CHEMISTRY COMMUNITY

Justin Yu 3H

Transition states are where two molecules hit each other with enough force to create a combined complex that can either form the products or separate back into the reactants. This requires a lot of energy because the products might be more thermodynamically stable in the product form, but a lot of e...

Justin Yu 3H

If this for showing a reaction mechanisms that fits the experimental rate law? Or is this to find the rate law of an unknown reaction? If it's the first, you would need to split the reaction into elementary reactions such that the rate law you find fits the experimental rate law. If the reaction is ...

Justin Yu 3H

He talked a little bit about OChem (functional groups and stuff) in class so maybe that?

Justin Yu 3H

You would use enthalpy of products minus enthalpy of reactants if the enthalpies given are enthalpies of formation; however, the enthalpies provided in the problem are enthalpies of combustion, meaning you would take the enthalpy of reactants minus the enthalpy of the products.

Justin Yu 3H

In what situations? Your question is a little vague.

Justin Yu 3H

There's probably a very complex reason why it is similar to the first order integrated rated law, but I just kinda see it as putting temperature on the reactant side of the chemical equation. As a result, provided everything else is constant, you can model that reaction as a pseudo first order react...

Justin Yu 3H

When have you seen that in problems? I've never seen that before.

Justin Yu 3H

If everything else is standardized, then I think it is safe to assume concentrations of 1M/bar. For your second question, there wouldn't be enough information to find the value of Q; you can assume that if something like that shows up on the test, more information will be provided.

Justin Yu 3H

We know the correct orientation, or at least have a general idea of how the molecules will be oriented because we know what parts of each molecule will react with what parts of the other molecule i.e what bonds are being broken and formed, and then orient them accordingly. For example, in part (c), ...

Justin Yu 3H

For the proposed activated complex, you have to consider the reactant the products in what bonds are broken and formed. The activated complex for (a) would be drawing CH3CHO where the bond between the carbons is partly dissociated. I'm not entirely sure, but I'm pretty confident that (b) would be te...

Justin Yu 3H

If you're referring to how they got the net rate of decomposition for O3 after substituting in O, they honestly just added everything on that side together. The term k1[O3], when multiplied by the denominator of the other terms, will cancel the middle term out and double the last term, resulting in ...

Justin Yu 3H

Well, I don't think there's a "set" method of drawing them. You just examine the molecules that are reacting, and see which bonds are being broken between what atoms and which bonds are being formed. You then draw a complex combining the molecules, where the bonds that would form if the re...

Justin Yu 3H

I'm not too sure what oxygens you are referring to. If you just use the enthalpy of combustion to calculate the reaction enthalpy, you get -1300 - 2(-286) - (-1300) = -312 kJ mol ^-1 I looked up some stuff though; the standard enthalpy of combustion is defined as the change in enthalpy when one mole...

Justin Yu 3H

The reaction between the cathode and anode needs to be spontaneous for the cell to work and have a positive voltage, which implies a negative gibbs free energy. Since the formation of Ti3+ has a positive gibbs free energy, you have to reverse it, then plug it into G = -nFE to find the standard poten...

Justin Yu 3H

Firstly, I will assume that the 2 you are referring to is the exponent of [O3]. If you use pre-equilibrium, you first get that K = [O2][O]/[O3] => [O] = K[O3]/[O2] Plugging [O] into the slow step yields (I don't know how to do subscripts so the "k" below is supposed to be a k sub 2) k[O][O...

Justin Yu 3H

I agree with the logic you used for part (b) of the problem. I also agree that the work done in (a) is 0; however I am not sure how you are certain that the change in energy is positive, as dissolution reactions can be either exothermic or endothermic.

Justin Yu 3H

To find K' in a pseudo reaction, like a second order reaction, you would solve for k without changing the concentration of one of the reactants that is first order. That reactant must be in so much surplus that its concentration would not change through the reaction. Thus, the reaction becomes a fir...

Justin Yu 3H

Yes, you are correct. Because there is no coefficient on O2, its rate of formation is the same as the unique rate of the reaction.

Justin Yu 3H

That second element, which I assume is [C], has no effect on the initial rate, so it is zero order. You can see this by comparing experiments 1 and 4. After realizing that, you can then disregard all changes in concentration of [C] and only look at the first two columns to find the order of the othe...

Justin Yu 3H

You are right in that elementary reactions always follow the format where bimolecular elementary reactions are second order reactions, but you cannot assume that for all reactions.

Justin Yu 3H

No, it is not possible to have a pseudo zero order reaction. If we consider the pseudo first order reaction for a second order reaction, we essentially hold one of the reactants at such a high concentration that changes are negligible and we can ignore its effects. Theoretically, we could model a fi...

Justin Yu 3H

Catalysts just speed up the rate at which reactions reach equilibrium. Since we can reach equilibrium from an excess of products or reactants, it makes logical sense that they would thus work both ways.

Justin Yu 3H

Furthermore, a catalyst lowers the free energy of activation by lowering the energy required to break bonds (decreases standard enthalpy of activation) and puts the reactants closer together and in their correct orientations (increases standard entropy of activation). How does that increase the sta...

Justin Yu 3H

Yes. Because the reaction with a higher activation energy proceeds at a slower pace than the other, it is the rate determining step and dictates the speed of reaction.

Justin Yu 3H

Thanks for sharing this!

Justin Yu 3H

The oxidizing agent acts to oxidize something else. In that sense, the "something else" becomes oxidized, while the oxidizing agent accepts the electrons and becomes reduced.

Justin Yu 3H

Argon is very very similar to the ideal gas. The molar heat capacity at constant volume for argon is the same as 3/2 R, that of a monoatomic ideal gas. So even though they write argon, you can just interpret that as viewing oxygen as an ideal gas.

Justin Yu 3H

Yes. For all equations like these, you generally assume that all other factors are held constant.

Justin Yu 3H

The scalar you multiply R by corresponds with the degrees of freedom of the molecule. Each degree of freedom corresponds to R/2, and there are both directional degrees of freedom and rotational degrees of freedom. The heat capacity at constant volume is the same as the sum of the degrees of freedom ...

Justin Yu 3H

You are completely correct! When calculating the enthalpy of formation of a molecule where the product has a different phase than the reactants, the change in enthalpy due to phase change would be negative, and in moving from a liquid to a gas, it would be vice versa.

Justin Yu 3H

What's the chemical formula of a banana?
BaNa2

Justin Yu 3H

What do you do with a sick chemist?
If you can't helium and you can't curium, then you might as well barium.

Justin Yu 3H

First off, although trigonal planar and t-shaped molecules both consist of 4 atoms, they differ in the amount of electron pairs on the central atom. Trigonal planar molecules have 4 electron pairs, 1 lone and 3 pairs with the other 3 atoms, while t-shaped molecules have 5 electron pairs, 2 lone and ...

Justin Yu 3H

Another important thing to note is that resonance stabilizes molecules. Since the electrons are more delocalized, they occupy a greater volume and the energy of the molecule is lower, making it more stable. Also, like the previous poster mentioned, the resonance structures you draw would be potentia...

Justin Yu 3H

Hello,
I know that problem 2.3 isn't a required problem, but I was working on it and I didn't really understand how they used the chart to answer the questions. Could anyone help me with this?

Justin Yu 3H

In Section 1.2, it states that "[the light ray's] electric field pushes the electron first in one direction and then pulls it in the opposite direction, over and over again." This denotes the frequency of the radiation. However, in space, where it is almost all empty space, would the wave ...

CHEMISTRY COMMUNITY

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Re: Transition States

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Problem 2.3

Oscillation of Light