CHEMISTRY COMMUNITY

Justin Chang 2K

This is the steps:
NO+1/2 O^2-->NO^2
NO^2--> NO + O (under UV light)
O + O^2 --> O^3

Overall: 3/2 O^2--> O^3

NO is the catalyst because it is first consumed, and then produced.
NO2 is the intermediate it is first produced, and then consumed.

Justin Chang 2K

NO is the catalyst. The reaction needs to occur under UV light.

Justin Chang 2K

We are only expected to be able to recognize functional groups. I assume he will give us a compound (for example, CH3CH2CH2COOH) and ask us to identify functional groups (in this case, carboxylic acid). I don't think naming organic compounds will be on the final.

Justin Chang 2K

I am not completely sure, but I think catalysts can appear in the rate law. If there is more of the catalyst, then it makes sense that the rate would be faster since more catalysts are able to catalyze more reactions and thus the rate is faster.

Justin Chang 2K

If you have, say, A+B-->C, the rate would be in the form of k[A]^n[B]^m. In order to determine n, you need to see how [A] changes over time. This can be done by making the concentration of [B] extremely high, so it can be regarded as a constant. If it is regarded as a constant, then that means the r...

Justin Chang 2K

It is important to not think that the rate is based on the stoichiometric coefficients. Usually, the rate will NOT agree with the stoichiometric coefficients of the balanced equation. If they do, that's merely coincidence.

Justin Chang 2K

I'm not sure, but I would understand how to do it incase it comes up on the final.

Justin Chang 2K

Yes, the units for the rate constants will change depending on what the order of the reaction is. I find it helpful to always include units in your calculations so you know what the units for the rate constant should be.

Justin Chang 2K

Yeah, zero and first order reactions both have a -kt, but second order reactions have +kt.

Justin Chang 2K

Yeah, since both differential and integrated rate laws are provided on the equations, I would just worry about memorizing which equation relates to zero order, first order, and second order.

Justin Chang 2K

From what I understand, MnO4- reacts with something and produces MnO2. This shows that Mn goes from 7+ to 2+. I am not quite sure why it does this, but the example of MnO4- was in Dr. Lavelle's first lecture on electrochemistry and in examples 14.1 and 14.2 in the textbook, so you can look at those ...

Justin Chang 2K

E^o=0V because the standard potentials inside and outside the cell are the same. The reaction only involves potassium going in and out of the cell. Therefore, if you do E=E(cathode)-E(anode), you will get 0 because the same number minus the same number is 0. However, keep in mind that this is referr...

Justin Chang 2K

I'm not completely sure, but I think that confirms that the reaction is done in acidic conditions (which for this reaction makes sense since there is H+ on one side of the reaction).

Justin Chang 2K

In this reaction, the potassium and chloride ions do not participate in the reaction. Mn is being reduced from +7 to +2, and Fe is being oxidized from +2 to +3. Because K and Cl remain +1 and -1 charge throughout, respectively, they do not participate in the redox reaction and are just spectator ions.

Justin Chang 2K

yes, here is the problem! Ethanol: what is the entropy of vaporization of 1 mol of ethanol at 285K? C(l)=111J/Kmol C(g)=78.3J/Kmol B.P=351.5K deltaHvap=43.5kJ/mol 1) deltaS=n*C(l)*ln(351.5/285)= 23.28J/K 2) deltaS=(43500J/mol) / 351.5= 123.755J/K 3) deltaS=n*C(g)*ln(285/351.5)= -16.42J/K deltaStotal...

Justin Chang 2K

I agree, always make sure that you check your units of everything else before deciding what value of R to use.

Justin Chang 2K

I think it's C(p)= 7/2R

Justin Chang 2K

Are you talking about 11.17, or another problem? 11.17 is about the formation of ammonia, not a reaction about carbon dioxide.

Justin Chang 2K

S^o (m) is the standard molar entropy (the values in the back of the textbook). I think S^o (f) is similar to the S^o (m). I don't quite know what S^o (r) is though.

Justin Chang 2K

Here's the way I think about it: If you heat up the reaction, the equilibria will shift towards the side that does NOT have heat on it (I like to think that the reaction does not want to produce more heat, so it shifts towards the side with no heat). If you cool down the reaction, the equilibria wil...

Justin Chang 2K

Yes, deltaS(total)=deltaS(system)+deltaS(surroundings). We also talked about how deltaS(surroundings)= -Hsys/T. This is helpful because it tells us an exothermic reaction is usually spontaneous because of the increase in total entropy. The two negatives will cancel, and deltaS(surroundings) will be ...

Justin Chang 2K

I don't think Dr. Lavelle has told us which one is most/least accurate of the three methods, so I wouldn't worry about it unless he says it in lecture sometime. You will just have to calculate deltaG based on the information that is given to you (if you have deltaH, deltaS, and a temperature, then u...

Justin Chang 2K

Yes. Also, because in reversible reactions, you increase/decrease the pressure by small increments, that explains why we must use w=-nRTln(V2/V1) since pressure will not be constant. However, because in irreversible reactions, the external pressure remains constant, we use w=-PdeltaV because the ext...

Justin Chang 2K

Adding more of a system increases the internal energy. Also it is good to know you cannot add/remove something from a system in a closed and isolated system.

Justin Chang 2K

Yes. As long as you know the reaction is performed under constant pressure (if it's in an open beaker) then q=deltaH(rxn)

Justin Chang 2K

Yes, you would make the sign negative.

The rule also applies to chemical equations in general. If you found that deltaH of A-->B is -50 kJ, then that means deltaH of B-->A is +50 kJ.

Justin Chang 2K

For me at least, drawing the Lewis structures gives me a really good visual as to what bonds are broken and what bonds are being formed. So I would personally recommend reviewing the Lewis structures to help prepare you for the tests.

Justin Chang 2K

Yeah, the definition of enthalpy is the heat absorbed or released at a constant pressure. q with a subscript p means the heat absorbed or released at a constant pressure, so the two are interchangeable. If there is no subscript p after the q, however, then the two are not interchangeable.

Justin Chang 2K

I think because 8.53 says the reaction took place in a bomb calorimeter, so you would use q=C(cal)*deltaT, which relates the heat to the specific heat capacity of the calorimeter.

Justin Chang 2K

Yeah, you can use Hess’s Law indefinitely to calculate enthalpies, since enthalpy is a state function you can just keep on adding and adding enthalpies. However, I guess for time’s sake, 2-4 equations is enough to test our knowledge of how enthalpy works and how chemical reactions work (for example,...

Justin Chang 2K

Yeah, for heat transfer you need to know every single individual step that is involved in the reaction. However, for enthalpy you only need to know the initial and final amounts since it is a state function; it doesn’t matter the pathway that is taken in order to reach the final amount.

Justin Chang 2K

It honestly depends on what’s given to you in the problem. For example, if you’re given the bond enthalpies of the compounds, then you should definitely use the bonds broken- bonds formed way of calculating enthalpy.

Justin Chang 2K

I would still just use the ice table method. That way, incase the final answer you get is wrong, you at least get partial credit by setting up the ice table and showing that you understand the concept.

Justin Chang 2K

Yeah, more electronegative elements can help stabilize and delocalize the charge.

Justin Chang 2K

Yeah, you wouldn't know whether or not to add cis or trans unless they give you the structure.

Justin Chang 2K

I would just look at the moles in the actual chemical reaction.

Justin Chang 2K

The sigma bond refers to the bond along the internuclear axis, so perhaps trying to fit 2 pairs of electrons along the same internuclear axis would not work since the electrons would want to repel each other as much as possible. Not sure if this reasoning is entirely correct but this is how I think ...

Justin Chang 2K

oh okay, thank you so much!

Justin Chang 2K

Hey guys!
So in the textbook on table 17.4, it says to use cyanido, fluorido, chlorido, bromido, etc... but in lecture I wrote down "use cyano, chloro, etc. when naming ligands". Maybe I just saw Dr. Lavelle's slides wrong, I'm not sure, but can anyone clarify the exact naming? Thank you!

Justin Chang 2K

Yeah, I would definitely be prepared to answer both, especially considering the fact that Dr. Lavelle is continuously asking us how many regions of e- density there are, the shape associated with that, and the molecular geometry as well during class all the time.

Justin Chang 2K

Yeah the bonds between carbon and chlorine are polar, but the whole molecule itself is non polar because the dipole moments cancel out since it is a linear molecule.

Justin Chang 2K

He will probably give it to us in the problem since we do not normally have the values for electronegativity to determine for sure if it is an ionic bond or covalent bond.

Justin Chang 2K

I would probably just draw the lone pairs in as if it were for a normal Lewis structure; for example, with the nitrate ion, put dotted lines for the bonds and write in all the lone pairs on the oxygen atoms.

Justin Chang 2K

I believe the 3d will come first since the d shell is now filled, so it overall is in a lower energy state than the 4s orbital. can someone else confirm this?

Justin Chang 2K

bigger anions have higher polarizability and cations with greater charge have higher polarizing power correct? Or is it the other way around

Justin Chang 2K

I'm not completely sure, but I think that if the atom is in the excited state, then the electron configuration would show an electron or two in a higher orbital (higher energy) than its ground state. For example, the electron configuration for oxygen is 1s2 2s2 2p4, which is the ground state of oxyg...

Justin Chang 2K

Yeah, all of the elements in those 2 groups have the same exception with the s and d shell, but for the purposes of this class, just memorize copper and chromium.

Justin Chang 2K

Yeah, especially when you discuss energy levels of the atom, if the photon does not have the exact amount of energy to go from n=1 to n=2, the electron won't be excited, even if the energy allows the electron to go from n=1 to n=2.5. It must be whole numbers in this case, hence why numbers are quant...

Justin Chang 2K

Yep! I did:

1m/1,650,763.73 wavelengths = x m/1 wavelength in order to figure out the length of only one wavelength, if looking at the mathematical representation helps at all!

Justin Chang 2K

Yeah, as long as you have 400-700 nm, 400 is around purple, and 700 in around red all memorized, I think you should be fine.

Justin Chang 2K

Yeah, I was also confused on that, but the solutions manual does indeed say 150 pm, so you should be good.

Justin Chang 2K

Yeah, you reference the catalyst by putting it above the arrow, but you wouldn't include it in the actual chemical equation.

Justin Chang 2K

What I did is I found out how many grams of carbon (0.682g * (1 mol CO2/44.01g) * (1 mol C/1 mol CO2) * molar mass of carbon to get grams), hydrogen (same process as carbon), and nitrogen. Then I did 0.376-(mass carbon+mass hydrogen+mass nitrogen) to get the grams of oxygen, and then divided that by...

Search found 53 matches