CHEMISTRY COMMUNITY

Jarrett Sung 3B

I think it's important for us to know, but it's easy if you're given the units for each axis.

Jarrett Sung 3B

All the half-life equations are based off the order equations on the equation sheet. Know which integrated equation goes with each order, and you can easily find the half-life equation.

Jarrett Sung 3B

It is possible! This happens when the products are at a lower free energy, so are thermodynamically favorable, but have a higher activation energy.

Jarrett Sung 3B

The log equation can only be used under standard conditions, so the units balance out to 0.0592. Under any other conditions, you have to use E = Enaught - (RT/nF)ln Q, but remember you can also use that equation under standard conditions.

Jarrett Sung 3B

The n refers to the electrons exchanged in a redox reaction rather than moles of a substance. Definitely easy to mix up!

Jarrett Sung 3B

I think they're generally the same. A quick online search said that EMF is the voltage induced from a source, but it's measured in volts as is cell potential.

Jarrett Sung 3B

A reducing agent reduces a reactant while being oxidized itself, and an oxidizing agent oxidizes a reactant while being reduced itself. When looking for each, think about how a reducing agent is oxidized and how an oxidizing agent is reduced.

Jarrett Sung 3B

We use the Van't Hoff Equation when we're either looking for a second equilibrium constant at a second temperature or a second temperature to go with a second equilibrium constant. It can also be used to find the enthalpy change given the two K's and T's.

Jarrett Sung 3B

I think the above post is correct. I think I read the same thing in the textbook, that non-expansion applies because of the lack of volume change with the electron transfer.

Jarrett Sung 3B

Atoms with larger mass tend to have more particles, which makes their entropies higher. Molecules with high mass will also tend to have more atoms, which means more positional entropy.

Jarrett Sung 3B

I also was wondering this, and the answers help a lot! It's easy to get tripped up on problems like this that involve a change in two different units.

Jarrett Sung 3B

Don't think degeneracy has units. Tried to look back at my notes and on the internet and it doesn't seem like there is a unit.

Jarrett Sung 3B

You're right! Irreversible free expansions, when isothermal as well, have no change in enthalpy.

Jarrett Sung 3B

Since q = 0 in an adiabatic reaction, there is no temperature change and the final temp = initial temp

Jarrett Sung 3B

It's actually the opposite. The small increments done in a reversible reaction need more work than a sudden, large change in pressure.

Jarrett Sung 3B

I don't think that we need to know this, because the gas constant is really only for ideal gases. One thing that is important to remember is to use the right gas constant based on the units provided.

Jarrett Sung 3B

Open systems have a constant pressure. This is caused by the environment, which unless the system is placed in a very specific place will be relatively the same.

Jarrett Sung 3B

Think of qsys = -qsurr as the heat lost by the system is gained by the surroundings or vice versa. This is assuming a perfect system, but unless otherwise stated I think we should assume a perfect system.

Jarrett Sung 3B

Cp is at constant pressure, while Cv would be at constant volume. In this equation specifically, Cp couldn't be at constant temperature because of the deltaT. If deltaT was 0, Qp would also be zero and Cp wouldn't even be necessary.

Jarrett Sung 3B

We add them up because when bonds are formed the deltaH is negative (energy released), while deltaH is positive (energy absorbed) when bonds break.

Jarrett Sung 3B

Enthalpy is more about heat transferred, while entropy is about disorder and the number of possible positions for atoms/molecules. It's important to note that they're both state functions.

Jarrett Sung 3B

Oxidation number is the number of electrons gained or lost due to forming a bond. A negative oxidation number means the atom gained electrons when forming the bond because it is based on charge.

Jarrett Sung 3B

Enthalpy just depends on the initial and final states, while heat is what appears when there is a change in enthalpy or work. Heat depends on the path taken instead of the start and end points.

Jarrett Sung 3B

Catalysts don't shift the equilibrium of the reaction, but that's a great question! All the catalyst does is lower the activation energy of the reaction, both forward and reverse.

Jarrett Sung 3B

Nothing really changes about the reaction other than the activation energy: the energy needed for the reaction to proceed. The state of the reaction wouldn't change.

Jarrett Sung 3B

It makes the most sense to change it from mmol to mol, because mol/L is the unit for concentration, which is incredibly important for Q or K values.

Jarrett Sung 3B

I think the best way is just to memorize them. I've tried to figure out patterns to remember them better, but with no luck. It sucks, but it's the best we can do.

Jarrett Sung 3B

You have to compare the Q value to the K equilibrium constant, and if Q < K, the reaction will proceed towards the products while if Q > K, the reaction will proceed towards the reactants. If Q = K, the reaction is at equilibrium.

Jarrett Sung 3B

Not exactly sure where it came from, but it is most likely due to what the x value ended up being. It's a little odd that they use the 1/(6.1 x 10^-3) rather than 1.6 x 10^2, but they're the same value.

Jarrett Sung 3B

Standard temperature is 298K, so SATP

Jarrett Sung 3B

Make sure you check to see what the units are for the question, because the R value depends on the units used.

Jarrett Sung 3B

It's definitely possible that it shows up, but regardless it's an easy conversion to atm or mmHg using the equation sheet

Jarrett Sung 3B

The path that is taken matters with E, unlike H, S, or G. Since those three are all state functions, E is not a state function.

Jarrett Sung 3B

As those above have said, the salt bridge allows electron flow back to oxidation solution to maintain neutrality and prevent a buildup of electric charge.

Jarrett Sung 3B

I was also a little confused, but it seems like the anode tends to be positive while the cathode is negative.

Jarrett Sung 3B

You find Q and K the same way, by putting the concentrations of products over concentrations of reactants, taking into account the coefficients. If Q > K, the reaction favors the reactants and if Q < K, the reaction favors the products. Also, if Q = K, the reaction is at equilibrium.

Jarrett Sung 3B

The Van't Hoff equation is derived from deltaG = -RTlnK, but it's used when you find multiple temperatures and are trying to find the K value at one of those temps. Using either one just depends on the information given in a question.

Jarrett Sung 3B

K and Kc are pretty much the same thing, Kc just specifies equilibrium constant K for concentrations, while Kp is K for pressure. For all these versions of K, you can calculate them the same way as well.

Jarrett Sung 3B

Yup, you should assume the temperature is at standard conditions of 25 degrees C, or 298 K.

Jarrett Sung 3B

I would say that it's good to convert to Kelvin whenever given a temperature in Celsius. Sometimes a question will ask for the final temperature in Celsius, but it's easy to convert back to it and when plugging in temperature into equations it's going to be in Kelvin, especially with ideal gas law.

Jarrett Sung 3B

It's usually in Joules/Kelvin, but I find it a lot easier to look at what units the question is asking for so that I know to convert from KJ to J or J to KJ or to not change at all.

Jarrett Sung 3B

It helps for me to think about entropy as disorder, especially considering how molecules move around in different states. With gases, molecules have more room to move around as they are not as close together as a liquid, and same with liquids and solids. This goes in hand with Dr. Lavelle's discussi...

Jarrett Sung 3B

I think it's good to assume the system is closed unless otherwise stated, especially because we've dealt with mostly problems with closed systems so far.

Jarrett Sung 3B

I was also struggling with this a bit, like the post above said, make sure you keep in mind the units that are given and also the units that you're trying to find in the equation. The unit for pressure is the most important, but K is also good to look at.

Jarrett Sung 3B

Like the post above said, use ∆G = ∆G˚ + RT*lnQ. To find Q remember it is products/reactants using the values given.

Jarrett Sung 3B

The only difference is that in the lecture the negative is factored out and in the textbook is not, so it's (1/T1-1/T2) instead of -(1/T2-1/T1)

Jarrett Sung 3B

This problem is similar to the problems we did with enthalpy of formation that were on the last midterm. First you want to balance out the reaction so that there is 2D on each side, which would give A + B + 2D --> 2D + 2C. This cancels out to give the original reaction, A + B --> 2C. Because we flip...

Jarrett Sung 3B

Hydrogen bonds only occur with O, N, and F because of their high electronegativities that interact with the low electronegativity of the H. This results in the partial positive charge on the H interacting with the partial negative of another atom.

Jarrett Sung 3B

I think that the general order of the strength of bonds is: covalent, ionic, and hydrogen. What I have noticed though is that in some instances people say that ionic bonds are stronger than covalent bonds, but I think this depends on whether the bonds are intramolecular or intermolecular. I think i...

Jarrett Sung 3B

Like what's been posted above, non-metals tend to have a much higher ionization energy and electronegativity than metals, which often leads to the formation of ionic bonds, when the non-metal steals an electron from the metal. This usually leads to both atoms going back to a full outer electron shell.

Jarrett Sung 3B

Polarity in molecules is based on the difference in electronegativity between bonded atoms. Typically, polar molecules will have polar bonds where the two atoms have vastly different electronegativities, so the electron isn't shared equally. However, nonpolar molecules can also have polar bonds, but...

Jarrett Sung 3B

It's important to note the types of forces that occur with intramolecular forces and intermolecular forces, especially considering their strength. For example, individual intramolecular forces such as covalent bonds will be much stronger than individual intermolecular forces such as hydrogen bonds.

Jarrett Sung 3B

Yes, when atoms are ionized they fully lose their electron to another atom. In an ionic bond, the opposite charges of two atoms, one with an extra electron and one with a lost electron, attract to each other, creating the strength of the bond.

Jarrett Sung 3B

I don't think there's going to be a curve, but the C- policy just means if you get above a 50% you will get at least a C-.

Jarrett Sung 3B

Adding to the post above, each series is part of different regions of the electromagnetic spectrum. The Lyman series is part of the UV light region, the Balmer series is part of the visible light region, and the Paschen series is part of the infrared light region.

Jarrett Sung 3B

I've mostly been using textbook problems for practice, but going back to the textbook to look at things you may be having trouble with helps a lot too. If there's something in particular that I really struggle with, both the sample problems and the textbook problems are very useful. Going over notes...

Jarrett Sung 3B

Like those above have said, Dr. Lavelle said in the lecture that we're not going very in-depth on black body radiation, but he used the black body as an example of what something would be like if it could absorb all frequencies of light. If you end up taking physics, it might be an important topic.

Jarrett Sung 3B

Yup! When they increase the intensity of the light they are increasing the number of photons, which doesn't end up changing the energy, so in the photoelectric experiment the intensity change didn't increase the electrons released. This was how they realized the frequency changes the energy of the p...

Jarrett Sung 3B

Agreed with the post above. There's a minimum energy required from a photon to excite an electron to the next energy level, so if you have enough energy in the photon to meet multiple energy requirements, the electron can jump multiple energy levels with a single photon.

Jarrett Sung 3B

Thanks so much for this, it'll be so useful just to know this when trying to remember all of the prefixes.

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