CHEMISTRY COMMUNITY

David Minasyan 1C

The negative power brings the concentration of that species to the bottom of the rate so having more of it would lower the rate. So yeah it's much like an inhibitor.

David Minasyan 1C

I dont think so, but if someone would confirm that would be good.

David Minasyan 1C

n is the movement of electrons so if you do the oxidation and reduction half reactions you should find that n = 4. Also I think you can just look at the charge of iron going from 0 in the reactants to +4 in the products and determine that the iron is being oxidized and 4 electrons are leaving, but i...

David Minasyan 1C

Just to segue off this post, there's molar mass, the positional disorder and what other factors that determine a compound's entropy? And which should we look at first if we were to rank them?

David Minasyan 1C

As far as I know, yes the constant for the forward reaction, k, and the constant k used in the rate law are the same. And then for the others, the capital K is of course the concentration of products divided by concentration of reactants. And as you said, k' is the reverse rate law. So I guess the o...

David Minasyan 1C

More of the reactant means that the probability of collisions can occur more frequently which means the reaction is more likely to proceed faster.

David Minasyan 1C

AS far as I know, lowering the activation energy of the reaction means the reaction has less of a barrier to overcome so the reaction would proceed faster.

David Minasyan 1C

As far as I know it just helps with identifying what the rate constant is.

David Minasyan 1C

The slow elementary step is much like the concept of a limiting reactant from last quarter.

David Minasyan 1C

Know that zero order reactions are mol/(L*s) and then as you go up one order, multiply by L/mol each time.

David Minasyan 1C

Units of k start out at mol/(L*s) at the zero order and every time you go up one order you can multiply by L/mol to get the new units of k.

David Minasyan 1C

I'm pretty sure you're supposed to use H+ (at least thats what my TA said) but I guess as long as you balance both H and O atoms then you should be fine.

David Minasyan 1C

As far as I know, the anode is always on the left. I haven't seen it on the right - not yet at least.

David Minasyan 1C

If the standard potential is positive (it'll make delta G negative bc of delta G = -nFE) so it'll be spontaneous.

David Minasyan 1C

Just to add on, the S in the residual entropy is like the minimum energy of the molecule - it's the entropy that the molecule has without anything being done to it. That's about the only time we use S instead of delta S.

David Minasyan 1C

The spectator ions get cancelled out anyway when you go from the ionic reaction to the net ionic reaction, I think. I don't think we need to know that for the test, but they cancel out anyway and don't matter.

David Minasyan 1C

My TA was telling us that when we do the cell diagrams, the anode goes on the left and then the cathode on the right by convention.

David Minasyan 1C

I think that it would - at least that's what I did when I was doing this problem. Gas to liquid means less movement of molecules so that explanation should suffice.

David Minasyan 1C

I'm pretty sure you're just supposed to use the C value that corresponds to the phase of the reactant. So if it's a liquid, use the regular C value at 4.184, or if its solid then use that one and the same with vapor.

David Minasyan 1C

I'm pretty sure you can just look at the K value changing and come to a conclusion based off that. I don't really know how to solve this mathematically, so using K sufficed for me.

David Minasyan 1C

Oddly enough, water going from liquid to solid can also be exergonic given that the delta H being negative outweighs the fact that the entropy is decreasing.

David Minasyan 1C

Generally speaking, the entropy for gas > liquid > solid.

David Minasyan 1C

To put it another way, since b and c are bound they will have less disorder than the molecules that aren't so the entropy would be lower for those than say compared to the molecules that aren't bound to one another.

David Minasyan 1C

That wasn't mentioned in class nor was it mentioned in discussion so I'd imagine that you don't need to worry about superheating.

David Minasyan 1C

It seems like the product and reactant enthalpies are negative. And if we're doing products minus reactants the reactants are now a positive value (-products - (-reactants) is the same as reactants - products). Hope that helps

David Minasyan 1C

That's the amount of energy it takes for water to change phases

David Minasyan 1C

Reversible processes are incredibly slow whereas irreversible processes are done quickly and inefficiently. If there was all the time in the world where our car engine could turn 100% of the fuel it was given into the energy that's needed to drive the car then it would be reversible and it would tak...

David Minasyan 1C

This is a highly simplified version but: 1) go to appendix 2A and find the NO reaction and the enthalpy for it 2) using that reaction and the two reactions given, rearrange them so that things cancel out and all that's left on the product side is dinitrogen pentoxide. 3) Make sure you adjust the ent...

David Minasyan 1C

From what I know, specific heat is dependent on a unit mass to heat something up. So in terms of a gas, the volume and pressure are important because the smaller the volume, the less molecules of gas and when you heat it up, it would take more heat (thus a higher heat capacity).

David Minasyan 1C

An extensive property depends on the mass while intensive does not.

David Minasyan 1C

For Hess's law they give us the enthalpy for each equation anyway so the compounds being in their standard states shouldnt matter (because from what I understand we would only need them in their standard states to get the correct enthalpy from the enthalpy table), but I'm not entirely sure. If anyon...

David Minasyan 1C

I think so. I remember him saying something like that in class.

David Minasyan 1C

Yes. If the element bonded to a Hydrogen atom is really electronegative then the electrons are largely attracted to that element instead of it being equally distributed so its more likely that the Hydrogen "falls off." That makes the compound more electronegative.

David Minasyan 1C

You can always try drawing out lewis structures and seeing if the compound can give, receive, or both give and receive lone pairs. If it can do both then its amphoteric. That's how I'd go about it - I dont really know of any other way.

David Minasyan 1C

I look at the lone pairs on the atoms to see if it can bind to other atoms and then determine if its a mono, bi, tri,...poly-dentate.

David Minasyan 1C

I think the way you're thinking of it is fine. I actually think about it in more simpler terms; I just think that if you try to rotate two balls (representing atoms) when they're stuck together by 2 sticks, then the sticks are gonna break, the same way the pi bonds would break.

David Minasyan 1C

I'm pretty sure it would although it would only be slightly less and slightly more, respectively.

David Minasyan 1C

I'm not sure about the first question, but the exception with ionization energy is that Oxygen has a lower ionization energy than Nitrogen because Oxygen has an extra electron in its first orbital that it would much rather remove whereas Nitrogen has nicely balanced 3 electrons.

David Minasyan 1C

Yes (at least that's the way I understand it).

David Minasyan 1C

It depends. For something like water, you'd put the lone pairs next to each other on top of oxygen to get that bent shape but for something like XeF4, then you'd be putting the lone pairs on opposite sides to get the right molecular geometry.

David Minasyan 1C

Just based off the periodic trends, we see that going down a group electronegativity decreases and chlorine is lower than oxygen. Also, since the chlorine atom is larger than the oxygen atom, it will be less electronegative because the electrons will be farther away from the pull of the nucleus. Oxy...

David Minasyan 1C

The carbon is the central atom (as given) and the CH2's come off the sides of carbon with each C atom having 2 Hydrogens attached to it. There are 24 valence electrons total so I started out by simply drawing it out with single bonds and then I made the central Carbon and another random carbon a dou...

David Minasyan 1C

A cool trick to knowing these trends by the way is only knowing the atomic radius one (which decreases across the table and increases down a group) and everything else is just the opposite of that.

David Minasyan 1C

Isn't it the other way around? Cl becomes the larger circle because it gains an electron and it's greater than its ground state because of that gained electron whereas Na loses that electron to Cl and becomes the smaller circle.

David Minasyan 1C

Yes that's exactly what that means. It's called the noble gas configuration if we we're being fancy and it's really a short hand way of writing the noble gas configuration up to the noble gas (in this case argon) instead the whole 1s2, 2s2, 2p6..and so on.

David Minasyan 1C

I don't think you'd have to do that unless it would better help you personally understand that the p sublevel has 3 orbitals (and therefore 6 electrons max).

David Minasyan 1C

Ionization energy is the energy to remove an electron, so I'm pretty sure that the energy to ionize an electron would be the energy that would remove it.

David Minasyan 1C

From what I know (and if someone would confirm) if the energy of the photon equals the amount of energy to eject the electron then that electron is ejected with a 0 kinetic energy.

David Minasyan 1C

I think this refers to the fact that for lyman series, for example, the energy levels are jump up from n=1 to the other energy levels or they all come down from other energy levels to n=1. Same applies to balmer and paschen except for n=2 and n=3, respectively. Also it could mean the fact that lyman...

David Minasyan 1C

Yes from what I was told today in my discussion, you can assume that n1 is 1 if it's not given to you.

David Minasyan 1C

Oh I thought they didn't. Okay cool

David Minasyan 1C

Conservation of momentum and conservation of energy imply that when a photon hits an electron on the surface of a metal with enough energy to eject it, the electron would conserve both the energy of momentum; however, if a photon has no mass - by which mv (momentum) and .5mv^2 (kinetic energy) equal...

David Minasyan 1C

I would think that the attraction between the nuclei of the atoms and the electrons of the other atom are stronger than the repulsion between the two nuclei of those atoms and that holds covalently bonded atoms together.

David Minasyan 1C

My TA was telling us that we'd probably need to know how to name molecules after the first test. We may go over it in class, but I'm unsure as to when. I guess the sooner you start learning them, the better off you'll be for the next test.

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