CHEMISTRY COMMUNITY

Humza_Khan_2J

From parts a and b, we can see that this reaction has K<1, and thus favors the reactants at equilibrium. We can also see that the forward reaction is endothermic. Raising the temperature of an endothermic reaction will favor more products, and thus will increase K. As for the rate constants, they wi...

Humza_Khan_2J

I believe the problem you're talking about asks for the rate of NO production, which would be 2 times the unique rate, which is what is observed.

Humza_Khan_2J

Check out the solution manual error guide (https://lavelle.chem.ucla.edu/wp-conten ... rs_6Ed.pdf). This problem is incorrect; you'd want to use Cv, which is 5/2R for a diatomic ideal gas.

Humza_Khan_2J

The total rxn chamber is .750 L and the gasses both occupy this whole volume. This is why we use it as the volume to find the concentration of H2 and I2.

Humza_Khan_2J

In general, temperature matters as the k value changes as the temperature does.

Humza_Khan_2J

The rate constants would change(if the equilibrium constant K does not equal 1), as could the order of the reaction depending on the actual rate of the reverse reaction. The laws will not change.

Humza_Khan_2J

Well, if something takes 8 seconds for half of it to decompose, then we know how many half cycles it'll take for any amount to decompose. The formula for this would be (1/2)^n=(x), where x is the amount remaining(as a percent) and n is the number of cycles the system goes through to decompose to the...

Humza_Khan_2J

It is possible that he asks for the derivations, but knowing how to utilize and understand the equations is much more important.

Humza_Khan_2J

The way you've described it works; a reaction can be first order with respect to its reactants A and B, but in total would be a 2nd order reaction.

Humza_Khan_2J

From some outside reading I've done(link posted below), certain complex reaction mechanisms with rate determining steps can result in fractional orders. They are indeed possible.

https://chemistry.stackexchange.com/que ... -mechanism

Humza_Khan_2J

These concepts are overarching within chemistry. If K is much larger than 1, then the reaction heavily favors the products(as written). If Q>K, then more reactants will be formed. Similarly, if Q<K then more product will be formed.

Humza_Khan_2J

Test 2 will be solely electrochem, according to Dr. Lavelle.

Humza_Khan_2J

In the solutions to cell diagrams in 17b, 23a, 23b, why is H+ included as a part of the cathode?

Humza_Khan_2J

Yes, this is an important part of balancing redox reactions. Conservation of charge dictates that there must be equal amounts of electrons on both sides.

Humza_Khan_2J

If there was no salt bridge, the reaction would simply stop as there would be no replenishment electrons on the anode side. A salt bridge is present in pretty much all the problems we deal with because it keeps the reaction going for us to analyze.

Humza_Khan_2J

Dr. Lavelle said he would be relatively lenient on accepting either kJ or kJ/mol. This is because, given a particular reaction, the delta G is generally the delta G for the reaction as a whole, as written. This could be said as "per mole equivalent" of the reaction.

Humza_Khan_2J

For each problem, you would need to use nC(for constant pressure or volume)*ln(T2/T1). The only thing that would change is if you use the molar heat capacity for constant volume or for constant pressure. We utilize this because the process occurs with a temperature change(check the textbook for the ...

Humza_Khan_2J

Can someone explain why the lower pressure gas has the higher entropy? Is it simply due to a volume difference(which I thought would be offset by the change in T)?

Humza_Khan_2J

For molecules positions, I think drawing it out is best. For example, on 9.25, it would not be possible to utilize the formula.

Humza_Khan_2J

A higher energy density means that, per mole, the substance releases a large amount of energy relative to another substance with lower energy density.

Humza_Khan_2J

All of the functions you mentioned are state functions, as well as Gibbs free energy. Heat and work depend on the path taken and thus require more analysis to calculate, as opposed to simply final minus initial states.

Humza_Khan_2J

It would probably help to put this in context of the equation, ∆G=∆H-T∆S. In a process with a positive ∆H and a positive ∆S, having a high T(for specific calculations, T>(∆H/∆S)) would make ∆G negative, and thus the reaction would be spontaneous and favorable.

Humza_Khan_2J

Since the system is at constant temperature, we can take into account the fact that q=-w. Therefore, deltaU must be 0, so I suppose that the solution equation deltaE and deltaU.

Humza_Khan_2J

Within the text(and the equation sheet on Prof. Lavelle's website), we can find that the olar heat capacity of a monoatomic ideal gas at constant pressure to be 5/2R. Since R is 8.314, 5/2R is the 20.8 J/gC you were looking for.

Humza_Khan_2J

There is no practice test; the homework is probably going to be your best bet as to what problems are going to look like. In my opinion, the best exercises to gauge your knowledge of the subject matter are the Integrated Exercises that were assigned in the problem set.

Humza_Khan_2J

Furthermore, in all of these examples, matter cannot escape either. Both matter and heat must be trapped within the system for it to be isolated.

Humza_Khan_2J

Basically, the most important thing to note here is that the process is adiabatic. This means that q must be 0. From this fact, you can derive all of these things. Be careful, what you marked as true are only sometimes true. For example, in a), if work was done on the system, the change in internal ...

Humza_Khan_2J

Bouncing off of the last commenter, the problem should give you which phase the water is in. If it it transitioning from phase to phase, you would have to use either the enthalpy of fusion or the enthalpy of vaporization.

Humza_Khan_2J

You would need to use the fact that work of expansion against opposing constant pressure is equal to -PdeltaV. This yields (-1.00 atm)(6.52-4.29 L). The second part would require the isothermal expansion equation, which is also in the book. To get n in this case for the aforementioned equation, one ...

Humza_Khan_2J

From what I understand, q=cdeltaT regardless of constant V or P; however, the C constants would differ in these two distinguishable scenarios.

Humza_Khan_2J

This question wholeheartedly depends on q. W is 0 because dV is 0. Therefore, it should be easy using basic qcal calculations to figure this out!

Humza_Khan_2J

How do we know the answer in 12.57 will only be one sig fig? I used two and it gave a pretty different answer (.10 versus the book's .09). Thanks!

Humza_Khan_2J

If the attached atoms to an O-H bond have a higher electronegativity, they delocalize the charge on the O- and make it more stable. If anion of an acid post-ionization is more stable, then it will be more acidic because this molecular with ionize more than others, simply due to the anion being more ...

Humza_Khan_2J

In what scenarios do elements use their bronsted definition and when do they use their lewis definition? In other words, are there certain types of reactions that use one of the definitions more often? Generally, we're just asked if a certain molecule or compound in a reaction is a Lewis/Bronsted a...

Humza_Khan_2J

How do we go about determining whether a particular ligand is a monodentate, bi, tri, etc. ligand? I've drawn out Lewis Dot Structures for all molecules and see that, for example, CO3^2- has a CO double bond and 2 CO single bonds, all of which have the appropriate number of lone pairs. How do we go ...

Humza_Khan_2J

I believe so, if it is covered in the HW and lecture.

Humza_Khan_2J

This would be a particular circumstance, such as a decomposition, where the increase/decrease in products' total pressure would not affect the pressure of one particular compound's partial pressure.

Humza_Khan_2J

For question 73, I believe it's more qualitative. We know the base geometry, but then have to change it depending on the number of lone pairs(bond angles generally get lesser as lone pairs increase). We can assume that ones with the lesser "initial" bond angles based on the base geometry(a...

Humza_Khan_2J

Another way to think about it is that pi bonds are unhybridized orbitals lying side by side(generally, from my knowledge, two p orbitals).

Humza_Khan_2J

Is the reason Be doesn't need an octet because of the amount of valence electrons it has?

Humza_Khan_2J

I agree with the above user, but I think it's much more about conceptualizing based on the number of lone pairs, bonds, and how they repel each other.

Humza_Khan_2J

Are there any main exceptions to the electronegativity rule of increasing across a period and decreasing down a group?

Humza_Khan_2J

Hey guys, as Dr. Lavelle said in lecture, I started this problem by putting the atom with the lowest IE at the center(which is O because O has a lower IE than N as an exception to the IE general rule). I got the bonds and lone pairs right in the problem, but apparently N should be in the center. Why...

Humza_Khan_2J

As for M sub l, or the angular momentum quantum number, a negative value is simply describing the particular orientation of an orbital in a subshell.

Humza_Khan_2J

As for why the 4s has more energy than 3d, this is simply because it's in a higher energy shell. A higher n implies a higher energy level.

Humza_Khan_2J

Is an orbital simply the l value then? Or are there different orbitals in each s, p, d, f subshell?

Humza_Khan_2J

Delocalized Lewis dot structures are more stable simply because the resonant partial double bonds are the lowest possible energy of the structure.

Humza_Khan_2J

Anything with momentum has a wavelength. However, since momentum is related to mass and velocity, something with a high mass can make the wavelength relatively undetectable.

Humza_Khan_2J

Angular momentum is based on velocity, ergo I would believe that it does have a level of uncertainty. However, there's a set angular momentum quantum number for each electron(but this determines shape). I believe these are two different concepts. Does anyone know more about a potential distinction b...

Humza_Khan_2J

The h is different in the Uncertainty Equation, it's called "h bar." "H bar" is h/2pi. This is why it's "h bar"/2 instead of h/4pi.

Humza_Khan_2J

You can use it to determine the wavelength of any moving particle. This seems a little trippy, and it is, but everything that's moving has a wavelength and exhibits some wavelike properties. The DeBroglie equation simply inversely relates the wavelength of the particle(or any mass, for that matter) ...

Humza_Khan_2J

How do we know, in this problem, that n=1 is the initial energy level? I can get n=1 based on the wavelength provided(Lyman series), but how do we know where in the calculation to put it? I understand it would yield a negative answer if not, but how do we know where to put it in the first place?

Humza_Khan_2J

You say that as the excited electron goes back to n=2 it will "releas[e] light energy with wavelengths that lie somewhere in the Balmer series," but isn't there only supposed to be one wavelength of light for these transitions? Or do the Balmer series lines represent all the transitions fr...

Humza_Khan_2J

With regards to this problem, you have to use the molar ratio between Sodium ions and Sodium carbonate, which is 2:1. You can appropriate calculations based on this ratio.

Humza_Khan_2J

Generally, if temperature is part of the values in your calculation, yes.

Humza_Khan_2J

To provide a more conceptual view on the topic, we can see that the oxygen used in the reaction accounts for the added mass(.4 g) in the crucible along with the metallic tin. You can then find how many moles of oxygen that is, and compare it to how many moles are in 1.5 g of tin. By seeing this rati...

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