Search found 36 matches
- Sat Mar 17, 2018 12:10 am
- Forum: *Nucleophilic Substitution
- Topic: Transition States
- Replies: 3
- Views: 1133
Re: Transition States
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...
- Fri Mar 16, 2018 11:57 pm
- Forum: General Rate Laws
- Topic: Coefficients and Exponents
- Replies: 2
- Views: 428
Re: Coefficients and Exponents
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 ...
- Fri Mar 16, 2018 11:53 pm
- Forum: Student Social/Study Group
- Topic: what to study for the final
- Replies: 1
- Views: 365
Re: what to study for the final
He talked a little bit about OChem (functional groups and stuff) in class so maybe that?
- Fri Mar 16, 2018 11:51 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.57
- Replies: 2
- Views: 307
Re: 8.57
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.
- Fri Mar 16, 2018 11:48 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: flipping half reactions
- Replies: 2
- Views: 284
Re: flipping half reactions
In what situations? Your question is a little vague.
- Fri Mar 16, 2018 11:56 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Arrhenius eq and integrated rate law
- Replies: 1
- Views: 168
Re: Arrhenius eq and integrated rate law
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...
- Fri Mar 16, 2018 11:51 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Nernst Equation
- Replies: 2
- Views: 216
Re: Nernst Equation
When have you seen that in problems? I've never seen that before.
- Fri Mar 16, 2018 11:35 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Final 2008, Q3
- Replies: 1
- Views: 225
Re: Final 2008, Q3
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.
- Fri Mar 16, 2018 11:28 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 15.85
- Replies: 1
- Views: 162
Re: 15.85
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), ...
- Fri Mar 16, 2018 11:22 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: 15.85
- Replies: 1
- Views: 178
Re: 15.85
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...
- Fri Mar 16, 2018 11:15 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Example 15.7
- Replies: 1
- Views: 166
Re: Example 15.7
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 ...
- Fri Mar 16, 2018 11:08 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: drawing activated complexes
- Replies: 1
- Views: 197
Re: drawing activated complexes
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...
- Tue Mar 13, 2018 5:07 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.57
- Replies: 2
- Views: 230
Re: 8.57
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...
- Tue Mar 13, 2018 4:54 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: 14.33
- Replies: 2
- Views: 376
Re: 14.33
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...
- Tue Mar 13, 2018 4:37 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: example 15.7
- Replies: 1
- Views: 168
Re: example 15.7
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...
- Tue Mar 13, 2018 4:15 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 8.18
- Replies: 1
- Views: 170
Re: 8.18
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.
- Mon Mar 05, 2018 4:56 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: K'
- Replies: 7
- Views: 518
Re: K'
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...
- Mon Mar 05, 2018 4:42 pm
- Forum: General Rate Laws
- Topic: 15.3
- Replies: 1
- Views: 179
Re: 15.3
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.
- Mon Mar 05, 2018 4:38 pm
- Forum: Experimental Details
- Topic: 15.17A
- Replies: 1
- Views: 514
Re: 15.17A
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...
- Mon Mar 05, 2018 4:35 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Rate Laws of Elementary Reactions
- Replies: 4
- Views: 240
Re: Rate Laws of Elementary Reactions
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.
- Mon Mar 05, 2018 4:22 pm
- Forum: Zero Order Reactions
- Topic: Pseudo Zero Order Reactions
- Replies: 3
- Views: 925
Re: Pseudo Zero Order Reactions
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...
- Mon Mar 05, 2018 3:54 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Catalyst in a Equilibria
- Replies: 2
- Views: 205
Re: Catalyst in a Equilibria
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.
- Mon Mar 05, 2018 3:48 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Catalyst
- Replies: 6
- Views: 769
Re: Catalyst
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...
- Mon Mar 05, 2018 1:52 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Kinetics [ENDORSED]
- Replies: 2
- Views: 222
Re: Kinetics [ENDORSED]
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.
- Mon Mar 05, 2018 1:48 pm
- Forum: General Rate Laws
- Topic: Useful Resource
- Replies: 1
- Views: 179
Re: Useful Resource
Thanks for sharing this!
- Sun Feb 18, 2018 4:03 pm
- Forum: Balancing Redox Reactions
- Topic: agents
- Replies: 3
- Views: 265
Re: agents
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.
- Mon Feb 12, 2018 9:40 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Example 9.5
- Replies: 3
- Views: 277
Re: Example 9.5
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.
- Mon Feb 12, 2018 9:30 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: General Question on these equations
- Replies: 1
- Views: 109
Re: General Question on these equations
Yes. For all equations like these, you generally assume that all other factors are held constant.
- Mon Jan 29, 2018 11:03 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: R value in entropy equation [ENDORSED]
- Replies: 2
- Views: 198
Re: R value in entropy equation [ENDORSED]
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 ...
- Wed Jan 17, 2018 3:42 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.67 moving from a gas to a liquid
- Replies: 1
- Views: 106
Re: 8.67 moving from a gas to a liquid
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.
- Sat Dec 09, 2017 7:20 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 8832
- Views: 1496847
Re: Post All Chemistry Jokes Here
What's the chemical formula of a banana?
BaNa2
BaNa2
- Sun Dec 03, 2017 3:04 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 8832
- Views: 1496847
Re: Post All Chemistry Jokes Here
What do you do with a sick chemist?
If you can't helium and you can't curium, then you might as well barium.
If you can't helium and you can't curium, then you might as well barium.
- Wed Nov 15, 2017 11:44 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: trigonal planar vs. t-shaped
- Replies: 5
- Views: 1664
Re: trigonal planar vs. t-shaped
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 ...
- Wed Nov 08, 2017 3:48 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Resonance Structure
- Replies: 4
- Views: 282
Re: Resonance Structure
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...
- Sun Oct 29, 2017 9:58 pm
- Forum: *Shrodinger Equation
- Topic: Problem 2.3
- Replies: 1
- Views: 197
Problem 2.3
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?
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?
- Tue Oct 17, 2017 5:04 pm
- Forum: Properties of Light
- Topic: Oscillation of Light
- Replies: 3
- Views: 233
Oscillation of Light
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 ...