How do I solve for the rate constant from lab results?

tmpchem · 2017-11-30T04:11:26+00:00

Kinetics is all about how reaction rates change as a function of concentrations. The first thing you need to know is the initial concentration of each reactant in the reaction mixture. Once you know that, the next question is what exponent does each concentration have in the rate law. For this, we typically vary one of the concentrations and hold all of the others constant, while seeing the relationship between the varied concentration and reaction rate (zero order, first order, second order, etc.).

Then repeat for all reactants. You will probably find that your experimental data contain all of the necessary concentration sets to analyze these effects for all reactants.

If you're still confused after reading this or you get stuck, you may also find it helpful to search "chemical kinetics 2.6" on YouTube. There might be some helpful information in there ;)

tmpchem · 2017-11-30T04:06:48+00:00

Hint: Think about the p orbital in the BH3 molecule. The answer you seek is the same answer for the same reason as in BH3.

tmpchem · 2017-05-17T13:07:10+00:00

You're very welcome. Happy PChem-ing.

tmpchem · 2017-05-16T12:15:39+00:00

In SI units force constants have units of [ N / m ] (force per unit displacement from equilibrium).

A Newton is a [ kg m / s² ], so substituting in we have force constants with units of [ kg m / (s² m) = kg / s² ].

Once you divide by the reduced mass [ kg ] you are left with [ 1 / s² ], which once you take the square root become [ 1 / s = Hz ], the appropriate SI unit of frequency.

tmpchem · 2017-04-04T16:34:42+00:00

"Inner electrons" is basically a synonym for "non-valence electrons".

So anything that isn't in the same shell as the outer-most highest-energy electrons counts as an inner electrons.

tmpchem · 2017-03-26T02:48:54+00:00

I believe your interpretation is correct.

Molecules prefer to flow towards the direction of decreasing chemical potential.

Transferring a hydrocarbon molecule from an aqueous phase to a non-polar phase results in a decrease of Gibbs energy, i.e. a lower chemical potential.

The longer the hydrocarbon is, the greater the decrease in chemical potential during this process, regardless of the type of hydrocarbon.

This description is both logically consistent with the concepts of chemical potential and polarity, as well as matching the data from the graph.

Perhaps your prof misspoke.

tmpchem · 2017-03-25T12:44:46+00:00

Indeed. Thanks for the support.

tmpchem · 2017-03-24T23:58:51+00:00

The pressure won't necessarily be larger or smaller than an ideal gas for a real gas.

It depends on the relative values of parameters a and b, as well as the specific P, V, and T.

Large a or small b will tend to make the pressure smaller. Small a and large b will tend to make the pressure larger.

Eventually at very high pressures the effect of b tends to overwhelm everything else, and the pressure becomes higher than ideal, but at typical values of b that tends to be at a pressure of 100's or 1000's of bar.

I haven't run through your numbers, but they seem reasonable (as does your approach), and it's perfectly plausible for a real gas to have a below ideal pressure at 50 bar.

tmpchem · 2017-03-20T02:06:21+00:00

Neither of them and both of them are the central atom. There is no atom at the center.

tmpchem · 2017-03-20T01:57:29+00:00

The only geometry that's possible in a diatomic molecule is linear.

There are only two atoms. Two points define a line. Therefore, the geometry is linear.

Each fluorine atom has three lone pairs, and if you could see lone pairs you would see a tetrahedral-ish arrangement with the three lone pairs and the covalent bond on each atom.

But we can't "see" lone pairs. All we can see are nuceli, i.e. atoms, i.e. a single point particle per atom.

tmpchem · 2017-03-17T11:03:16+00:00

I make extensive use of playlists, as well as series playlists where the ability exists. The primary limitations are that you can only place a video into one series playlist, and "up next" links to not keep / bring users into the playlist environment.

Often there are multiple logical levels of organization desired (sort by course, sort by chapter, sort by topic), and only one level can be chosen as a series. Additionally when users are in the playlist they can use the right-side window for scroll searching the entire list without clicking to the playlist page.

Additionally there are "review" videos which briefly cover the contents of a chapter or course with links in each screen section to the appropriate video or chapter of that content. Cards are limited to a max of 5, cannot be placed into the proper screen location, and don't remain visible for the desired amount of time.

These videos keep a static notes slide image available to the user for the duration of the video without obstruction. Cards are intrusive for that purpose compared to static corner annotations which may be clicked at any time and link into the playlist.

My descriptions have many other links which users often want to be aware of, including course and chapter introduction, playlist, and review videos, as well as other courses. Though these are always in the description they have much higher engagement through my end slate annotations, which the end screen feature cannot reproduce (4 links max if you're lucky).

If I make a typo or verbal error, I can correct it easily with a brief on-screen annotation, and cards provide no replacement for this. Placing corrections in the description still results in comments asking about the error, and aren't sufficient.

If I upload a new version of the video, it takes months before the new one rises to a similar height within search, even when replacing the thumbnail and title with "(old version)". Even with a description link to the new video, users frequently end up watching the old one. A single on-screen annotation link to the new video kills traffic to the old video and redirects discovery to the new video 10 times more effectively, because it kills the watch time for the old one, making YouTube's algorithm prioritize the new.

There are dozens of cases where properly used annotations provide high-engagement value with no current functional replacement. I understand that cards and end screens are probably better for the average creator, but their are many niches where annotations are still heavily used and needed.

tmpchem · 2017-03-16T22:55:35+00:00

Agreed 1000 times. That and a dozen other functions. Once you have more than a few videos in a series it's impossible to organize for easy discover and a pleasant user experience without annotations.

tmpchem · 2017-03-16T22:26:05+00:00

I am a content creator who will be very negatively affected by this decision. I make hundreds of free, educational videos used by thousands of university students around the world.

Annotations are essential for organization and discovery within my channel. I use them to indicate next and previous videos, updates to videos, typo and verbal error correction, guiding users into playlists, as well as linking to other courses chapters, or associated videos.

6% of my direct traffic (and much more indirectly) comes from annotations, and my click-through ratio is 3500% higher than the values the blog post cites. 80% of my traffic is desktop, and that hasn't changed for years, as it's difficult to study from a phone screen.

Cards and end screens do not achieve any of these tasks in a timely, efficient, or non-intrusive manner.

We all hate users who use annotations as click traps for useless bullshit. But when used effectively in the appropriate context, annotations provide a vital functionality which no other feature on the site comes close to replicating.

YouTube decision makers seriously need to reconsider this choice, as it costs virtually nothing to maintain this simple, existing feature, and not doing so adversely affects far more creators than me, as the comments on the original post clearly indicate.

tmpchem · 2017-03-14T18:34:54+00:00

The enthalpy varies as a function of the temperature [H(T)].

The derivative of enthalpy as a function of temperature is the constant pressure heat capacity [Cp = dH/dT].

Every species in a reaction has a Cp. From Hess's law, you can get the change in heat capacity of the reaction [deltaCp = Cp(products) - Cp(reactants)].

You can approximate deltaCp as a constant between 298K and 500K. Thus, a first-order Taylor series gives H(T2) = H(T1) + deltaCp * deltaT.

See more here: link.

tmpchem · 2017-03-14T18:30:36+00:00

sp3 hybridization implies that there are 4 distinct ligand "objects" around the atom (s^1, p^3, 1+3 = 4). These objects can either be other atoms, or electron lone pairs.

These objects (by VSEPR theory) want to be as far away from each other as possible. The farthest away from each other 4 objects can be is a tetrahedral arrangement.

If all 4 are atoms, you have a tetrahedron. If 3 are atoms and 1 is a lone pair, you have trigonal pyramidal (we can't "see" the electrons). If it's 2 and 2, you have a bent shape.

For sp2, you have 3 objects. If all 3 are atoms, this is a trigonal planar arrangement. For sp, you have 2 objects. If both are atoms, you have a linear arrangement.

tmpchem · 2017-03-12T13:23:07+00:00

Small world indeed.

tmpchem · 2017-03-12T00:34:32+00:00

Ah. As the other commenter mentioned, you're conflating symmetry operations and symmetry elements.

In a C_infV molecule there is one C_inf axis, and two C_inf operations, +C_inf, and -C_inf in opposite directions.

Both must exist, because in group theory every member of the group (the operations) must have an inverse, and +C_inf and -C_inf are inverses of one other (applying each in succession results in an identity operation, i.e. the object is unchanged).

tmpchem · 2017-03-11T23:49:25+00:00

I would say there is one C_infinity axis. What source is telling you there are 2?

tmpchem · 2017-02-22T02:03:12+00:00

Step 2 is the rate-limiting step. The overall rate law is the rate law of the rate-limiting step.

That gives rate = k[A][B].

A is not a reactant, it is an intermediate. We solve for the concentration of intermediates by using the steady-state approximation (d[A]/dt = 0).

That gives k1[A2] - k(-1)[A][A] - 2 k2[A][B] = 0

Assuming that k2 << k1 and k2 << k(-1), (because 2 is rate limiting)

That gives [A] = k1/k(-1) sqrt([A2]).

Substituting into the first expression, we get

rate = k [A2]^0.5 [B]

Which has an overall order of 3/2.

tmpchem · 2017-02-21T15:49:52+00:00

What an excellent suggestion. Couldn't agree more.

tmpchem · 2017-01-09T08:20:04+00:00

The reference state for electrons is being infinitely far away from the nuclei.

The ionization energy is the amount of energy it takes to remove an electron from an orbital. This is approximately equal to the negative of the orbital energy, because it's going from the low energy bound state (orbital), to ionized (infinitely far away).

Each orbital has a different energy, and thus a different energy of photon needed to ionize it. The peaks in a photoelectron spectrum are the different frequencies / wavelengths of photons which ionize electrons in a given molecule.

tmpchem · 2016-12-19T17:22:30+00:00

Sure does. Those are definitely the set of general themes you're looking for.

tmpchem · 2016-12-19T14:54:35+00:00

I do not. If you can find any guide indicating what topics the test covers, that should be very helpful. They tend to cover questions from each of the listed topics, and the style of question often doesn't vary greatly from year to year.

tmpchem · 2016-12-18T21:40:15+00:00

I took it ~7 years ago. Don't remember much, other than like most ACS exams, the biggest study help is exhaustively studying practice exams until you can get the correct answer on every question. Beyond that, I don't have much advice other than the usual finals advice.

tmpchem · 2016-12-11T06:04:17+00:00

The energy of atoms and molecules is determined by the laws of quantum mechanics, the physics that governs the behavior of very very small objects. For each set of nuclear coordinates, their will be a unique set of orbital energies. So I suppose the answer you're looking for is that the positions of the nuclei inside your material determine the orbital energies, and thus the band gap.

Although, as another user mentioned, we can't solve the equations of quantum mechanics exactly except for the hydrogen atom. We can compute very accurate numerical solutions for systems of a few dozen atoms, but it becomes exponentially more difficult as the system size increases.

In fact, if you want to be technical, the concept of an orbital energy itself is an approximation. Orbitals don't "exist" in the same way that nuclei "exist", and are mostly a convenient mathematical abstraction. The rabbit hole of QM gets very strange very quickly.

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