the same solutio...

Kaitlyn Ang 1J

As everyone above said, Gibbs free energy is how you determine if a reaction is spontaneous or not. However, it might be important to remember that the reason Gibbs free energy is so clear cut is that it’s because it’s based off total entropy which is always increasing.

Kaitlyn Ang 1J

You can also calculate deltaG the same way. I think for any state function you can calculate it this way because all we care about is the change from initial to final so in reactions, the “final” stage is the products and the “initial” stage is the reactants, so you would do products - reactants, w...

Kaitlyn Ang 1J

The definition given in class is that Gibbs free energy is the energy given off that’s now available (“free”) to do work. Basically, if deltaG is negative, that means it gives off energy that is now “free” to do work, which results in a spontaneous reaction. The important thing to know is that when ...

Kaitlyn Ang 1J

What I did was I found Ka and used the pH given to find the concentration of H+. Since the equation for Ka is [H+][A-]/[HA], you can find the ratio of conjugate base to acid by dividing Ka by the concentration of H+.

Kaitlyn Ang 1J

I don't know if this is the right section to post this question under, but I know state functions is a topic in thermodynamics and this was the closest category I could find in asking this question. In Lyndon's review session, he mentioned that the answer to 3H "Which of the following are not s...

Kaitlyn Ang 1J

For this problem, you have 3 mini reactions. The first is the rise in temperature of the reactants. The second is the change from reactants to products at the higher temperature. The third is the decrease in temperature of the products. In other words, the three steps are 1) the enthalpy change of t...

Kaitlyn Ang 1J

Not necessarily. Isothermal means that there is no temperature change (aka deltaT = 0) whereas reversibility has to do with entropy and if total entropy is 0 (aka in a reversible system) or if entropy of the surroundings is 0 (aka in an irreversible system).

Kaitlyn Ang 1J

We cannot have negative entropies because entropy is defined as the energy needed to break a bond, which is always positive (aka it always requires energy to break a bond)

Kaitlyn Ang 1J

If you have L.atm and want to get to joules, you wouldn't use a gas constant, but you would multiply by 101.325 J.L^-1.atm^-1 to get the L.atm to cancel out with only J left

Kaitlyn Ang 1J

Benzene is more stable and less reactive than would be predicted from its Kekulé structures. Use the data in Table 4E.3 to calculate the lowering in molar energy when resonance is allowed between the Kekulé structures of benzene. When we calculate the molar energy without resonance, we get 2880 kJ,...

Kaitlyn Ang 1J

Adding to this, you know work = 0 when $\Delta{V} = 0$ because $w = -P\Delta{V}$ so 0 for $\Delta{V}$ (aka there is no change in volume, aka the volume is constant) makes the entire work equation 0

Kaitlyn Ang 1J

For equations with constant pressure, you can assume that $q_{p} = \Delta{H}$ , and for equations with constant volume, $\Delta{V} = 0$ so w = 0 (since $w = -P\Delta{V}$ )

Kaitlyn Ang 1J

You can think of \Delta{U} = q + w where q = heat and w = work. Since you know q_{p} (heat with constant pressure) can also be considered \Delta{H} , you now have \Delta{U} = \Delta{H} + w . To make \Delta{U} = \Delta{H} , you have to make work equal 0. Since w = -P\Delta{V} , work equals 0 when \De...

Kaitlyn Ang 1J

Another example is the difference between specific heat capacity and heat capacity. Specific heat capacity is the heat required to raise the temperature of an object per mole. Since the "per mole" is there, regardless of how many moles are added, the heat required per mole is the same, the...

Kaitlyn Ang 1J

It also provides context on why we add and subtract bond enthalpies to calculate the overall change in enthalpy. When we draw Lewis structure, we show that we know that it's because of the bonds breaking and forming that we get changes in enthalpy.

Kaitlyn Ang 1J

Standard enthalpy of formation ( $\Delta H^{o}_{f}$ ) is the standard reaction enthalpy ( $\Delta H^{o}$ ) to form one mole of the substance, which explains why the units for $\Delta H^{o}_{f}$ is kJ/mol

Kaitlyn Ang 1J

However, if they give you the standard state of the entire reaction (i.e. if they said that $\Delta H^{o} = +212kJ$ ) you don't have to worry about calculating each individual element's standard state, because they've shortened your work for you by giving you the overall standard state.

Kaitlyn Ang 1J

Because enthalpy is a state function, meaning it depends on the physical state of the molecule, enthalpy changes are additive.

Kaitlyn Ang 1J

Tiffany Chao 2H wrote:Here: download/file.php?id=5559

Thanks :))

Kaitlyn Ang 1J

Hi!

Where can we find the practice test ws?

Kaitlyn Ang 1J

We are given .20 M of C6H5NH2 which has a Kb of 4.3 x 10^-10 and we have to find the pH. The first way I tried it was finding Ka by 1.0x10^-14/4.3x10^-10 and then finding the concentration of H30+ and then calculating pH from it with -log[H3O+]. However, this got me the wrong answer. The second way ...

Kaitlyn Ang 1J

If one of the products in the reaction is H3O+, use Ka because the original compound given was an acid.
If one of the products in the reaction is OH-, use Kb because the original compound given was a base.

Kaitlyn Ang 1J

I believe so. It's Topic 1 and 2 on the syllabus.

Kaitlyn Ang 1J

Yes, increasing the pressure by decreasing the volume makes the side of the reaction (products or reactants) with less moles of gas favorable. The reason is that the concentrations of everything changes by the amount of volume lowered (aka if V halves, the concentration would double) which alters Q,...

Kaitlyn Ang 1J

I also got 3.2 x 10^-15 M

Kaitlyn Ang 1J

I understand the difference between pKa and pH in terms of the formula, but I don't really understand how they differ conceptually. Adding to this, I don't understand why when pKa > pH, it's acidic and when pKa < pH, it's basic. Does it have something to do with the conjugate base? It has to be righ...

Kaitlyn Ang 1J

Adding to this, for example, if you were asked to find K and you were given that N2 had an initial concentration of .1 mol and 60% of it turned into NH3 at equilibrium, you know that there is a final concentration of .06 mol NH3. And by doing ICE calculations, you can calculate the equilibrium molar...

Kaitlyn Ang 1J

I think Lavelle mentioned this in lecture, but yeah, in an experiment, Q wouldn't be greater than K, however, if you were taking a sample of a river let’s say, you don’t know the exact step in the reaction that your sample is in, so there could be more products than reactants because of you writing ...

Kaitlyn Ang 1J

I think that since you’re compressing the entire system, both the pressure of the reactants and products are increasing by the same amount, so in a sense, the ratios aren’t changing so there is little/no effect on the system.

Kaitlyn Ang 1J

They’re extremely helpful for learning material as most Undergraduate Assistants (UAs) make their own practice problems for you to do and worksheets. They’re also available to go to for almost any time of day and any day, so if you ever have free time, you can see which ones are available (which can...

Kaitlyn Ang 1J

Most of the time, if you’re dealing with aqueous solutions, you use Kc. Also, a lot of time, the problems just have K without the subscript c or p, so if you want to be safe, you can just solve for K.

Kaitlyn Ang 1J

Yes it is

Kaitlyn Ang 1J

Yes, but according to what I've found online, I think the main difference is that hemoglobin has 4 heme complexes

Kaitlyn Ang 1J

As a follow up question, what are the general rules for determining the strengths of bases? For acids, it's through bond length and electronegativity, but I don't think Lavelle explicitly mentioned the criteria for bases.

Thanks!

Kaitlyn Ang 1J

The question asks about which acid is stronger in a pair of acids. My question is about part d where it asks which is stronger, HClO4 or H3PO4. How can we figure this out? My original thought process was that H3PO4 would be stronger because it's polyprotic, but the answer key says it's HClO4. Thanks!

Kaitlyn Ang 1J

Original meme content

Kaitlyn Ang 1J

Nevermind, I understand it now

Kaitlyn Ang 1J

So in terms of IMFs, more specifically H-bonds, we're taking into consideration the possible bonds between the H atom and the F, O, or N atom. When a compound has a F, O, or N atom, it is automatically a hydrogen bonding site (it's lone pairs of electrons each count as 1 site). Regardless if an H a...

Kaitlyn Ang 1J

Nikki Razal 4E wrote:How would you know that Bi2O3 is amphoteric?

Since Bi is one of the amphoteric elements (which you can tell because it’s near the metalloids), that means Bi2O3 is amphoteric.

Kaitlyn Ang 1J

B12 helps make DNA, helps make the myelin in neurons, and helps in the production of red blood cells.

Kaitlyn Ang 1J

Similarly, if there are atoms with more electronegativity, they result in a stronger acid because the more electronegative atoms are “better at pulling” the electron density away from the oxygens and spread more evenly across the molecule.

Kaitlyn Ang 1J

On test 2, there was a question (7a) about how many hydrogen bonds can form in paraxanthine. According to my TA, the correct answer should be 11 H-bonds, but I don’t understand why. I thought H-bonds were between H and N, O, F so shouldn’t we only count the areas where there are lone pairs on NOF or...

Kaitlyn Ang 1J

(a) 2 sigma bonds (b) 2 sigma and 2 pi bonds You can tell based on the lewis structure - single bonds are one sigma bond, double bonds are one sigma and one pi bond, and triple bonds are one sigma and two pi bonds. Note that there are 2 ways to draw SO2, one with 2 double bonds linking S with the 2...

Kaitlyn Ang 1J

There are 2 sigma bonds in H2S which are the single bonds between sulfur and the two hydrogens There are 2 sigma bonds and 1 pi bond in SO2 because there is a double bond connecting the sulfur to one oxygen (double bond = 1 sigma and 1 pi bond) and a single bond connecting the sulfur to the other ox...

Kaitlyn Ang 1J

I don't think it matters too much, but in general, I think it's preferred to write the hybridizations in the order of s, p, d, then f, so for your example, it would be sp^3d

Kaitlyn Ang 1J

Adding to this, Lavelle mentioned in class that the -ate adds an emphasis that it is an anion because normally, the compounds are cations

Kaitlyn Ang 1J

Yes, SnCl2 is not a radical, but you are correct in that it doesn't fill the octet, so Sn has a single bond with the 2 Cl and has only 1 lone pair of electrons

Kaitlyn Ang 1J

Adding to this, make sure to recognize that just because a Lewis structure might look like it's symmetrical, you have to take into account the VSEPR shape to see if the molecule truly is symmetrical (i.e. tetrahedrals such as CH2Cl2 in a Lewis structure looks symmetrical, but in VSEPR, it's not)

Kaitlyn Ang 1J

And along the same thread, if there is no E, you don't have to write E with a subscript of 0, you can just write AX4 or whatever

Kaitlyn Ang 1J

Lone pairs do influence the bond angles and contribute to molecular shape (i.e. H2O is a bent shape instead of linear because of the 2 lone pairs) but when we are asked to find bond angles or draw dipole moments, we focus on the bonded atoms and not the lone pairs. And though we know that H2O has te...

Kaitlyn Ang 1J

3F and 2E

2F is about sigma and pi bonds so we'll probably be covering that tomorrow.

Kaitlyn Ang 1J

After browsing chem community further, I think I have a better grasp on IPE. Since IPE is the amount of energy released when two atoms interact (which is always negative because it always /takes/ energy for atoms to interact) so when two atoms are more naturally attractive (aka more polarizable), th...

Kaitlyn Ang 1J

The two symbols are alpha1 and alpha2 which refer to the polarizability of each individual atom that is interacting with one another.

Kaitlyn Ang 1J

While London dispersion forces are the go-to for nonpolar molecules, make sure to also check for dipole-induced dipole intermolecular forces which is the interaction of a polar molecule and a nonpolar molecule.

Kaitlyn Ang 1J

In lecture, Lavelle said that "increasing size or molar mass results in stronger attractive interactions" so that means IPE gets closer to 0 because the denominator becomes larger, right? So I think it means that values closer to 0 are more attracted to each other. Saying this, I'm not 100...

Kaitlyn Ang 1J

I think that if there was a question with interaction potential energy, they would have to give you the radius and charges in order for you to calculate the IPE, or they would ask more conceptually and ask you given two IPEs, which sets of molecules have the greater radius which you don't need to ac...

Kaitlyn Ang 1J

I think this is what happens. When alpha increases, that means the polarizability increases, and since polarizability depends on the size of the atom (and how many e- there are), r increases too so the denominator "grows" faster than the alphas in the numerator so even though the numerator...

Kaitlyn Ang 1J

And when molecules with dipole moments are around other molecules with dipole moments, they can automatically arrange themselves so that their opposite dipoles are closer to each other (induced-dipoles) which is a type of attractive force that explains some of the patterns of how molecules arrange t...

Kaitlyn Ang 1J

Also, in context, this means that if the radius between the atoms increases, the denominator gets larger, meaning that the interaction potential energy gets less negative (so it has stronger attractive interactions).

Kaitlyn Ang 1J

You're correct in that IE increases from left to right, but Nitrogen has a higher IE because it has a half-filled p subshell which is more stable than Oxygen's 4 e- p subshell, so it actually requires more energy to take away that electron from Nitrogen than Oxygen

Kaitlyn Ang 1J

Since there's a negative sign in front of the equation, that means that when r increases, the denominator increases meaning it gets closer to 0, which means that Ep actually increases (from more negative to less negative).

Kaitlyn Ang 1J

The more polarizable an atom is, the more likely the atom's electron cloud can be distorted. Meaning that when an atom has a higher polarizability, the electrons are more likely to be attracted to the other atom (aka the original atom is more likely to lose or share it's electrons).

Kaitlyn Ang 1J

I don't know that there is a certain "max" amount of bonds those elements can have, but some of the most common molecules are PCl5 and SF6 which you can see has the same amount of bonds as they have valence electrons initially (aka before bonding). I think you just have to be aware it's po...

Kaitlyn Ang 1J

All atoms have some covalent character, but the atoms can have more ionic character (aka less covalent character) if there is a greater difference in electronegativity between the atoms. This also means that atoms with more covalent character (aka less ionic character) has a smaller difference in el...

Kaitlyn Ang 1J

In addition, if the element is in the d blocks, they are more likely to lose their "s" electrons first in order to fill a full or half shell of "d" electrons.

Kaitlyn Ang 1J

The book says to draw the dipole arrow with the head pointing to the positive side and the tail at the negative side. However, it does acknowledge that it can be reversed. What does Dr. Lavelle prefer?

Kaitlyn Ang 1J

It's used as a tool to draw Lewis Structures in their most stable form. To do this, you manipulate L and S until FC = 0 (as best as possible).

Kaitlyn Ang 1J

I agree with everyone else that you don't have to write out the px, py, and pz, but for example, if the question asks you how many unpaired electrons are there for an element, writing px, py, and pz out can be helpful.

Kaitlyn Ang 1J

Also, in general, if there is a problem that asks you to draw the Lewis Structure for a molecule, the element in the middle of the molecular formula normally actually IS the element in the middle. Of course, this doesn't work all the time, and that's where checking electronegativity is important, bu...

Kaitlyn Ang 1J

Electron affinity has the same periodic trend as ionization energy, so if you remember IE, you can remember EA

Kaitlyn Ang 1J

As the atoms go down a group, there are more electron levels added and the valence electrons (the outer shell of electrons) gets farther away from the nucleus, which not only increases the radius because there are more electron levels in between, but the attraction that binds the electrons and nucle...

Kaitlyn Ang 1J

As there are more valence electrons (aka from left to right on the periodic table), the attraction between the valence electrons and the nucleus grows stronger, so the electrons gravitate closer, making the radius smaller

Kaitlyn Ang 1J

I agree that salt water is a mixture, because you can separate the two (salt and water) from each other simply by boiling away the water. The salt crystals from the salt water don't chemically combine/react with the water molecules, so it wouldn't be a compound

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