CHEMISTRY COMMUNITY

Vincent Tse 1K

I think you can distinguish between reversible and irreversible on a PV diagram, for the most part. Reversible processes should be entirely differentiable (i guess a good way to put it is "curved", such that you can only integrate it to find the area under) while irreversible processes sho...

Vincent Tse 1K

I think you're talking about the equipartition theorem. Someone else asked about it not too long ago and I gave a short explanation here: https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=75&t=29894&sid=698fe4496d8474f226e4f407bf8fd120 The syllabus doesn't explicitly say we need to know, a...

Vincent Tse 1K

The equipartition theorem is just a way for us to PREDICT heat capacities of something by factoring in its rotational, translational, and vibrational energy. When you heat something up, some of the energy is stored in these three categories rather than actually heating it up. Therefore, its heat cap...

Vincent Tse 1K

In a general sense, it doesn't particularly matter if you use definite or indefinite integrals. All a definite integral does is specify the bounds of integration. You should be able to come to the same equations using both--I think Dr. Lavelle did it in class with indefinite, and the textbook uses d...

Vincent Tse 1K

It sorta depends on the problem, I guess, depending on what information you're given from the start. As long as you adjust your reactant concentrations accordingly, you should be fine. For example, if A -> 2B and they ask you to find k, given initial [A] and final [A], you don't really need to mind ...

Vincent Tse 1K

I don't really see a real need to convert to log, unless that makes the problem easier/more convenient. You should get the same answer either way if you're doing it right! Whatever's more convenient for you and your calculator/head, I suppose.

Vincent Tse 1K

If a reaction is "first order", it simply means that the rate is proportional to the concentrations of one of the reactants. If a reaction is "second order", it simply means that the rate is proportional to the SQUARE of the concentrations of that reactant. A reaction being first...

Vincent Tse 1K

Yeah, Eli's right! Unless, of course, the intermediate IS part of the rate-determining step. It kind of depends on the reaction, really--for a coupled reaction, just look for the slowest one. Unless they themselves are part of the rate-determining step, intermediates have little function on the reac...

Vincent Tse 1K

Zero order just means the rate of the reaction doesn't depend on the concentration of the reactant(s), so no, not necessarily.

Vincent Tse 1K

I'm not entirely sure about all the scenarios, but I suppose you would need an inert electrode whenever you don't want the electrode messing up the reaction, as in, you only want it to transfer electrons instead of exchanging ions. That's typically the situation for gas/ion electrode reactions. As f...

Vincent Tse 1K

Not entirely sure I understand your question, but it shows that I- ions are oxidized (lose electrons) to become I2.

Vincent Tse 1K

I believe n is the # of electrons involved in the balanced redox reaction. I don't particularly know if you necessarily HAVE to find the balanced equation to figure out n, but it's certainly one of the easiest ways to. Just write out the half reactions!

Vincent Tse 1K

I think the confusion oftentimes comes from a multistep process. Say you want to calculate the entropy change of an ideal gas that expands by a certain amount while it is simultaneously heated to a certain amount. You'd think the volume isn't constant, since the problem says the gas expands. But, if...

Vincent Tse 1K

Is the difference between using ΔS = q(rev)/T and ΔS = nCln(T2/T1) that you can only use the former in isothermal conditions, while the latter can be used in any other reversible scenario where the temperature isn't constant?

Vincent Tse 1K

ΔU = 3/2 nRΔT for monoatomic ideal gases, so anytime ΔT = 0, that's true. Even if it weren't monoatomic, only the 3 changes to some other number, so the ΔT = 0 condition still holds true. ΔU also stays the same if a multi-step process (the PV diagrams!) returns to its original starting point, if tha...

Vincent Tse 1K

Everybody above said it well!

If you haven't already, I recommend trying problems 9.21, 9.23, and especially 9.25 in the textbook; those really helped me learn the concept when I was doing them last week :)

Vincent Tse 1K

Heat of formation of a certain element is defined to be 0 for that element in its standard state (which is usually the state it is found in naturally, the diatomic one).

I suppose ΔG has a similar definition and entropy doesn't, but I'm not too entirely sure about those.

Vincent Tse 1K

I think it's b/c internal energy is directly related to change in temp.

Forgot where exactly, but there's a derivation somewhere in the book that uses the equipartition theorem and says that molar internal energy is U = 3/2 RT.
Therefore, ΔU = 3/2 RΔT. If ΔT is 0, then ΔU is also 0.

Vincent Tse 1K

Yep, Liz's right. The formal definition of thermal equilibrium between two objects is that there is no net heat transfer between them. Since heat flows from stuff with higher temperature to stuff with lower temperature, if there is no heat flow, then they must be at the same temperature. In fact, th...

Vincent Tse 1K

Nicole's right. To add on to significance: Remember that during last friday's lecture Dr. Lavelle showed us two different processes--reversible and irreversible. In reversible expansion, the temperature is constant, the volume increases, and the pressure decreases. The work done is the area under th...

Vincent Tse 1K

Entropy is a thermodynamic quantity denoted by ΔS, and refers to the "disorder" (an potentially confusing term, if you remember from lecture) of something. In essence, most reactions tend to favor going from low entropy to high entropy (you won't need to put in work). A physical example wo...

Vincent Tse 1K

Not sure I entirely understand your question, but here's my take: Derivatives are super small parts of something. Like a "small piece of X". It's helpful to use derivatives in the sciences to refer to an infinitesimally small part of something that we don't know much about, or something th...

Vincent Tse 1K

I think it's liquid water. At least, that's what I've mostly seen in the problems I've done. Have you found a problem that uses gaseous water for combustion?

Vincent Tse 1K

As the delta in front of U suggests, U is a measure of the change of energy within the system; it's simply a measure of what it started as vs. what it ended as. q and w, tho, depends on the path taken within the process itself. They are NOT measures of initial and final states; that's why there's no...

Vincent Tse 1K

To add on/consider: Enthalpy is a state function in which when you know the beginning and end states, you'll know the change in enthalpy; whatever happens in the "middle" doesn't matter. On the other hand, heat does depend on the "path" taken between those beginning and end state...

Vincent Tse 1K

Um, what problem is this specifically?

I guess for the most part, Q = mCΔT is a good start. You'll want to look at what the "system" is, and then what the "surrounding" is, and note any energy changes that occur.

Vincent Tse 1K

Cv = 3/2 * R is the relationship that you derive for monoatomic gases. Cv is the heat capacity at a constant volume. The derivation involves using ΔE = (3/2)nRΔT (for monoatomic gases) and the first law of thermodynamics, Q = ΔE + W. W is 0 at constant volume, so you plug in ΔE on the right, the sub...

Vincent Tse 1K

I mean, the general trend of bond enthalpy exists and resonance shouldn’t "affect" the listed enthalpy since that one is found experimentally (whatever you see in those tables). I'm not sure if this is entirely true, but theoretically the enthalpies listed should be the MEAN bond enthalpy ...

Vincent Tse 1K

Steam "burns" a lot more b/c it typically carries way more energy than boiling water does, and that's due to phase changes, mostly. You might want to view a water phase diagram as a reference, but here goes: So liquid water at say, room temp, first needs to heat up to 100 degrees C. It nee...

Vincent Tse 1K

Not entirely sure, but my take: Internal energy (U) is basically all the energy in the system. Enthalpy (H) is the thermodynamic potential, which is internal energy and the work done on/from its surroundings. That latter part can be represented by PV. There's the first law of thermodynamics (#7 on t...

CHEMISTRY COMMUNITY

Search found 30 matches

Re: PV Diagrams

Re: Origin of Internal Energy

Re: Equipartition theorem

Re: Deriving the Rate Laws

Re: Stoichiometric Coefficients

Re: Log vs ln

Re: First Order Reactions [ENDORSED]

Re: Intermediates [ENDORSED]

Re: Zero Order Reactions [ENDORSED]

Re: Inert Electrodes

Re: Cell Diagram [ENDORSED]

Re: n

Re: Cp and Cv

ΔS = q/t vs. ΔS = nCln(T2/T1) [ENDORSED]

Re: Delta U as 0

Re: Degeneracy (W)

Re: Why is deltaG of formation 0 for diatomic molecules?

Re: Isothermal expanision and deltaU [ENDORSED]

Re: Heat Capacity

Re: q(REV)

Re: Define Entropy

Re: Integrals [ENDORSED]

Re: Combustion Equations [ENDORSED]

Re: Change in internal energy as a state function

Re: Heat vs. Enthalpy

Re: Calorimeter

Re: Constant pressure/volume

Re: Bond Enthalpies

Re: Steam Burning

Re: Enthalpy and Internal Energy